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3 years ago

The ratios are ___; therefore, the points in the table form a direct variation.

Mathematics

1 answer:

barxatty [35]3 years ago

8 0

Are not constant
Do not

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Find the equation of an exponential function in the form y = ab^x, given the points (0, 3) and (2, 108/25). Please simplify your

lilavasa [31]

We have the equation:

$y=a\cdot b^x$

We know two points and we will use them to calculate the parameters a and b.

The point (0,3) will let us know a, as b^0=1.

$\begin{gathered} y=a\cdot b^x \\ 3=a\cdot b^0=a \\ a=3 \end{gathered}$

Now, we use the point (2, 108/25) to calcualte b:

$\begin{gathered} y=3\cdot b^x \\ \frac{108}{25}=3\cdot b^2 \\ 3\cdot b^2=\frac{108}{25} \\ b^2=\frac{108}{25\cdot3}=\frac{108}{3}\cdot\frac{1}{25}=\frac{36}{25} \\ b=\sqrt[]{\frac{36}{25}} \\ b=\frac{\sqrt[]{36}}{\sqrt[]{25}} \\ b=\frac{6}{5} \end{gathered}$

Then, we can write the equation as:

$y=3\cdot(\frac{6}{5})^x$

5 0

1 year ago

What’s the slope of the line?

Dmitry [639]

The slope of the line is 1. the equation would be y=x-3

3 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>

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

8 0

3 years ago

I need help show your work

Assoli18 [71]

The answer is 71 and 180
have a nice day!!

3 0

3 years ago

Please helppppppppppppp

KengaRu [80]

Answer:

f(x)=33√x2(3x−16)xf(x)=3x23(3x-16)x

am not sure wait for the next answer sorry if it didn't help

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3 years ago

The ratios are _________; therefore, the points in the table ______ form a direct variation.

The ratios are ___; therefore, the points in the table form a direct variation.