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2 years ago

Do number 7 plz finding QR

Mathematics

1 answer:

Elina [12.6K]2 years ago

6 0

Answer:

QR = sqrt(147)

Step-by-step explanation:

Since this is a right triangle, we can use the Pythagorean theorem

a^2 + b^2 = c^2

7^2 + QR^2 = 14^2

49 + QR^2 = 196

QR^2 = 196-49

QR^2 = 147

Taking the square root of each side

QR = sqrt(147)

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What is x cubed in standard form

Gelneren [198K]

Answer:

x^3

Step-by-step explanation:

It stays the same.

Hope this helps.

3 0

3 years ago

How do you do this and what are the answers

scoray [572]

I can answer 1 to 4...
The area of a circle can be found by $\pi r^{2}$ . Reminder: $\pi = 3.14$ (usually used when finding out the area of a circle)

1. $4^{2} = 16$
$\pi *16$ or $3.14 * 16 = 50.24$
$50.24/4=12.56$

2. $6^{2} = 36$
$\pi *36$ or $3.14 * 36 = 113.04$
$113.04/10=11.304$

(Not really sure if this is correct) 3. $2^{2}=4$
$\pi *4$ or $3.14 * 4$
$12.56/2.66666667=11.70999999$

4. $6^{2}=36$
$\pi *36= 113.04$
$113.04/6=18.84$

5 0

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,

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