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3 years ago

How can you change the equation y = mx + b to shift the graph down?

1 answer:

8 0

Step-by-step explanation: To shift the graphs up and down on the coordinate system, you need to manipulate or change the y-intercepts.

The y-intercept is just the constant term in the equation.

As long as the slopes remain the same, the graphs can be

translated up or down by changing the constant value.

Take a look at these lines graphed below.

Notice that y = x + 11 is the same as y = x + 4 but

y = x + 11 is translated 7 units down to get y = x + 4.

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The position function of a particle in rectilinear motion is given by s(t) = 2t3 – 21t2 + 60t + 3 for t ≥ 0 with t measured in s

Norma-Jean [14]

The positions when the particle reverses direction are:

$s(t_1)=55ft\\\\s(t_2)=28ft$

The acceleraton of the paticle when reverses direction is:

$a(t_1)=-18\frac{ft}{s^{2}}\\ \\a(t_2)=a(5s)=18\frac{ft}{s^{2}}$

Why?

To solve the problem, we need to remember that if we derivate the position function, we will get the velocity function, and if we derivate the velocity function, we will get the acceleration function. So, we will need to derivate two times.

Also, when the particle reverses its direction, the velocity is equal to 0.

We are given the following function:

$s(t)=2t^{3}-21t^{2}+60t+3$

So,

- Derivating to get the velocity function, we have:

$v(t)=\frac{ds}{dt}=(2t^{3}-21t^{2}+60t+3)\\\\v(t)=3*2t^{2}-2*21t+60*1+0\\\\v(t)=6t^{2}-42t+60$

Now, making the function equal to 0, to find the times when the particle reversed its direction, we have:

$v(t)=6t^{2}-42t+60\\\\0=6t^{2}-42t+60\\\\0=t^{2}-7t+10\\(t-5)*(t-2)=0\\\\t_{1}=5s\\t_{2}=2s$

We know that the particle reversed its direction two times.

- Derivating the velocity function to find the acceleration function, we have:

$a(t)=\frac{dv}{dt}=6t^{2}-42t+60\\\\a(t)=12t-42$

Now, substituting the times to calculate the accelerations, we have:

$a(t_1)=a(2s)=12*2-42=-18\frac{ft}{s^{2}}\\ \\a(t_2)=a(5s)=12*5-42=18\frac{ft}{s^{2}}$

Now, substitutitng the times to calculate the positions, we have:

$s(t_1)=2*(2)^{3}-21*(2)^{2}+60*2+3=16-84+120+3=55ft\\\\s(t_2)=2*(5)^{3}-21*(5)^{2}+60*5+3=250-525+300+3=28ft$

Have a nice day!

3 0

3 years ago

Find sin A and sec B given a=2 b=11

Sonbull [250]

If there is a triangle with Sid a and b then sina b/✓a2+b2and seven=✓a2+b2/a

7 0

3 years ago

Read 2 more answers

Pls help asap

VMariaS [17]

Answer: Addition

Step-by-step explanation: You can represent that as 10^5*10^-5

If you add the exponent values, you get 10^0 which is equal to 1. If you multiply them, you get 10^-25 which is wrong.

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2 years ago

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3 years ago

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Dahasolnce [82]

Answer:

228

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2 years ago