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BARSIC [14]
3 years ago
8

If y = x - 6 were changed to y = x + 8, how would the graph of the new function compare with the first one?

Mathematics
2 answers:
11111nata11111 [884]3 years ago
8 0
The first one would have a y-intercept of (0,-6) and the second one would have a y-intercept of (0,8) <span>and x (the slope) would still be the same.
</span><span>y=x-6 would have points with (0,-6) (2,-4) (4,-2) (6,0)
y=x+8 would have points with (0,8) (-2,6) (-4,4) (-6,2)
</span><span>In other words shifted up. Hope this helps!</span>

wlad13 [49]3 years ago
8 0

Answer: it would be shifted up

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Determine whether the statement is true or false. If it is true, explain why. If it is false, explain why or give an example tha
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Answer:

False

Step-by-step explanation:

the <u>dot product</u> between two vectors A and B is:

A·B=ABcos\theta

where \theta is the angle between the vectors, if they are parallel, this angle is zero. so  \theta=0

and so the dot product is:

A·B =ABcos(0)

and since  cos (0)=1

the dot product is equal to

A·B=AB

The dot product of parallel vectors is NOT zero

5 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°

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2 years ago
the radius if a sphere r is given by the formula below where s is the surface area if the sphere/ aspherical ballon has a maxium
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Solution :

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The p-value for the variable "Number of bidders" is 0.1940

Since $\text{p-value}$  is not less than $0.05$, so it is significant to the model.

c). We cannot say that model is significant because variable " the number of bidders" is not significant.

But as both variables have positive coefficient so as the variable increases the price received for the item also increased.

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frozen [14]
A hole 8 feet deep because it’s down making it negative
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2 years ago
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