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3 years ago

What are the coordinates of the vertex of the parabola described by the

Mathematics

2 answers:

mixas84 [53]3 years ago

5 0

Did you mean y = (2x +5)^2 - 3 ?
The answer would then be (-2.5, -3)

ICE Princess25 [194]3 years ago

3 0

Step-by-step explanation:

please type the question properly

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The table shows values for functions f(x) and g(x). x f(x)=2x−3 g(x)=73x−2 −1 −52 −133 0 −2 −2 1 −1 13 2 1 83 3 5 5 4 13 223 5 2

Verizon [17]

0 and 2

f(x) = g(x) will be the input or x value at which f and g have the same output or y value. Look in the table where two numbers repeat right next to each other.

−1 −7/2 −9/2

0 −3 −3

1 −2 −3/2

2 0 0

3 4 3/2

4 12 3

5 28 9/2

There are two solutions to f(x) = g(x) which are x=0 and x=2.

6 0

3 years ago

Which expression is equivelent to (-2)(a+6) <br> -2a+6 <br> 2a+12 <br> -2a-12 <br> -2a+12

pogonyaev

Answer:

-2a -12

Step-by-step explanation:

a minus * a plus is a negative therefore -12

7 0

3 years ago

4sin²<img src="https://tex.z-dn.net/?f=%5Cfrac%7Bx%7D%7B2%7D" id="TexFormula1" title="\frac{x}{2}" alt="\frac{x}{2}" align="absm

raketka [301]

Answer:

$\displaystyle x=\left \{\frac{2\pi}{3}+2\pi k,\frac{4\pi}{3}+2\pi k, \frac{8\pi}{3}+2\pi k, \frac{10\pi}{3}+2\pi k\right \}k\in \mathbb{Z}$

Step-by-step explanation:

Hi there!

We want to solve for $x$ in:

$4\sin^2(\frac{x}{2})=3$

Since $x$ is in the argument of $\sin^2$ , let's first isolate $\sin^2$ by dividing both sides by 4:

$\displaystyle \sin^2\left(\frac{x}{2}\right)=\frac{3}{4}$

Next, recall that $\sin^2x$ is just shorthand notation for $(\sin x)^2$ . Therefore, take the square root of both sides:

$\displaystyle \sqrt{\sin^2\left(\frac{x}{2}\right)}=\sqrt{\frac{3}{4}},\\\sin\left(\frac{x}{2}\right)=\pm \sqrt{\frac{3}{4}}$

Simplify using $\displaystyle \sqrt{\frac{a}{b}}=\frac{\sqrt{a}}{\sqrt{b}}$ :

$\displaystyle \sin\left(\frac{x}{2}\right)=\pm \sqrt{\frac{3}{4}},\\\sin\left(\frac{x}{2}\right)=\pm \frac{\sqrt{3}}{\sqrt{4}}=\pm \frac{\sqrt{3}}{2}$

Let $\phi = \frac{x}{2}$ .

<h3><u>Case 1 (positive root):</u></h3>

$\displaystyle \sin(\phi)=\frac{\sqrt{3}}{2},\\\phi = \frac{\pi}{3}+2\pi k, k\in \mathbb{Z}, \\\\\phi =\frac{2\pi}{3}+2\pi k, k\in \mathbb{Z}$

Therefore, we have:

$\displaystyle \frac{x}{2}=\phi = \frac{\pi}{3}+2\pi k, k\in \mathbb{Z}, \\\\\frac{x}{2}=\phi =\frac{2\pi}{3}+2\pi k, k\in \mathbb{Z},\\\\\begin{cases}x=\boxed{\frac{2\pi}{3}+2\pi k, k\in \mathbb{Z}},\\x=\boxed{\frac{4\pi}{3}+2\pi k , k \in \mathbb{Z}}\end{cases}$

<h3><u>Case 2 (negative root):</u></h3>

$\displaystyle \sin(\phi)=-\frac{\sqrt{3}}{2},\\\phi = \frac{4\pi}{3}+2\pi k, k\in \mathbb{Z}, \\\\\phi =\frac{5\pi}{3}+2\pi k, k\in \mathbb{Z},\\\begin{cases}x=\boxed{\frac{8\pi}{3}+2\pi k, k\in \mathbb{Z}},\\x=\boxed{\frac{10\pi}{3}+2\pi k , k \in \mathbb{Z}}\end{cases}$