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blagie [28]
1 year ago
9

Which of the following statements best describes the lines?

Mathematics
1 answer:
kiruha [24]1 year ago
7 0
Line p has a greater slope. the higher the line the more yhe slope
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A video store charges $4 to rent the first movie and x for each additional movie. What expression would represent the cost for r
V125BC [204]
Since it costs $4 to rent the first movie, and $X for each additional movie, the correct equation is y=4x+4.
The reason it's 4x instead of 5x is because the first movie costs $4, and the remaining 4 movies cost $X each.
4 0
3 years ago
Simplify the answer<br><br> 1 - 5/6?
ludmilkaskok [199]
The answer is:  " \frac{1}{6}"  .
______________________________________________
<u>Note</u>:
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          " 1 - (5/6)  = (6/6) - (5/6) = (6-5)/6 = 1/6 " . 
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5 0
3 years ago
Select the correct answer from each drop-down menu.
olchik [2.2K]

For the linear function in the graph, we have that:

  • The slope is of 3.
  • The y-intercept is of -4.
  • The equation of the line is: y = 3x - 4.

<h3>What is a linear function?</h3>

A linear function is modeled by:

y = mx + b

In which:

  • m is the slope, which is the rate of change, that is, by how much y changes when x changes by 1.
  • b is the y-intercept, which is the value of y when x = 0, and can also be interpreted as the initial value of the function.

The line goes through point (0,-4), hence:

The y-intercept is of -4.

The function also goes through point (2,2), hence the slope is:

m = (2 - (-4))/(2 - 0) = 6/2 = 3.

Then, the equation of the line is:

y = 3x - 4.

More can be learned about linear functions at brainly.com/question/24808124

#SPJ1

8 0
2 years ago
What's the answer please let me know asap
velikii [3]
2d is flat 3D isn't
7 0
2 years ago
A 1300 kg car moving at 5.2 m/s is initially traveling north in the positive y direction. After completing a 90° right-hand turn
stich3 [128]

Answer:

a. 6760(i - j) kgm/s b. -6760i kgm/s c. 2172.74 N d. 19314.29 N e. 45°

Step-by-step explanation:

Given that mass of car, m = 1300 kg

initial velocity in the y-direction, v₁ = 5.2j m/s

time taken for 90° turn to positive x-direction, t₁ = 4.4 s

time taken for collision in positive x-direction, t₂ = 350 ms

a. Impulse, J₁  on car during turn.

impulse, J =mv₂ - mv₁

v₁= initial velocity and v₂ = final velocity

v₁ = 0i + 5.2j m/s since it was initially moving in the positive y-direction.

v₂ = 5.2i + 0j m/s since it was initially moving in the positive x-direction.

So, J₁ = m(5.2i + 0j -(0i + 5.2j))

          = m(5.2i - 5.2j)

         = 1300× 5.2(i-j)

         = 6760(i-j) kgm/s

b. Impulse after collision J₂

v₁= initial velocity=5.2i + 0j m/s since it was initially moving in the positive x-direction.

v₂= 0m/s since the car stops after collision

So, J₂= mv₂ - mv₁

         = m(0 - (5.2i + 0j)) m/s

         =-5.2mi kgm/s

         = -1300 × 5.2i kgm/s

         =-6760i kgm/s

c. Average force, F₁ during turn

Impulse J = Ft

From (a) the impulse J₁ = 6760(i-j) kgm/s. The time taken for the turn t₁ = 4.4 s. So, F₁ = J₁/t₁ = 6760(i-j)/4.4 = 1536.36(i-j) N.

Magnitude of F₁ = F₁ = average force during turn=1536.36√2= 2172.74 N

d. Average force, F₂ after collision

From J=Ft, F=J/t

From (b) above, our impulse during collision is J₂ = -6760i kgm/s. The time taken for the impulse or collision to occur is t₂ = 350 ms.

So, F₂ = J₂/t₂ = -6760i /(350 × 10⁻³) N= - 19314.29i N.

So magnitude of F₂= F₂=average force during collision = 19314.29 N

e. Angle between average force in (c) and the positive x- direction.

We know that F₁= 1536.36(i-j) N = 1536.36i - 1536.36j N. The unit vector in the positive x-direction is i.

For the angle between two vectors, we have that cosθ= a.b/ab where a,b are vectors and a,b their magnitudes respectively.

So, cosθ = (1536.36i - 1536.36j).i/(1536.36√2) =1536.36/1536.36√2=1/√2

cosθ = 1/√2

θ=cos⁻¹(1/√2)

θ=45°

4 0
3 years ago
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