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3 years ago

HELP!!!!

Mathematics

1 answer:

Ahat [919]3 years ago

7 0

The answer is C. y=3x

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Drag the tiles to the boxes to form correct pairs. Not all tiles will be used.

xz_007 [3.2K]

Answer:

Part 1) $f(x)=\frac{2x-1}{x+2}$ -------> $f^{-1}(x)=\frac{-2x-1}{x-2}$

Part 2) $f(x)=\frac{x-1}{2x+1}$ -------> $f^{-1}(x)=\frac{-x-1}{2x-1}$

Part 3) $f(x)=\frac{2x+1}{2x-1}$ -----> $f^{-1}(x)=\frac{x+1}{2(x-1)}$

Part 4) $f(x)=\frac{x+2}{-2x+1}$ ----> $f^{-1}(x)=\frac{x-2}{2x+1}$

Part 5) $f(x)=\frac{x+2}{x-1}$ -------> $f^{-1}(x)=\frac{x+2}{x-1}$

Step-by-step explanation:

Part 1) we have

$f(x)=\frac{2x-1}{x+2}$

Find the inverse

Let

y=f(x)

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

Exchange the variables x for y and t for x

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

Isolate the variable y

$x=\frac{2y-1}{y+2}\\ \\ xy+2x=2y-1\\ \\xy-2y=-2x-1\\ \\y[x-2]=-2x-1\\ \\y=\frac{-2x-1}{x-2}$

Let

$f^{-1}(x)=y$

$f^{-1}(x)=\frac{-2x-1}{x-2}$

Part 2) we have

$f(x)=\frac{x-1}{2x+1}$

Find the inverse

Let

y=f(x)

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

Exchange the variables x for y and t for x

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

Isolate the variable y

$x=\frac{y-1}{2y+1}\\ \\2xy+x=y-1\\ \\2xy-y=-x-1\\ \\y[2x-1]=-x-1\\ \\y=\frac{-x-1}{2x-1}$

Let

$f^{-1}(x)=y$

$f^{-1}(x)=\frac{-x-1}{2x-1}$

Part 3) we have

$f(x)=\frac{2x+1}{2x-1}$

Find the inverse

Let

y=f(x)

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

Exchange the variables x for y and t for x

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

Isolate the variable y

$x=\frac{2y+1}{2y-1}\\ \\2xy-x=2y+1\\ \\2xy-2y=x+1\\ \\y[2x-2]=x+1\\ \\y=\frac{x+1}{2(x-1)}$

Let

$f^{-1}(x)=y$

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

Part 4) we have

$f(x)=\frac{x+2}{-2x+1}$

Find the inverse

Let

y=f(x)

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

Exchange the variables x for y and t for x

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

Isolate the variable y

$x=\frac{y+2}{-2y+1}\\ \\-2xy+x=y+2\\ \\-2xy-y=-x+2\\ \\y[-2x-1]=-x+2\\ \\y=\frac{-x+2}{-2x-1} \\ \\y=\frac{x-2}{2x+1}$

Let

$f^{-1}(x)=y$

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

Part 5) we have

$f(x)=\frac{x+2}{x-1}$

Find the inverse

Let

y=f(x)

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

Exchange the variables x for y and t for x

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

Isolate the variable y

$x=\frac{y+2}{y-1}\\ \\xy-x=y+2\\ \\xy-y=x+2\\ \\y[x-1]=x+2\\ \\y=\frac{x+2}{x-1}$

Let

$f^{-1}(x)=y$

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

7 0

3 years ago

Find the value of x, y, and z in the image below.

Yakvenalex [24]

Answer:

Step-by-step explanation:

y=18°

x=64°

x+y+z=180

64+18+z=180

82+z=180

z=180-82=98°

7 0

3 years ago

Read 2 more answers

Use implicit differentiation to find an equation of the tangent line to the curve at the given point. 7x2 + xy + 7y2 = 15, (1, 1

riadik2000 [5.3K]

Assume $y=y(x)$ .

$7x^2+xy+7y^2=15$
$\dfrac{\mathrm d}{\mathrm dx}[7x^2+xy+7y^2]=\dfrac{\mathrm d}{\mathrm dx}[15]$
$14x+y+x\dfrac{\mathrm dy}{\mathrm dx}+14y\dfrac{\mathrm dy}{\mathrm dx}=0$
$\dfrac{\mathrm dy}{\mathrm dx}(x+14y)=-(14x+y)$
$\dfrac{\mathrm dy}{\mathrm dx}=-\dfrac{14x+y}{x+14y}$

The slope of the tangent line to the curve at $(a,b)$ is then $-\dfrac{14a+b}{a+14b}$ , so the tangent to $(1,1)$ has slope

$\dfrac{\mathrm dy}{\mathrm dx}\bigg|_{x=1,y=1}=-\dfrac{15}{15}=-1$

The point-slope form of the tangent line is then

$y-1=-(x-1)\iff y=-x+2$

7 0

3 years ago

Simplify <br> 3x+4.4-2(6.6+x)

Taya2010 [7]

Im not 100% sure but I think it’s X-8.8

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2 years ago

Give the quotient and remainder <br><br>196÷5

max2010maxim [7]

Answer:

Q: 39 R: 1 Hope this helps you out!

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3 years ago