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miss Akunina [59]

3 years ago

5

he graph represents the function f(x) = x2 + 3x + 2. If g(x) is the reflection of f(x) across the x-axis, g(x) = . (Write the fu

nction in standard form. Use ^ to indicate an exponent.)

1 answer:

kifflom [539]3 years ago

3 0

if g(x) is the reflection of f(x) across the x-axis that would mean g(x)= - f(x)

so that would mean you have to change all the signs of f(x) to negative

Ans: g(x)= - x^2 - 3x - 2

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Sedaia [141]

Answer:

A

Step-by-step explanation:

Hey There!

So these triangles are congruent so segment CD is equal to segment DB

so to find x we use the equation

7x+10=10x+7

now we solve for x

step 1 subtract each side by 7

10-7=3

7-7 cancels out

now we have 7x+3=10x

step 2 subtract each side by 7 x

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step 3 divide each side by 3

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Please help me with this question thank you

gulaghasi [49]

Answer:

$\mathrm{The\:solution\:is}$ :

$x=\frac{-y^2-1}{2-y^2}\space\left\{y\ge \:0\right\}$

Step-by-step explanation:

Given

$y=\sqrt{\frac{2x+1}{x-1}}$

Taking square of both sides

$y^2=\left(\sqrt{\frac{2x+1}{x-1}}\right)\:^2$

$\mathrm{Subtract\:}\left(\sqrt{\frac{2x+1}{x-1}}\right)^2\mathrm{\:from\:both\:sides}$

$y^2-\left(\sqrt{\frac{2x+1}{x-1}}\right)^2=\left(\sqrt{\frac{2x+1}{x-1}}\right)^2-\left(\sqrt{\frac{2x+1}{x-1}}\right)^2$

$\mathrm{Simplify}$

$y^2-\left(\sqrt{\frac{2x+1}{x-1}}\right)^2=0$

As we know that $\mathrm{Apply\:Difference\:of\:Two\:Squares\:Formula:\:}x^2-y^2=\left(x+y\right)\left(x-y\right)$

$\mathrm{Factor\:}y^2-\left(\sqrt{\frac{2x+1}{x-1}}\right)^2:\quad \left(y+\sqrt{\frac{2x+1}{x-1}}\right)\left(y-\sqrt{\frac{2x+1}{x-1}}\right)$

so

$\left(y+\sqrt{\frac{2x+1}{x-1}}\right)\left(y-\sqrt{\frac{2x+1}{x-1}}\right)=0$

$\mathrm{Using\:the\:Zero\:Factor\:Principle:\quad \:If}\:ab=0\:\mathrm{then}\:a=0\:\mathrm{or}\:b=0\:\left(\mathrm{or\:both}\:a=0\:\mathrm{and}\:b=0\right)$

$\mathrm{Solve\:}\:y+\sqrt{\frac{2x+1}{x-1}}=0$

$\mathrm{Subtract\:}y\mathrm{\:from\:both\:sides}$

$y+\sqrt{\frac{2x+1}{x-1}}-y=0-y$

$\sqrt{\frac{2x+1}{x-1}}=-y$

$\mathrm{Square\:both\:sides}$

$\left(\sqrt{\frac{2x+1}{x-1}}\right)^2=\left(-y\right)^2$

$\mathrm{Expand\:}\left(\sqrt{\frac{2x+1}{x-1}}\right)^2$

$\left(\sqrt{\frac{2x+1}{x-1}}\right)^2$

$\mathrm{Apply\:radical\:rule}:\quad \sqrt{a}=a^{\frac{1}{2}}$

$=\left(\left(\frac{2x+1}{x-1}\right)^{\frac{1}{2}}\right)^2$

$=\frac{2x+1}{x-1}$

so equation $\left(\sqrt{\frac{2x+1}{x-1}}\right)^2=\left(-y\right)^2$ becomes

$\frac{2x+1}{x-1}=y^2$

now

$\mathrm{Solve\:}\:\frac{2x+1}{x-1}=y^2$

$\frac{2x+1}{x-1}=y^2$

$\mathrm{Multiply\:both\:sides\:by\:}x-1$

$\frac{2x+1}{x-1}\left(x-1\right)=y^2\left(x-1\right)$

$2x+1=y^2\left(x-1\right)$

$2x+1=xy^2-y^2$ ∵ $y^2\left(x-1\right):\quad xy^2-y^2$

$2x=xy^2-y^2-1$

$2x-xy^2=-y^2-1$

$x\left(2-y^2\right)=-y^2-1$ ∵ $\mathrm{Factor}\:2x-xy^2:\quad x\left(2-y^2\right)$

$\mathrm{Divide\:both\:sides\:by\:}2-y^2$

$\frac{x\left(2-y^2\right)}{2-y^2}=-\frac{y^2}{2-y^2}-\frac{1}{2-y^2}$

$x=\frac{-y^2-1}{2-y^2}$

so

$y+\sqrt{\frac{2x+1}{x-1}}=0:\quad x=\frac{-y^2-1}{2-y^2}\space\left\{y\le \:0\right\}$

similarly

$y-\sqrt{\frac{2x+1}{x-1}}=0:\quad x=\frac{-y^2-1}{2-y^2}\space\left\{y\ge \:0\right\}$

$\mathrm{Verify\:Solutions}:\quad x=\frac{-y^2-1}{2-y^2}$

$\mathrm{Check\:the\:solutions\:by\:plugging\:them\:into\:}y^2=\left(\sqrt{\frac{2x+1}{x-1}}\right)^2$

$\mathrm{Remove\:the\:ones\:that\:don't\:agree\:with\:the\:equation.}$

$\mathrm{Plug}\quad x=\frac{-y^2-1}{2-y^2}$

$y^2=\left(\sqrt{\frac{2\left(\frac{-y^2-1}{2-y^2}\right)+1}{\left(\frac{-y^2-1}{2-y^2}\right)-1}}\right)^2$

$\mathrm{Subtract\:}\left(\sqrt{\frac{2\frac{-y^2-1}{2-y^2}+1}{\frac{-y^2-1}{2-y^2}-1}}\right)^2\mathrm{\:from\:both\:sides}$

$y^2-\left(\sqrt{\frac{2\cdot \frac{-y^2-1}{2-y^2}+1}{\frac{-y^2-1}{2-y^2}-1}}\right)^2=\left(\sqrt{\frac{2\cdot \frac{-y^2-1}{2-y^2}+1}{\frac{-y^2-1}{2-y^2}-1}}\right)^2-\left(\sqrt{\frac{2\cdot \frac{-y^2-1}{2-y^2}+1}{\frac{-y^2-1}{2-y^2}-1}}\right)^2$

$y^2-\left(\sqrt{\frac{2\cdot \frac{-y^2-1}{2-y^2}+1}{\frac{-y^2-1}{2-y^2}-1}}\right)^2=0$

$\mathrm{Factor\:}y^2-\left(\sqrt{\frac{2\frac{-y^2-1}{2-y^2}+1}{\frac{-y^2-1}{2-y^2}-1}}\right)^2:\quad \left(y+\sqrt{\frac{2\cdot \frac{-y^2-1}{2-y^2}+1}{\frac{-y^2-1}{2-y^2}-1}}\right)\left(y-\sqrt{\frac{2\cdot \frac{-y^2-1}{2-y^2}+1}{\frac{-y^2-1}{2-y^2}-1}}\right)$

so

$\left(y+\sqrt{\frac{2\cdot \frac{-y^2-1}{2-y^2}+1}{\frac{-y^2-1}{2-y^2}-1}}\right)\left(y-\sqrt{\frac{2\cdot \frac{-y^2-1}{2-y^2}+1}{\frac{-y^2-1}{2-y^2}-1}}\right)=0$

$\mathrm{Using\:the\:Zero\:Factor\:Principle:\quad \:If}\:ab=0\:\mathrm{then}\:a=0\:\mathrm{or}\:b=0\:\left(\mathrm{or\:both}\:a=0\:\mathrm{and}\:b=0\right)$

$\mathrm{Solve\:}\:y+\sqrt{\frac{2\cdot \frac{-y^2-1}{2-y^2}+1}{\frac{-y^2-1}{2-y^2}-1}}=0:\quad y\le \:0$

$\mathrm{Solve\:}\:y-\sqrt{\frac{2\cdot \frac{-y^2-1}{2-y^2}+1}{\frac{-y^2-1}{2-y^2}-1}}=0:\quad y\ge \:0$

$\mathrm{True\:for\:all}\:y$

Therefore, $\mathrm{The\:solution\:is}$ :

$x=\frac{-y^2-1}{2-y^2}\space\left\{y\ge \:0\right\}$

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3 years ago

In a certain triangle, the hypotenuse is 8 cm longer than the first leg and 1 cm longer than the second. Find the lengths of the

anzhelika [568]

Hyp: 8+x
Leg1: x
Leg2: x+7

Pythagorean triplet 5,12 &13 works

5 cm, 12 cm and 13 cm(Hypotenuse)<span />

7 0

3 years ago

Choose 1 answer

Likurg_2 [28]

It’s c got perfect from this

4 0

3 years ago