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2 years ago

3(x -y) + 4 if X = 3 and y = 2

Mathematics

1 answer:

Alchen [17]2 years ago

7 0

Answer:

Step-by-step explanation:

3(3-2)+4

3(1)+4

3+4

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Please help!!! Find the value of the variable and the measures of the angles.

e-lub [12.9K]

Answer:

x= 38°

<P = 57°

<R = 38

<Q = 85°

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1 year ago

Could you please Factor 12 + 16

-BARSIC- [3]

If you add that together you get 28

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3 years ago

X^2 -9X = 0. Factor and use the Zero property to solve

Veseljchak [2.6K]

The first step of factoring is to try to factor out a common factor.
The terms x^2 and -9x have the factor x in common.
Factor out x from both terms.

x^2 - 9x = 0

x(x - 9) = 0

Now you have a product of fully factored terms equaling zero, so you can apply the zero product property to solve.

x = 0 or x - 9 = 0

x = 0 or x = 9

Answer: x = 0 or x = 9

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3 years ago

The graph of f(x) = x is given. Which equation will translate the graph 3 units down?

Rudiy27

Answer:

g(x)=x-3

Step-by-step explanation:

4 0

3 years ago

Read 2 more answers

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,

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

We have

$f=-\dfrac ba\cot\theta\implies\dfrac{\mathrm df}{\mathrm d\theta}=\dfrac ba\csc^2\theta$

and so the second derivative is

$\dfrac{\mathrm d^2y}{\mathrm dx^2}=\dfrac{\frac ba\csc^2\theta}{-a\sin\theta}=-\dfrac b{a^2}\csc^3\theta$

4 0

3 years ago