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3 years ago

9

What is the slope of this equation, v=3x+60

2 answers:

aliina [53]3 years ago

5 0

Answer:

slope=3

Step-by-step explanation:

Tju [1.3M]3 years ago

3 0

The slope is 3.

In y=mx+b, the number before x (or m) is the slope.

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Width of an MP3 player

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1.73 inches wide
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3 years ago

Jeremy flew 8 times last week for work. If each flight was 2 hours long, how much time did Jeremy spend in the air?

tatiyna

Answer:

16

Step-by-step explanation:

8 0

2 years ago

Help!! 50 points and brainliest!

Viktor [21]

Answer:

Second choice:

$x=2t$

$y=4t^2+4t-3$

Fifth choice:

$x=t+1$

$y=t^2+4t$

Step-by-step explanation:

Let's look at choice 1.

$x=t+1$

$y=t^2+2t$

I'm going to subtract 1 on both sides for the first equation giving me $x-1=t$ . I will replace the $t$ in the second equation with this substitution from equation 1.

$y=(x-1)^2+2(x-1)$

Expand using the distributive property and the identity $(u+v)^2=u^2+2uv+v^2$ :

$y=(x^2-2x+1)+(2x-2)$

$y=x^2+(-2x+2x)+(1-2)$

$y=x^2+0+-1$

$y=x^2$

So this not the desired result.

Let's look at choice 2.

$x=2t$

$y=4t^2+4t-3$

Solve the first equation for $t$ by dividing both sides by 2:

$t=\frac{x}{2}$ .

Let's plug this into equation 2:

$y=4(\frac{x}{2})^2+4(\frac{x}{2})-3$

$y=4(\frac{x^2}{4})+2x-3$

$y=x^2+2x-3$

This is the desired result.

Choice 3:

$x=t-3$

$y=t^2+2t$

Solve the first equation for $t$ by adding 3 on both sides:

$x+3=t$ .

Plug into second equation:

$y=(x+3)^2+2(x+3)$

Expanding using the distributive property and the earlier identity mentioned to expand the binomial square:

$y=(x^2+6x+9)+(2x+6)$

$y=(x^2)+(6x+2x)+(9+6)$

$y=x^2+8x+15$

Not the desired result.

Choice 4:

$x=t^2$

$y=2t-3$

I'm going to solve the bottom equation for $t$ since I don't want to deal with square roots.

Add 3 on both sides:

$y+3=2t$

Divide both sides by 2:

$\frac{y+3}{2}=t$

Plug into equation 1:

$x=(\frac{y+3}{2})^2$

This is not the desired result because the $y$ variable will be squared now instead of the $x$ variable.

Choice 5:

$x=t+1$

$y=t^2+4t$

Solve the first equation for $t$ by subtracting 1 on both sides:

$x-1=t$ .

Plug into equation 2:

$y=(x-1)^2+4(x-1)$

Distribute and use the binomial square identity used earlier:

$y=(x^2-2x+1)+(4x-4)$

$y=(x^2)+(-2x+4x)+(1-4)$

$y=x^2+2x+-3$

$y=x^2+2x-3$ .

This is the desired result.

3 0

3 years ago

Read 2 more answers

Subtract<br> 505,000 - 427,619 -

yawa3891 [41]

Answer:

77381 is your answer

Step-by-step explanation:

5 0

2 years ago

Is x=3 a solution for the equation: 3x + 7 = 13<br> Why or why not ?

expeople1 [14]

Answer:

No.

Step-by-step explanation:

There are two different ways to do this - solve for x in the given equation or substitute the value for x and see if you get a true statement.

3x + 7 = 13 subtract 7 from both sides

3x = 6 divide by 3 on both sides

x = 2

3(3) + 7 = 13

9 + 7 = 13

16 ≠ 13

Therefore, x = 3 is not a solution for 3x + 7 = 13.

5 0

2 years ago