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Nat2105 [25]
3 years ago
14

Write a quadratic equation whose solutions are 7i and -7i

Mathematics
1 answer:
aliya0001 [1]3 years ago
4 0
Because -7i and 7i are the solutions to the quadratic equation, you can first write it in factored form (x-7i)(x+7i). The you multiply this out and get x^{2} +49.
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Give the exponential function A(x)=P(1+r)x, what value for r will make the function a growth function?
wolverine [178]

for this to be a growth function it has to be any number that is latger than zero

FOR EXAMPLE :-

50% which equals to 0.5

4 0
3 years ago
A(0, 1), B(4,1), C(1, 2), D(1,6): AB and CD
Ivanshal [37]
The answer is D because I took the test
8 0
3 years ago
You are an office manager and you need to order supplies for your office. you need 3 boxes of computer diskettes at $8.99 per bo
DENIUS [597]
3 boxes of computer diskettes at $8.99 per box = (3 * 8.99 = 26.97 ) 
2 packages of pens at $ 2.50 per package = (2 * 2.50 = 5.00 ) 
6 boxes of paper at $ 22.95 per box = (6 * 22.95 = 137.70) 
1 printer ribbon at $25 each. = (1 * 25 = 25) 
Add up (26.97+5.00+137.70+25 ) and the answer is = 194.67
5 0
3 years ago
The quadrilateral shown is rotated 90° clockwise about the origin. In which quadrant is the image of the quadrilateral located?
Viktor [21]

Answer:

Option (2). 1

Step-by-step explanation:

Coordinates of point A, B, C and D are,

A(-4, 4), B(-2, 4), C(-2, 1) and D(-4, 3).

Quadrilateral ABCD when rotated 90° clockwise about the origin,

Rule for the rotation of the vertices,

(x, y) → (y, -x)

Following the rule of rotation coordinates of the image points,

A(-4, 4) → A'(4, 4)

B(-2, 4) → B'(4, 2)

C(-2, 1) → C'(1, 2)

D(-4, 3) → D'(3, 4)

Since all image points have the positive coordinates (x and y coordinates), image quadrilateral A'B'C'D' will be located in 1st quadrant.

Option (2) is the correct option.

4 0
2 years ago
Solve the given initial-value problem. the de is of the form dy dx = f(ax + by + c), which is given in (5) of section 2.5. dy dx
shutvik [7]

\dfrac{\mathrm dy}{\mathrm dx}=\cos(x+y)

Let v=x+y, so that \dfrac{\mathrm dv}{\mathrm dx}-1=\dfrac{\mathrm dy}{\mathrm dx}:

\dfrac{\mathrm dv}{\mathrm dx}=\cos v+1

Now the ODE is separable, and we have

\dfrac{\mathrm dv}{1+\cos v}=\mathrm dx

Integrating both sides gives

\displaystyle\int\frac{\mathrm dv}{1+\cos v}=\int\mathrm dx

For the integral on the left, rewrite the integrand as

\dfrac1{1+\cos v}\cdot\dfrac{1-\cos v}{1-\cos v}=\dfrac{1-\cos v}{1-\cos^2v}=\csc^2v-\csc v\cot v

Then

\displaystyle\int\frac{\mathrm dv}{1+\cos v}=-\cot v+\csc v+C

and so

\csc v-\cot v=x+C

\csc(x+y)-\cot(x+y)=x+C

Given that y(0)=\dfrac\pi2, we find

\csc\left(0+\dfrac\pi2\right)-\cot\left(0+\dfrac\pi2\right)=0+C\implies C=1

so that the particular solution to this IVP is

\csc(x+y)-\cot(x+y)=x+1

5 0
2 years ago
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