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1 year ago

Which graph shows the solution to this system of inequalities? y<-x+1 y≤2x-3

Mathematics

1 answer:

professor190 [17]1 year ago

5 0

Answer:

graph C

Step-by-step explanation:

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What is the semiannual interest payments of a 30 year bond with a face value of $50 million and a stated annual interest rate of

Snowcat [4.5K]

Based on the stated annual interest rate and the face value of the bond, the semiannual payments will be $1,000,000.

<h3>How can the semiannual interest payment be found?</h3>

The formula to find the semiannual payment is:

= (Face value x Stated annual interest rate) / 2 semi-annual periods per year

Solving gives:

= (50,000,000 x 4%) / 2

= 2,000,000 / 2

= $1,000,000

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1 year ago

Two numbers are between 20 and 30. their greatest common factor is 4.Which two numbers could they be?

Andrew [12]

24 and 28
These are the only two numbers (other than 20 itself) that have a factor of 4. And if you were to find all of their common factors, 4 would be the greatest!

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2 years ago

Find the value of f(x) = -3x +2 <br> When x=2

Liono4ka [1.6K]

Answer:

it is -4

Step-by-step explanation:

-3 * 2 = -6 + 2 = -4

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2 years ago

If x = a cosθ and y = b sinθ , find second derivative

Olin [163]

I'm guessing the second derivative is for <em>y</em> with respect to <em>x</em>, i.e.

$\dfrac{\mathrm d^2y}{\mathrm dx^2}$

Compute the first derivative. By the chain rule,

$\dfrac{\mathrm dy}{\mathrm dx}=\dfrac{\mathrm dy}{\mathrm d\theta}\dfrac{\mathrm d\theta}{\mathrm dx}=\dfrac{\frac{\mathrm dy}{\mathrm d\theta}}{\frac{\mathrm dx}{\mathrm d\theta}}$

We have

$y=b\sin\theta\implies\dfrac{\mathrm dy}{\mathrm d\theta}=b\cos\theta$

$x=a\cos\theta\implies\dfrac{\mathrm dx}{\mathrm d\theta}=-a\sin\theta$

and so

$\dfrac{\mathrm dy}{\mathrm dx}=\dfrac{b\cos\theta}{-a\sin\theta}=-\dfrac ba\cot\theta$

Now compute the second derivative. Notice that $\frac{\mathrm dy}{\mathrm dx}$ is a function of $\theta$ ; so denote it by $f(\theta)$ . Then

$\dfrac{\mathrm d^2y}{\mathrm dx^2}=\dfrac{\mathrm df}{\mathrm dx}$

By the chain rule,

$\dfrac{\mathrm d^2y}{\mathrm dx^2}=\dfrac{\mathrm df}{\mathrm d\theta}\dfrac{\mathrm d\theta}{\mathrm dx}=\dfrac{\frac{\mathrm df}{\mathrm d\theta}}{\frac{\mathrm dx}{\mathrm d\theta}}$