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2 years ago

What is #5 I don’t know the answer.

Mathematics

1 answer:

denpristay [2]2 years ago

4 0

If the axis of symmetry is x=-1, the vertex will have x-coordinate -1.

The point (0, 5) cannot be the vertex. The friend is not correct.

_____

If the point (1, 5) is on the graph so is the image of that point when it is reflected across the axis of symmetry. That image point is (-3, 5). A parabola that goes through points (1, 5) and (-3, 5) will not go through point (0, 5).

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Look at the attachment pls!!

Naddika [18.5K]

The terms $-12x^2y$ and $4x^2y$ can be added to $3x^2y$ to result in a monomial.

Step-by-step explanation:

Given term is;

$3x^2y$

A monomial is an algebraic expression that consists of only one term.

So in the given expressions, we will add the terms which have same variables as given terms.

Given options are;

$3xy\\-12x^2y\\2x^2y^2\\7xy^2\\-10x^2\\4x^2y\\3x^3$

The terms $-12x^2y$ and $4x^2y$ can be added to $3x^2y$ to result in a monomial.

4 0

3 years ago

Simplify the following expression. show your work.<br> (-y + 5.3) + (7.2y - 9)

Kitty [74]

Answer:

6.2y - 3.7

Step-by-step explanation:

−y+5.3+7.2y−9

Subtract 9 from 5.3 to get −3.7.

−y−3.7+7.2y

Combine −y and 7.2y to get 6.2y.

6.2y−3.7

6 0

2 years ago

What is 10/3 as a mixed number

Fudgin [204]

<span>10/3 as a mixed number = 3 1/3
------------------------------------------------</span>

7 0

3 years ago

Read 2 more answers

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

3 years ago

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Follow the steps and finish the solution. 7 (x minus 3) = 28 Distributive property: 7 x minus 21 = 28. Addition property of equa

Nookie1986 [14]

Answer:

Step-by-step explanation:

7(x - 3) = 28 Given

7x - 21 = 28 Distributive property

7x - 21 + 21 = 28 + 21 Addition property of equality

7x = 49 Division Property of equality

7x/7=49/7

x = 7

6 0

3 years ago

Read 2 more answers