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ivanzaharov [21]
3 years ago
10

Y=x+3 2x-y=-1 write solution as ordered pair

Mathematics
1 answer:
dsp733 years ago
5 0

The first equation tells you that y=x+3. If you substitute this expression for y in the second, we have

2x-(x+3)=-1 \iff x-3=-1 \iff x=2

And we can deduce y=x+3=2+3=5

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With one<br> do I pick<br> the right on e
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B) 2n - 5

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8 0
3 years ago
I'm confused. Can someone help me out with this problem step-by-step using the P = 2L + 2W formula?
german

We can't use P = 2L + 2W because  the sides have different lengths.

P = a + b + c + d

a = x

b = 2x

c = 2x - 2

d = x + 5

P = x + 2x + (2x - 2) + (x + 5)

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3 0
4 years ago
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Alexeev081 [22]
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6 0
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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
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