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LuckyWell [14K]

3 years ago

14

What is the equation of the line that is parallel to y=3x+2 and goes through the points (5,8)?

1 answer:

ira [324]3 years ago

6 0

Parallel lines has equal slopes.

Line y = 3x + 2 has a slope of 3.

Required equation is y - 8 = 3(x - 5)
y = 3x - 15 + 8
y = 3x - 7

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valerie has 10 Video games. 7 of her video games are racing games. what is valerie's racing games written as a decimal?

svet-max [94.6K]

Answer:

7/10.

Step-by-step explanation:

If theres a total of 10 games and 7 out of the 10 are racing games, it means that 7/10 of the games are racing games.

6 0

3 years ago

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Can someone plz help Me?

hammer [34]

Answer:

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2 years ago

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Find f(x) and g(x) so the function can be expressed as y = f(g(x)). (1 point)

Rina8888 [55]

Answer:

$g(x)=x^2+4$ and $f(x)=\frac{8}{x}$

Step-by-step explanation:

If we want to express $y$ as $y=f(g(x))$ , this means that put the function $g(x)$ into $x$ of $f(x)$ to get $y$ .

The function $y$ is given as $y=\frac{8}{x^{2}+4}$

<u>So we need to figure out a function $g(x)$ that we can put into another $f(x)$ to get $y$ </u>

If we let $g(x)=x^2+4$ and $f(x)=\frac{8}{x}$ , we can clearly see that putting g(x) into x of f(x) will give us $\frac{8}{x^2+4}$ , which is y.

Hence $g(x)=x^2+4$ and $f(x)=\frac{8}{x}$

3 0

3 years ago

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The surface area of this rectangular prism is 34 square centimeters. What is the volume is the width is 1 cm, the length is 2 cm

Bezzdna [24]

Answer: $10\ cm^3$

Step-by-step explanation:

Given

The surface area of a rectangular prism is $34\ cm^2$

Width of prism $w=1\ cm$

Length of prism $l=2\ cm$

height of prism $h=x\ cm$

Surface area is given by

$S.A.=2[lw+wh+hl]$

Put values

$34=2[2\times 1+1\times x+2x]\\17=2+3x\\15=3x\\x=5\ cm$

Now, the volume of Prism is given by

$V=l\times w\times h$

$\Rightarrow V=2\times1\times 5=10\ cm^3$

8 0

2 years ago

(b) dy/dx = (x - y+ 1)^2

Elanso [62]

Substitute $v(x)=x-y(x)+1$ , so that

$\dfrac{\mathrm dv}{\mathrm dx}=1-\dfrac{\mathrm dy}{\mathrm dx}$

Then the resulting ODE in $v(x)$ is separable, with

$1-\dfrac{\mathrm dv}{\mathrm dx}=v^2\implies\dfrac{\mathrm dv}{1-v^2}=\mathrm dx$

On the left, we can split into partial fractions:

$\dfrac12\left(\dfrac1{1-v}+\dfrac1{1+v}\right)\mathrm dv=\mathrm dx$

Integrating both sides gives

$\dfrac{\ln|1-v|+\ln|1+v|}2=x+C$

$\dfrac12\ln|1-v^2|=x+C$

$1-v^2=e^{2x+C}$

$v=\pm\sqrt{1-Ce^{2x}}$

Now solve for $y(x)$ :

$x-y+1=\pm\sqrt{1-Ce^{2x}}$

$\boxed{y=x+1\pm\sqrt{1-Ce^{2x}}}$

3 0

3 years ago