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Illusion [34]
2 years ago
7

HELP PLS I WILL MARK BRAINLIEST, DUE TODAYY 100 POINTS!!!!!!!!!

Mathematics
1 answer:
Flauer [41]2 years ago
5 0

Answer:

Y1 intercept = 2

(0,2) (3,-7)

m= (-7 - 2) / (3-0) =-9/3 = - 3

Then y = - 3x + 2

Y2 intercept = - 8

(0,-8) (3,-7)

m = (-7 - - 8) / (3-0) = 1/3

Then y= (1/3)x-8

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Find the value of ax 4 ; a = 2, x = 1. <br><br> Select one: a. 2 b. 4 c. 1 d. 8
alexgriva [62]

For this case we have the following expression:

ax ^ 4

We must find the value of the expression when:

a = 2\\x = 1

Substituting the values in the expression we have:

2 (1) ^ 4 = 2 * 1 = 2

Thus, the value of the expression is 2.

Answer:

The value of the expression is 2

Option A

8 0
3 years ago
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Write an algebraic expression that has three terms. Two of the terms should be variable terms, and one of the terms should be th
andrey2020 [161]

Answer:

<em>Hello.</em>

An expression does not have an equal sign in it. Here are three examples:

2x + 7y + 6

x + 3y - 6

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6 0
3 years ago
Line parallel to y-1=4(x+3) and passes through (4,32)
Lady_Fox [76]

Answer:

y - 32 = 4(x - 4)

Step-by-step explanation:

The equation of a line in point- slope form is

y - b = m(x - a)

where m is the slope and (a, b) a point on the line

y - 1 = 4(x + 3) ← is in point- slope form

with slope m = 4

Parallel lines have equal slopes.

Using m = 4 and (a, b) = (4, 32), then

y - 32 = 4(x - 4) ← equation of parallel line

6 0
3 years ago
The kittens, Annie and Josie, are pushing a ball, Annie with a force magnitude of 80 N in a direction of 133 degrees, and Josie
kvv77 [185]

Answer: Then the magnitude of the force is 37.86N, and the direction is 54.35°

Step-by-step explanation:

We can write the forces as vectors.

We know that Annie pushes with a magnitude of 80N in a direction of 133° (Remember that the angles are always measured from the x-axis)

The components of this force, (Ax, Ay), are then:

Ax = 80N*cos(133°)

Ay = 80N*sin(133°)

And we know that Josie pushes with a magnitude of 95N in direction of 290°

The components of this force, (Jx, Jy), are:

Jx = 95N*cos(290°)

Jy = 95N*sin(290°)

When we add these forces, the total force acting on the ball is:

F = (80N*cos(133°) ,  80N*sin(133°)) + (95N*cos(290°), 95N*sin(290°))

F = (80N*cos(133°) + 95N*cos(290°), 80N*sin(133°) + 95N*sin(290°))

Now, the third kitten wants to do a force K, in a direction θ, such that the net force acting on the ball is zero.

Then we must have that, each component of the force of the third cat (K*cos(θ)  on the x-axis and k*sin(θ) on the y-axis), is such that:

K*cos(θ) + 80N*cos(133°) + 95N*cos(290°) = 0

k*sin(θ)  + 80N*sin(133°) + 95N*sin(290°) = 0

Now we need to solve that system for k and θ

if we simplify the equations we get:

k*cos(θ) - 22.07N = 0

k*sin(θ)  -30.76N  = 0

Now we can rewrite them as:

k*cos(θ) = 22.07N

k*sin(θ)   = 30.76N  

Now we can take the quotiet between both equations to get:

(k*sin(θ))/(k*cos(θ)) = 30.76N/22.07N

Tan(θ) = 1.394

θ = Atan(1.394) = 54.35°

Now that we know the angle, we can find the value of the magnitude k, by using one of the two equations of the system:

k*cos(54.35°) = 22.07N

k = 22.07N/cos(54.35°) = 37.86N

Then the magnitude of this force is 37.86N, and the direction is 54.35°

7 0
2 years ago
Is (2,3) a solution of the system
seropon [69]

Answer:

Yes

Step-by-step explanation:

3 0
3 years ago
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