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2 years ago

Simplify h^6 x h ASAP please

Mathematics

1 answer:

ozzi2 years ago

8 0

Answer:1/2

Step-by-step explanation:

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SOMEONE HELP ME PLEASE ILL GIVE BRAINLY

e-lub [12.9K]

Answer:

Step-by-step explanation:

In order to determine the information you're being asked for, you need to complete the square on that quadratic. The first step is to move the constant over to the other side of the equals sign:

$x^2-2x=3$

Here would be the step where, if the leading coefficient isn't a 1, you'd factor it out. But ours is a 1, so we're good there. Now take half the linear term (the term with the single x on it), square it, and add it to both sides. Our linear term is a -2. Half of -2 is -1, and -1 squared is +1. We add +1 to both sides giving us this:

$x^2-2x+1=3+1$

Now we'll clean it up a bit. The right side becomes a 4, and the left side is written as its perfect square binomial, which is the whole reason we did this. That binomial is

$(x-1)^2=4$ (set equal to the 4 here). Now we'll move the 4 back over and set the whole thing back equal to y:

$y=(x-1)^2-4$

From this it's apparent what the vertex is: (1, -4),

the axis of symmetry is x = 1, and

the y-intercept is found by setting the x's equal to 0 in the original equation and solving for y. So the y-intercept is (0, -3).

Your choice for the correct answer is the very last one there.

8 0

2 years ago

Find the circumference and area of a circle with radius 3 yd. use the value 3.14 for pie

just olya [345]

Answer: circumference = 2Πr²

Where r= 3

Π=3.14

Therefore Circumference = 2×3.14× 3²

= 2×28.26

Circumference=56.52

Formula:

Subtitute in 3 for r.

Solve and the answer is 28.26

4 0

2 years ago

Read 2 more answers

In a game a player draws and replaces a card from a deck 2 times. The possible outcomes and payouts are shown. What is the

Goshia [24]

<u>Given</u>:

Given that in a game a player draws and replaces a card from a deck 2 times.

The possible outcomes and payouts are given.

We need to determine the expected value for someone playing the game.

<u>Expected value:</u>

The expected value for someone playing the game can be determined by

$EV=(\frac{26}{52})(\$ 20)+(\frac{52}{52})(\$4)+(\frac{52}{52})(\$ 0)+(\frac{26}{52})(-\$12)$

Simplifying the values, we have;

$EV=(\frac{1}{2})(\$ 20)+(1)(\$4)+(1)(\$ 0)+(\frac{1}{2})(-\$12)$

Dividing the terms, we get;

$EV=\$ 10+\$4+\$ 0+-\$6$

Adding, we have;

$EV=\$ 8$

Thus, the expected value for someone playing the game is $8

3 0

2 years ago

99 POINT QUESTION, PLUS BRAINLIEST!!!

VladimirAG [237]

First, we have to convert our function (of x) into a function of y (we revolve the curve around the y-axis). So:

$y=100-x^2\\\\x^2=100-y\qquad\bold{(1)}\\\\\boxed{x=\sqrt{100-y}}\qquad\bold{(2)} \\\\\\0\leq x\leq10\\\\y=100-0^2=100\qquad\wedge\qquad y=100-10^2=100-100=0\\\\\boxed{0\leq y\leq100}$

And the derivative of x:

$x'=\left(\sqrt{100-y}\right)'=\Big((100-y)^\frac{1}{2}\Big)'=\dfrac{1}{2}(100-y)^{-\frac{1}{2}}\cdot(100-y)'=\\\\\\=\dfrac{1}{2\sqrt{100-y}}\cdot(-1)=\boxed{-\dfrac{1}{2\sqrt{100-y}}}\qquad\bold{(3)}$

Now, we can calculate the area of the surface:

$A=2\pi\int\limits_0^{100}\sqrt{100-y}\sqrt{1+\left(-\dfrac{1}{2\sqrt{100-y}}\right)^2}\,\,dy=\\\\\\= 2\pi\int\limits_0^{100}\sqrt{100-y}\sqrt{1+\dfrac{1}{4(100-y)}}\,\,dy=(\star)$

We could calculate this integral (not very hard, but long), or use (1), (2) and (3) to get:

$(\star)=2\pi\int\limits_0^{100}1\cdot\sqrt{100-y}\sqrt{1+\dfrac{1}{4(100-y)}}\,\,dy=\left|\begin{array}{c}1=\dfrac{-2\sqrt{100-y}}{-2\sqrt{100-y}}\end{array}\right|= \\\\\\= 2\pi\int\limits_0^{100}\dfrac{-2\sqrt{100-y}}{-2\sqrt{100-y}}\cdot\sqrt{100-y}\cdot\sqrt{1+\dfrac{1}{4(100-y)}}\,\,dy=\\\\\\ 2\pi\int\limits_0^{100}-2\sqrt{100-y}\cdot\sqrt{100-y}\cdot\sqrt{1+\dfrac{1}{4(100-y)}}\cdot\dfrac{dy}{-2\sqrt{100-y}}=\\\\\\$

$=2\pi\int\limits_0^{100}-2\big(100-y\big)\cdot\sqrt{1+\dfrac{1}{4(100-y)}}\cdot\left(-\dfrac{1}{2\sqrt{100-y}}\, dy\right)\stackrel{\bold{(1)}\bold{(2)}\bold{(3)}}{=}\\\\\\= \left|\begin{array}{c}x=\sqrt{100-y}\\\\x^2=100-y\\\\dx=-\dfrac{1}{2\sqrt{100-y}}\, \,dy\\\\a=0\implies a'=\sqrt{100-0}=10\\\\b=100\implies b'=\sqrt{100-100}=0\end{array}\right|=\\\\\\= 2\pi\int\limits_{10}^0-2x^2\cdot\sqrt{1+\dfrac{1}{4x^2}}\,\,dx=(\text{swap limits})=\\\\\\$

$=2\pi\int\limits_0^{10}2x^2\cdot\sqrt{1+\dfrac{1}{4x^2}}\,\,dx= 4\pi\int\limits_0^{10}\sqrt{x^4}\cdot\sqrt{1+\dfrac{1}{4x^2}}\,\,dx=\\\\\\= 4\pi\int\limits_0^{10}\sqrt{x^4+\dfrac{x^4}{4x^2}}\,\,dx= 4\pi\int\limits_0^{10}\sqrt{x^4+\dfrac{x^2}{4}}\,\,dx=\\\\\\= 4\pi\int\limits_0^{10}\sqrt{\dfrac{x^2}{4}\left(4x^2+1\right)}\,\,dx= 4\pi\int\limits_0^{10}\dfrac{x}{2}\sqrt{4x^2+1}\,\,dx=\\\\\\=\boxed{2\pi\int\limits_0^{10}x\sqrt{4x^2+1}\,dx}$

Calculate indefinite integral:

$\int x\sqrt{4x^2+1}\,dx=\int\sqrt{4x^2+1}\cdot x\,dx=\left|\begin{array}{c}t=4x^2+1\\\\dt=8x\,dx\\\\\dfrac{dt}{8}=x\,dx\end{array}\right|=\int\sqrt{t}\cdot\dfrac{dt}{8}=\\\\\\=\dfrac{1}{8}\int t^\frac{1}{2}\,dt=\dfrac{1}{8}\cdot\dfrac{t^{\frac{1}{2}+1}}{\frac{1}{2}+1}=\dfrac{1}{8}\cdot\dfrac{t^\frac{3}{2}}{\frac{3}{2}}=\dfrac{2}{8\cdot3}\cdot t^\frac{3}{2}=\boxed{\dfrac{1}{12}\left(4x^2+1\right)^\frac{3}{2}}$

And the area:

$A=2\pi\int\limits_0^{10}x\sqrt{4x^2+1}\,dx=2\pi\cdot\dfrac{1}{12}\bigg[\left(4x^2+1\right)^\frac{3}{2}\bigg]_0^{10}=\\\\\\= \dfrac{\pi}{6}\left[\big(4\cdot10^2+1\big)^\frac{3}{2}-\big(4\cdot0^2+1\big)^\frac{3}{2}\right]=\dfrac{\pi}{6}\Big(\big401^\frac{3}{2}-1^\frac{3}{2}\Big)=\boxed{\dfrac{401^\frac{3}{2}-1}{6}\pi}$

Answer D.

6 0

2 years ago

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You provide a dog-walking service, and you report to the owners on the amount of exercise their dogs get on the walks. You walke

Eva8 [605]

the answer would be E because if you round answer E to 3mph and divide by 2 you get 1.5 and a hour is 60 minutes divide that by 2 and you get 30 minutes however you walked slightly slower so it would take 5 minutes longer.

7 0

2 years ago

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