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3 years ago

Write the equation for the line that passes through the coordinates (2,7) and (0, 1) in slope-intercept form

Mathematics

1 answer:

Bumek [7]3 years ago

6 0

Answer:

y = 3x + 1

Step-by-step explanation:

y2 - y1 / x2 - x1

1 - 7 / 0 -2

-6 / -2

= 3

y = 3x + b

1 = 3(0) + b

1 = 0 + b

1 = b

y = 3x + 1

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The cost, in dollars, of a one-day car rental, is given by C (x) = 31 + 0.18x, where x is the number of miles driven. In this fu

julia-pushkina [17]

Answer:

Step-by-step explanation:

the function is c(x) = 31 + 0.18*x

where $31 is a fixed cost, and $0.18 is the cost per mile drive (where the number of miles driven)

So the fixed cost, $31, is the cost per day of rent (this price does not depend on the number x), and the linear cost, $0.18, is the cost per mile driven (because this number is multiplied by x in the function), then the right answer is B: "$31 is the cost per day to rent the car and $0.18 is the cost per mile."

5 0

3 years ago

Pls can u help me this is due now T-T

Artist 52 [7]

Answer:

SOLUTION: X=5

Step-by-step explanation:

x=5

y=27

vertical input= (0, 1/9)

Hope that helps!

5 0

3 years ago

Y=-2x+7 and draw a graph

Airida [17]

Y=-2+7 here’s the graph

8 0

4 years ago

Hey could you please help me and my new friend found out this answer?

Ludmilka [50]

Answer:

i will be you friend and help you but where the question

Step-by-step explanation:

8 0

3 years ago

Read 2 more answers

In a lab experiment, a student is trying to apply the conservation of momentum. Two identical balls, each with a mass of 1.0 kg.

muminat

Answer:

Trial- 2 shows the conservation of momentum in a closed system.

Step-by-step explanation:

Given: Mass of balls are $m= 1.0\ kg$

Conservation of momentum in a closed system occurs when momentum before collision is equal to momentum after collision.

Let initial velocity of ball $A\ is\ u_1$
Initial velocity of ball $B\ is\ u_2$
Final velocity of ball $A\ is\ v_1$
Final velocity of ball $B\ is\ v_2$
Momentum before collision $= mu_1+mu_2$
Momentum after collision $=mv_1+mv_2$

Now, According to conservation of momentum.

Momentum before collision = Momentum after collision

$mu_1+mu_2=mv_1+mv_2$

We will plug each trial to this equation.

Trial 1

$mu_1+mu_2=mv_1+mv_2\\1.0(1)+1.0(-2)=1.0(-2)+1.0(-1)\\1-2=-2-1\\-1=-3$

Trial 2

$mu_1+mu_2=mv_1+mv_2\\1.0(.5)+1.0(-1.5)=1.0(-.5)+1.0(-\.5)\\.5-1.5=-.5-.5\\-1=-1$

Trial 3

$mu_1+mu_2=mv_1+mv_2\\1.0(2)+1.0(1)=1.0(1)+1.0(-2)\\2+1=1-2\\3=-1$

Trial 4

$mu_1+mu_2=mv_1+mv_2\\1.0(.5)+1.0(-1)=1.0(1.5)+1.0(-1.5)\\.5-1=1.5-1.5\\-.5=0$

We can see only Trial 2 satisfies the princple of conservation of momentum. That is momentum before collison should equal to momentum after collision.

5 0

4 years ago