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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)?

Mathematics
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
Read 2 more answers
Can someone plz help Me?
hammer [34]

Answer:

A

^{}

3 0
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
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