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2 years ago

10

What is the equation for the line in slope-intercept form?

2 answers:

masha68 [24]2 years ago

5 0

Answer:

y=mx+b is slope-intercept form

where m is the slope and b is the y intercept.

Since the line crosses the y axis at 0,0 the intercept is +0 or just nothing.

now all we need to do is find the slope

to do that just go from the y intercept (the first point) y units up and x units over untill u cross at the next point. for examples from (0, 0) to (1, 8)-the next point- i need to go up 8 units up and 1 unit over. this is described as rise over run and that is your slope 8/1 rise/run. rise is how many units i go up (or down) from the y intercept until the next point that lies on the line and run is how far i need to go over from how many units i just went up. If u continue to go 8 up and 1 over from each point u will see that u get a point lying of the line. This is why 8/1 is your slope

8/1 is the slope and 0,0 is your y intercept so we put nothing

the equation is y=8x

Step-by-step explanation:

stiv31 [10]2 years ago

4 0

Answer:

y=8x

Step-by-step explanation:

Pick two points (0,0) (2,16)

Then 16-0/2-0= 8

8 is the slope

The y intercept is 0 so

y=8x

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3 years ago

50 points! I understand A. and B. but I would really appreciate help with C.

FromTheMoon [43]

Answer:

$51.72\text{ cells per hour}$

Step-by-step explanation:

So, the function, P(t), represents the number of cells after t hours.

This means that the derivative, P'(t), represents the instantaneous rate of change (in cells per hour) at a certain point t.

C)

So, we are given that the quadratic curve of the trend is the function:

$P(t)=6.10t^2-9.28t+16.43$

To find the <em>instanteous</em> rate of growth at t=5 hours, we must first differentiate the function. So, differentiate with respect to t:

$\frac{d}{dt}[P(t)]=\frac{d}{dt}[6.10t^2-9.28t+16.43]$

Expand:

$P'(t)=\frac{d}{dt}[6.10t^2]+\frac{d}{dt}[-9.28t]+\frac{d}{dt}[16.43]$

Move the constant to the front using the constant multiple rule. The derivative of a constant is 0. So:

$P'(t)=6.10\frac{d}{dt}[t^2]-9.28\frac{d}{dt}[t]$

Differentiate. Use the power rule:

$P'(t)=6.10(2t)-9.28(1)$

Simplify:

$P'(t)=12.20t-9.28$

So, to find the instantaneous rate of growth at t=5, substitute 5 into our differentiated function:

$P'(5)=12.20(5)-9.28$

Multiply:

$P'(5)=61-9.28$

Subtract:

$P'(5)=51.72$

This tells us that at <em>exactly</em> t=5, the rate of growth is 51.72 cells per hour.

And we're done!

7 0

2 years ago