Please explain! Math

Multiply and simplify: sq rt 7x ((sq rt x)-7sq rt. 7)

jarptica [38.1K]

7x(√x - 7√7) Multiply √7x to "√x" and "-7√7" respectively:
√7x*√x - 7√7*√7x
First is to simplify: √7x² - 7√7²x
Extract the squares: x√7 - 7*7√x

Then, we simplify: x√7 - 49√x
After simplifying the equation, I got the answer x√7 - 49√x.

3 0

2 years ago

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In the relation defined by the equation y = x2, for all x > 0, y is a function of x because

Slav-nsk [51]

X cannot be a negative number

7 0

2 years ago

PLEASE ANSWER!!!

valkas [14]

Answer:

Step-by-step explanation:

3 0

1 year ago

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Applying GCF and LCM to fraction Operations? . . PLZ HELP

I am Lyosha [343]

When multiplying fractions just multiply the numerator and denominators together, then simplify the fraction you get.

To solve the addition problem, first convert 2 and 1/2 to an improper fraction. It should be 5/2. Now 3/8 and 5/2 must be added together.

To add these, they must have the same denominator. Let's give 5/2 a denominator of 8, so it matches 3/8.

To give 5/2 a denominator of 8, you are starting with a denominator of 2. 2 × 4 = 8, so you must multiply 5/2 by 4.

You should get 20/8. Now you have to add 20/8 and 3/8. The only thing you add is the numerator, so you should get 23/8 as a final answer.

For the subtraction problem, do these same steps to make the fractions you must subtract have the same denominator, then subtract only the numerator to get the answer. Once you solve the subtraction problem, text me what you got so I can check it for you. :)

4 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 y with respect to x, 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

2 years ago

Please explain! Math​

Please explain! Math