WORTH 25 POINTS PLS ANSWER
In the diagram, JM¯¯¯¯¯¯¯¯≅PR¯¯¯¯¯¯¯¯, MK¯¯¯¯¯¯¯¯¯¯≅RQ¯¯¯¯¯¯¯¯,and KJ¯¯¯¯¯¯¯¯≅QP¯¯¯¯¯¯¯¯.
Drag a tile to each empty box to complete the sentences correctly.
Using transformations, such as a ____, it can be varified that △JKM is congruent to △PQR if all pairs of corresponding angles are congruent.
In any pair of triangles, if it is known that all pairs of corresponding sides are congruent, then the triangles ___ congruent.

Answer:

220 divided by it self (220) will get you 1

Step-by-step explanation:

220/220=1

Answer:

220

Step-by-step explanation:

A person can save $1,200 - $300 = $900 with an HMO if they had ten physical therapy sessions costing $150 each.

A person with an HMO (Health Maintenance Organization) can save a significant amount of money on physical therapy sessions compared to someone with a health insurance policy. Let's calculate the savings a person would have with an HMO for ten physical therapy sessions costing $150 each.

With an HMO, the cost per visit for physical therapy is $30. Therefore, the total cost of 10 physical therapy sessions would be 10 x $30 = $300.

On the other hand, with a health insurance policy, after a deductible of $600, the policy pays 80% of the physical therapy costs. Since each session costs $150, the total cost for ten sessions would be 10 x $150 = $1,500.

The person would have to pay the deductible of $600, which means the insurance will cover 80% of the remaining cost. Therefore, the person will pay $600 (deductible) + $900 (20% of the cost) = $1,200.

In comparison, with an HMO, the person would only have to pay $300 for the ten sessions.

Therefore, a person can save $1,200 - $300 = $900 with an HMO if they had ten physical therapy sessions costing $150 each.

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The expression csc²θ - 2cot²θ can be simplified to (1 - 2cos²θ) / sin²θ is obtained by using trignomentry expressions. This expression is equivalent to option (F) in the given choices.

To simplify the expression csc²θ - 2cot²θ, we can rewrite csc²θ and cot²θ in terms of sinθ and cosθ.

csc²θ = (1/sinθ)² = 1/sin²θ

cot²θ = (cosθ/sinθ)² = cos²θ/sin²θ

Substituting these values back into the expression:

csc²θ - 2cot²θ = 1/sin²θ - 2(cos²θ/sin²θ)

Now, we can combine the terms with a common denominator:

= (1 - 2cos²θ) / sin²θ

This simplification matches option (F) in the given choices.

Therefore, the expression csc²θ - 2cot²θ can be expressed as (1 - 2cos²θ) / sin²θ.

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a. Profit-maximizing quantity: 50 bicycles, Price: $15.

b. Deadweight loss represented by the red triangle before tax and the blue triangle after tax.

a. To find the profit-maximizing quantity and price for Bill's Bicycle Company, we start with the demand function:

Q = 200 - 10P

From this, we can derive the price equation:

P = 20 - Q/10

Next, we calculate the revenue function:

R(Q) = Q(20 - Q/10) = 20Q - Q^2/10

To find the profit function, we subtract the total cost function from the revenue function:

Π(Q) = R(Q) - TC = (20Q - Q^2/10) - (10 + 10Q) = -Q^2/10 + 10Q - 10

To maximize profit, we take the derivative of the profit function with respect to Q and set it equal to zero:

Π'(Q) = -Q/5 + 10 = 0

Solving this equation, we find Q = 50. Substituting this value back into the demand function, we can find the price:

P = 20 - Q/10 = 20 - 50/10 = 15

Therefore, the profit-maximizing quantity for Bill's Bicycle Company is 50 bicycles, and the corresponding price is $15.

b. Before the imposition of the tax, the equilibrium price is $15, and the equilibrium quantity is 50 bicycles. The deadweight loss is the area of the triangle between the demand curve and the supply curve above the equilibrium point. This deadweight loss is represented by the red triangle in the diagram.

After the imposition of the tax, the price of each bicycle sold will be $15 + $2 = $17. The quantity demanded will decrease, and we can calculate it using the demand function:

Q = 200 - 10(17) = 30 bicycles

The deadweight loss with the tax is represented by the blue triangle in the diagram. We can observe that the deadweight loss has increased after the imposition of the tax because the government revenue needs to be taken into account.

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The general solution of y" + y = cotx is cos⁡x+c_2sin⁡x-(ln|cos⁡x|+C)sin⁡x.

a) The general solution of y″+y=cot⁡x

We can find the general solution of y″+y=cot⁡x by finding the complementary solution of y″+y  and then apply the method of variation of parameters.

So, the complementary solution of y″+y=0 is given by

c = c_1cos⁡x+c_2sin⁡xwhere c1 and c2 are constants of integration.

Then the particular solution of y″+y=cot⁡x is given by

y_p = -(ln|cos⁡x|+C)sin⁡x

where C is the constant of integration.

The general solution of y″+y=cot⁡x is

y = y_c + y_p

= c_1

cos⁡x+c_2sin⁡x-(ln|cos⁡x|+C)sin⁡x

The above solution is in the form of implicit solution.

We cannot find the constants of integration until initial or boundary conditions are given.

b) Solve the given equation using the method of undetermined coefficients.

Here, the homogeneous equation is given byy″+4y=0and the characteristic equation is

r^2+4=0

r^2=-4r

=±2i

So, the complementary solution of y″+4y=0 is

y_c=c_1cos⁡(2t)+c_2sin⁡(2t)where c1 and c2 are constants of integration.

Now, we find the particular solution of y″+4y = 4cos⁡2tusing the method of undetermined coefficients.

Let's assume that the particular solution of

y″+4y = 4cos⁡2t is

y_p=Acos⁡(2t)+Bsin⁡(2t)

where A and B are constants.

Now,y_p'=−2Asin⁡(2t)+2Bcos⁡(2t)y_p''

=−4Acos⁡(2t)−4Bsin⁡(2t)

Therefore,y_p''+4y_p

=−4Acos⁡(2t)−4Bsin⁡(2t)+4Acos⁡(2t)+4Bsin⁡(2t)

=4(cos⁡2tA+sin⁡2tB)=4cos⁡2t

Let's compare the coefficients.

We have cos⁡2t coefficient equal to 4 and sin⁡2t coefficient equal to 0.

So, A=2 and B=0.

Substituting A=2 and B=0, the particular solution isy_p=2cos⁡(2t)

Therefore, the general solution of y″+4y=4cos⁡2t is given by

y=y_c+y_p

=c_1cos⁡(2t)+c_2sin⁡(2t)+2cos⁡(2t)

Simplifying this, we have

y= (c1+2)cos⁡(2t)+c2sin⁡(2t)

Therefore, the solution to the given differential equation with the initial conditions

y(0)=0 and

y′(0)=1 is

y = 2cos⁡(2t)−\dfrac{1}{2}sin⁡(2t)

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It must be true that Jordan is a proficient writer who can efficiently write essays and outline chapters. This suggests that Jordan possesses good time organisation skills and is able to balance his workload effectively.


Working efficiently, Jordan can write 3 essays and outline 4 chapters each week. To determine what must be true, let's break it down step-by-step:

1. Jordan can write 3 essays each week.
  This means that Jordan has the ability to complete 3 essays within a week. It indicates his writing capability and efficiency.

2. Jordan can outline 4 chapters each week.

  This means that Jordan can create an outline for 4 chapters within a week. Outlining chapters is a task that requires organizing and summarizing the main points of each chapter.

Given these two statements, we can conclude the following:

- Jordan has the skill to write essays and outline chapters.

- Jordan's writing efficiency allows him to complete 3 essays in a week.

- Jordan's ability to outline chapters enables him to outline 4 chapters in a week.

It must be true that Jordan is a proficient writer who can efficiently write essays and outline chapters. This suggests that Jordan possesses good time management skills and is able to balance his workload effectively.

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The symbolic notation of the given statement is S(x) → C(x), C(x) → M(x), M(x) → E(x), S(m) → E(m)Where S(x) denotes that x is a student of Chemistry. C(x) denotes that x has passed Math. M(x) denotes that x has a test this week. E(x) denotes that x has an exam.b)

The argument can be proved to be valid by using predicate logic. To prove the validity of the argument, you can use a truth table. In this case, since the statement is a conditional statement, the only time it is false is when the hypothesis is true and the conclusion is false.

The truth table for the statement is as follows: S(x)C(x)M(x)E(x)S(m)E(m)TTTTF Therefore, the argument is valid as per predicate logic.

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The equation of the line perpendicular to Line 1 which passes through (6, -3) is: y = -x + 3.

To find the equation of the line parallel to Line 1 that passes through (6, -3), we know that both lines have the same slope. Thus, the new line's slope is 1. To find the y-intercept, we can substitute the x and y coordinates of the given point (6, -3) into the equation and solve for b: -3 = (1)(6) + b-3 = 6 + b-9 = b

Therefore, the equation of the line parallel to Line 1 which passes through (6, -3) is: y = x - 9.

To find the equation of the line perpendicular to Line 1 that passes through (6, -3), we know that the new line's slope is the negative reciprocal of Line 1's slope. Line 1's slope is 1, so the new line's slope is -1. To find the y-intercept, we can substitute the x and y coordinates of the given point (6, -3) into the equation and solve for b: -3 = (-1)(6) + b-3 = -6 + b3 = b

Therefore, the equation of the line perpendicular to Line 1 which passes through (6, -3) is: y = -x + 3.

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The translation to an equation is 2x + 6 = 8

To translate the given sentence into an equation, we need to break it down into mathematical terms. The sentence states that "the sum of 2 times a number and 6 is 8." Let's assign the unknown number as x.

The first step is to express "2 times a number" mathematically, which can be written as 2x. The second step is to include the phrase "and 6," indicating that we need to add 6 to the expression 2x. Finally, the equation states that the sum of 2x and 6 is equal to 8.

Putting it all together, we get the equation 2x + 6 = 8. This equation can be used to solve for the unknown number x by simplifying and isolating x on one side of the equation.

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The resulting matrix T² represents the probabilities of individuals moving between intersections after two time steps.

To calculate T²,  to first understand what the matrix T represents. Let's define the matrix T:

T = | t11 t12 |

| t21 t22 |

In this context, T is a transition matrix that describes the movement of individuals between the four intersections: Owens Bridge on Parkway East (OE), Bay Bridge on Parkway East (BE), Bay Bridge on Parkway West (BW), and Owens Bridge on Parkway West (OW).

Each entry tij of the matrix T represents the probability of an individual moving from intersection i to intersection j. For example, t11 represents the probability of someone moving from Owens Bridge on Parkway East (OE) back to Owens Bridge on Parkway East (OE), t12 represents the probability of someone moving from Owens Bridge on Parkway East (OE) to Bay Bridge on Parkway East (BE), and so on.

The transition matrix T should be constructed based on the given information about the movement of individuals between these intersections. The entries should be probabilities, meaning they should be between 0 and 1, and the sum of each row should be equal to 1 since a person must move to one of the four intersections.

Once the matrix T is defined, calculating T² means multiplying T by itself:

T² = T × T

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The polynomial  x³+2x²+4x+3 is the quotient obtained when using synthetic division to divide  x⁴+3x³+3x²+4x+3 by x+1.

The dividend is x⁴+3x³+3x²+4x+3 and the divisor is x+1. The first step to use synthetic division is to write down the coefficients of the dividend in a horizontal manner:

1 | 1 3 3 4 3 ___ |

The coefficient of the highest degree is 1. To the left of the vertical line, we will write the coefficients of the dividend, which are:

1, 3, 3, 4, and 3. 1 | 1 3 3 4 3 ____ |

The first step is to bring down the first coefficient of the dividend, which is

1.1 | 1 3 3 4 3____ | 1

The next step is to multiply the first term of the divisor by the number that was brought down. In this situation,

1 × 1 = 1.1 | 1 3 3 4 3____ | 1 1

After multiplying the first term of the divisor by the number that was brought down, the product is entered beneath the next coefficient of the dividend:

1 | 1 3 3 4 3____ | 1 1    ↓  1

The next step is to add the product to the next coefficient of the dividend

1 | 1 3 3 4 3____ | 1 1    ↓  1  1

The sum of the previous two numbers in the dividend is written below the line:

1 | 1 3 3 4 3____ | 1 1    ↓  1  1  4

Then, repeat the process with the new number, multiply it by the divisor and add the product to the following coefficient of the dividend

1 | 1 3 3 4 3____ | 1 1  4    ↓  1  1  4  7

Finally, repeat the procedure once more:

1 | 1 3 3 4 3____ | 1 1  4  7    ↓  1  1  4  7  10

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If ax + by = 1 for some x, y = Z, then ged(a, b) = 1 because if d is not equal to 1, then d is a common divisor of a and b that is greater than 1. This contradicts the fact that d is the gcd of a and b. If m is not a multiple of 5, then m² is either congruent to 1 or −1 modulo 5.

(1) Let m be an integer, not divisible by 5.

Hence, we can write, m = 5k + r,

where k and r are integers, and 0 < r < 5

(as if r = 0, then m would be divisible by 5).

If r = ±1,

then m² = (5k ± 1)²

= 25k² ± 10k + 1

= 5(5k² ± 2k) + 1

≡ 1 (mod 5).

If r = ±2,

then m² = (5k ± 2)²

= 25k² ± 20k + 4

= 5(5k² ± 4k) + 4

≡ −1 (mod 5).

Thus, we see that if m is not a multiple of 5, then m² is either congruent to 1 or −1 modulo 5.

(2) Suppose that d is the gcd of a and b.

Then, there exist integers x' and y' such that d = ax' + by' .

Now, suppose that d is not equal to 1, i.e., d > 1.

Then, ax' and by' are both multiples of d, so d divides ax' + by' = d.

Thus, d = ad' for some integer d'.

Hence, b = (1 − ax')y', so b is a multiple of d.

Therefore, if d is not equal to 1, then d is a common divisor of a and b that is greater than 1. This contradicts the fact that d is the gcd of a and b.

So, we see that there cannot exist a common divisor of a and b that is greater than 1, so ged(a, b) = 1.

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The number of different finishing orders possible for the 20 teams in the English Premier League can be calculated using the concept of permutations.

In this case, since all the teams are distinct and the order matters, we can use the formula for permutations. The formula for permutations is n! / (n - r)!, where n is the total number of items and r is the number of items taken at a time.

In this case, we have 20 teams and we want to find the number of different finishing orders possible. So, we need to find the number of permutations of all 20 teams taken at a time. Using the formula, we have:

20! / (20 - 20)! = 20! / 0! = 20!

Therefore, there are 20! different finishing orders possible for the 20 teams in the English Premier League.

To put this into perspective, 20! is a very large number. It is equal to 2,432,902,008,176,640,000, which is approximately 2.43 x 10^18. This means that there are over 2 quintillion different finishing orders possible for the 20 teams.

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Answer:

To simplify the expression "X + x + y + y," you can combine like terms:

X + x + y + y = (X + x) + (y + y) = 2x + 2y

So, the simplified form of the expression is 2x + 2y.

To find the point that represents the solution to the equation f(x) = g(x), we need to find the x-coordinate at which the two functions intersect. We can do this by setting f(x) equal to g(x) and solving for x.

Given: f(x) = 3x - 1 g(x) = -2x + 4

Setting f(x) equal to g(x): 3x - 1 = -2x + 4

Now we can solve for x: 3x + 2x = 4 + 1 5x = 5 x = 1

To find the corresponding y-coordinate, we substitute the value of x into either f(x) or g(x).

Let's use f(x): f(1) = 3(1) - 1 f(1) = 3 - 1 f(1) = 2

Therefore, the point that represents the solution to the equation f(x) = g(x) is (1, 2).

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The dimensions of the garden are 8 feet by 10 feet.

Let's denote the width of the garden as "x" (in feet).

According to the problem, the length of the garden is 2 feet greater than its width, so the length can be expressed as "x + 2" (in feet).

The area of the garden is given as 80 square feet, so we can set up the equation:

Area = Length * Width

80 = (x + 2) * x

Expanding the equation:

80 = x^2 + 2x

Rearranging the equation to make it a quadratic equation:

x^2 + 2x - 80 = 0

Now, we can solve this quadratic equation by factoring, completing the square, or using the quadratic formula. Let's solve it by factoring:

(x + 10)(x - 8) = 0

This gives us two possible solutions: x = -10 and x = 8. Since the dimensions of a garden cannot be negative, we discard the solution x = -10.

Therefore, the width of the garden is x = 8 feet.

To find the length, we can substitute the value of x into the expression for the length: x + 2 = 8 + 2 = 10 feet.

Therefore, the correct answer is option A: 8ft by 10ft.

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John needs to make a 16 inches cut of the tiles along the median. The correct answer is option C. 16 inches.

When cutting the tile along the median, we need to find the length of the cut that divides the trapezoid into two equal areas.

The median of a trapezoid is the line segment connecting the midpoints of the two non-parallel sides. In this case, the top base of the trapezoid is 13 inches and the bottom base is 19 inches.

To find the length of the cut, we can take the average of the lengths of the top and bottom bases. The average of 13 inches and 19 inches is (13 + 19) / 2 = 32 / 2 = 16 inches.

Therefore, John will need to make a 16-inch cut along the median to cut the tiles in half and create the desired pattern on his floor.

Option C, 16 inches, correctly represents the length of the cut required to cut the tiles along the median.

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The simplified expression is (-1-i)/2

To simplify the expression, (2-3i) / (1+5i), we have to multiply the numerator and denominator by the complex conjugate of the denominator.

We know that the complex conjugate of (1+5i) is (1-5i).

Hence, we can multiply the numerator and denominator by (1-5i) to get:

$$\frac{(2-3i)}{(1+5i)}=\frac{(2-3i)\cdot(1-5i)}{(1+5i)\cdot(1-5i)}$$$$=\frac{2-10i-3i+15i^2}{1^2-(5i)^2}$$$$=\frac{2-10i-3i+15(-1)}{1-25i^2}$$$$=\frac{-13-13i}{26}$$$$=\frac{-1-i}{2}$$

Thus, the simplified expression is (-1-i)/2.

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There are infinitely many vectors u in R such that B = {u, u2, u3} is an orthonormal basis.

Consider the vectors u1 = [1/2] [1/2] [1/2] [1/2], u2 = [1/2] [1/2] [-1/2] [-1/2], and u3 = [1/2] [-1/2] [1/2] [-1/2].

There is a vector u in R that the B = {u, u2, u3} is an orthonormal basis. If so, how many such vectors are there?

Solution:

Let u = [a, b, c, d]

It is given that B = {u, u2, u3} is an orthonormal basis.

This implies that the dot products between the vectors of the basis must be 0, and the norms must be 1.i.e

(i) u . u = 1

(ii) u2 . u2 = 1

(iii) u3 . u3 = 1

(iv) u . u2 = 0

(v) u . u3 = 0

(vi) u2 . u3 = 0

Using the above, we can determine the values of a, b, c, and d.

To satisfy equation (i), we have, a² + b² + c² + d² = 1....(1)

To satisfy equation (iv), we have, a/2 + b/2 + c/2 + d/2 = 0... (2)

Let's call equations (1) and (2) to the augmented matrix.

[1 1 1 1 | 1/2] [1 1 -1 -1 | 0] [1 -1 1 -1 | 0]

Let's do the row reduction[1 1 1 1 | 1/2][0 -1 0 -1 | -1/2][0 0 -2 0 | 1/2]

On solving, we get: 2d = 1/2

=> d = 1/4

a + b + c + 1/4 = 0....(3)

After solving equation (3), we get the equation of a plane as follows:

a + b + c = -1/4

So there are infinitely many vectors that can form an orthonormal basis with u2 and u3. The condition that the norms must be 1 determines a sphere of radius 1/2 centered at the origin.

Since the equation of a plane does not intersect the origin, there are infinitely many points on the sphere that satisfy the equation of the plane, and hence there are infinitely many vectors that can form an orthonormal basis with u2 and u3.

So, there are infinitely many vectors u in R such that B = {u, u2, u3} is an orthonormal basis.

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Using Fermat's Little Theorem  83^98 mod 13 is 2.

Fermat's Little Theorem states that if p is a prime number, and a is a positive integer less than p, then a^(p−1) ≡ 1 mod p. Now we can use this theorem to compute 83^98 mod 13.

a = 83 and p = 13

Since 83 is not divisible by 13, we can use Fermat's Little Theorem. Here, we have to find the exponent (p-1), which is 12 because 13-1=12.Therefore, we can use a^(p-1) ≡ 1 mod p to simplify the expression:

83^(12) ≡ 1 mod 13

Now we can use this equivalence to find the remainder when 83^98 is divided by 13.83^(12) = 1 mod 1383^96 = (83^12)^8 = 1^8 = 1 mod 1383^98 = 83^96 * 83^2 = 1 * 83^2 mod 13

Now, we need to calculate the remainder when 83^2 is divided by 13.83^2 = 6889 = 13 * 529 + 2

Hence, 83^98 ≡ 83^2 ≡ 2 mod 13.

Therefore, 83^98 mod 13 is 2.

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The resulting matrix after performing the given elementary row operations is:

[2 0 -1|-7]

[0 4 -1|-1]

[0 8 -1|-0]

Performing the indicated elementary row operation(s), the given matrix can be transformed as follows:

[2 0 -1|-7]

[1 -4 0| 3]

[-2 8 0|-0]

2R₂ + R₁:

[2 0 -1|-7]

[0 4 -1|-1]

[-2 8 0|-0]

R₂ + R₁:

[2 0 -1|-7]

[0 4 -1|-1]

[0 8 -1|-0]

So, the resulting matrix after performing the given elementary row operations is:

[2 0 -1|-7]

[0 4 -1|-1]

[0 8 -1|-0]

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The value of d is sqrt(10), which is approximately 3.162.

To find the value of d, we need to determine the coordinates of point B on the line AB. We know that the gradient of the line AB is 3, which means that for every 1 unit increase in the x-coordinate, the y-coordinate increases by 3 units.

Given that point A has coordinates (5, 9), we can use the gradient to find the coordinates of point B. Since B lies on the line AB, it must have the same gradient as AB. Starting from point A, we move 1 unit in the x-direction and 3 units in the y-direction to get to point B.

Therefore, the coordinates of B can be calculated as follows:

x-coordinate of B = x-coordinate of A + 1 = 5 + 1 = 6

y-coordinate of B = y-coordinate of A + 3 = 9 + 3 = 12

So, the coordinates of point B are (6, 12).

Now, to find the value of d, we can use the distance formula between points A and B:

d = [tex]sqrt((x2 - x1)^2 + (y2 - y1)^2)[/tex]

= [tex]sqrt((6 - 5)^2 + (12 - 9)^2)[/tex]

= [tex]sqrt(1^2 + 3^2)[/tex]

= sqrt(1 + 9)

= sqrt(10)

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The result of (3+4i)^20 is -1,072,697,779,282,031 + 98,867,629,664,588i.

To find the value of (3+4i)^20, we can use the concept of De Moivre's theorem. According to De Moivre's theorem, (a+bi)^n can be expressed as (r^n) * (cos(nθ) + i*sin(nθ)), where r is the magnitude of a+bi and θ is the angle it forms with the positive real axis.

In this case, a = 3 and b = 4, so the magnitude r can be calculated as √(a^2 + b^2) = √(3^2 + 4^2) = √(9 + 16) = √25 = 5. The angle θ can be found using the inverse tangent function, tan^(-1)(b/a) = tan^(-1)(4/3) ≈ 53.13 degrees (or ≈ 0.93 radians).

Now, we can express (3+4i)^20 as (5^20) * [cos(20*0.93) + i*sin(20*0.93)]. Evaluating this expression, we get (5^20) * [cos(18.6) + i*sin(18.6)].

Since cos(18.6) ≈ -0.9165 and sin(18.6) ≈ 0.3999, we can simplify the expression to (5^20) * (-0.9165 + 0.3999i).

Finally, calculating (5^20) = 9,536,743,164,062,500, we can substitute this value back into the expression and obtain the final result of -1,072,697,779,282,031 + 98,867,629,664,588i.

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The earliest time the operation can continue is approximately 1.03 minutes. According to the given rational function C(t) = 3t/(t^2 + 3), the concentration of the antibiotic in the blood can be determined.

The operation can begin when the concentration reaches 0.5 g/mL. By solving the equation, it is determined that the earliest time the operation can continue is approximately 1.03 minutes.

To find the earliest time the operation can continue, we need to solve the equation C(t) = 0.5. By substituting 0.5 for C(t) in the rational function, we get the equation 0.5 = 3t/(t^2 + 3).

To solve this equation, we can cross-multiply and rearrange terms to obtain 0.5(t^2 + 3) = 3t. Simplifying further, we have t^2 + 3 - 6t = 0.

Now, we have a quadratic equation, which can be solved using factoring, completing the square, or the quadratic formula. In this case, let's use the quadratic formula: t = (-b ± √(b^2 - 4ac)) / (2a).

Comparing the quadratic equation to our equation, we have a = 1, b = -6, and c = 3. Plugging these values into the quadratic formula, we get t = (-(-6) ± √((-6)^2 - 4(1)(3))) / (2(1)).

Simplifying further, t = (6 ± √(36 - 12)) / 2, which gives us t = (6 ± √24) / 2. The square root of 24 can be simplified to 2√6.

So, t = (6 ± 2√6) / 2, which simplifies to t = 3 ± √6. We can approximate this value to t ≈ 3 + 2.45 or t ≈ 3 - 2.45. Therefore, the earliest time the operation can continue is approximately 1.03 minutes.

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Answer:

Graph 2

Step-by-step explanation:

On graph 2, the line goes slowly up along the y value, meaning that his speed is increasing. (Chip begins his ride slowly)

Then, it suddenly stops and does not increase for an interval of time. (Chip stops to talk to some friends)

The speed then gradually picks back up. (He continues his ride, gradually picking up his speed)

The first three nonzero terms in the Taylor polynomial approximation for the given initial value problem are: 1 - t^2/8 + t^4/128.

Given the initial value problem: x′′ + 8tx = 0; x(0) = 1, x′(0) = 0. To find the first three nonzero terms in the Taylor polynomial approximation, we follow these steps:

Step 1: Find x(t) and x′(t) using the integrating factor.

We start with the differential equation x′′ + 8tx = 0. Taking the integrating factor as I.F = e^∫8t dt = e^4t, we multiply it on both sides of the equation to get e^4tx′′ + 8te^4tx = 0. This simplifies to e^4tx′′ + d/dt(e^4tx') = 0.

Integrating both sides gives us ∫ e^4tx′′ dt + ∫ d/dt(e^4tx') dt = c1. Now, we have e^4tx' = c2. Differentiating both sides with respect to t, we get 4e^4tx' + e^4tx′′ = 0. Substituting the value of e^4tx′′ in the previous equation, we have -4e^4tx' + d/dt(e^4tx') = 0.

Simplifying further, we get -4x′ + x″ = 0, which leads to x(t) = c3e^(4t) + c4.

Step 2: Determine the values of c3 and c4 using the initial conditions.

Using the initial conditions x(0) = 1 and x′(0) = 0, we can substitute these values into the expression for x(t). This gives us c3 = 1 and c4 = -1/4.

Step 3: Write the Taylor polynomial approximation.

The Taylor approximation to three nonzero terms is x(t) = 1 - t^2/8 + t^4/128 + ...

Therefore, the starting value problem's Taylor polynomial approximation's first three nonzero terms are: 1 - t^2/8 + t^4/128.

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Answer:

Step-by-step explanation:

The correct option is D. 4

Result: the degree of a polynomial is the highest of the degrees of the polynomial equation  with non-zero coefficients.

Given,

[tex]12x^4-8x+4x^2-3[/tex]

Clearly it is polynomial in x with coefficient 12 and highest degree is 4.

Therefore the degree of the polynomial is 4.

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Answer:

3) Definition of angle bisector

4) Reflexive property (of congruence)

5) SAS

In this manner, ready to conclude that an < am for all positive integers n and a few positive numbers k.

To appear that an < am, we got to compare the values of the arrangements an and am for all positive integers n and a few positive numbers k.

Given:

an = 2n + 1

am = n + k*o(n)

where o(n) signifies the arrange of n, speaking to the number of digits in n.

Let's compare an and am by substituting the expressions for an and am:

an = 2n + 1

am = n + k*o(n)

We want to appear that an < am, so we got to demonstrate that 2n + 1 < n + k*o(n) holds for all positive integers n and a few positive numbers k.

Let's simplify the inequality:

2n + 1 < n + k*o(n)

Modifying the terms:

n < k*o(n) - 1

Presently, we ought to consider the behavior of the arrange work o(n). The arrange work o(n) counts the number of digits in n. For any positive numbers n, o(n) will be greater than or break even with to 1.

Since o(n) ≥ 1, able to conclude that k*o(n) ≥ k.

Substituting this imbalance back into the first disparity, we have:

n < k*o(n) - 1 ≤ k - 1

Since n could be a positive numbers, and k may be a positive numbers, we have n < k - 1, which holds for all positive integers n and a few positive numbers k.

In this manner, ready to conclude that an < am for all positive integers n and a few positive numbers k.

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The solution is an < m.

Here is a more detailed explanation of the solution:

The first step is to show that ko(n) is always greater than or equal to 0. This is true because k is a positive integer, and the order of operations dictates that multiplication is performed before addition.

Therefore, ko(n) = k * o(n) = k * (n + 1), which is always greater than or equal to 0.

The second step is to show that m = n + ko(n) is always greater than or equal to n.

This is true because ko(n) is always greater than or equal to 0, so m = n + ko(n) = n + (k * (n + 1)) = n + k * n + k = (1 + k) * n + k.

Since k is a positive integer, (1 + k) is always greater than 1, so (1 + k) * n + k is always greater than n.

The third step is to show that an = 2n + 1 is always less than m.

This is true because m = (1 + k) * n + k is always greater than n, and an = 2n + 1 is always less than n.

Therefore, an < m.

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Answer: A

Step-by-step explanation: