What is the midpoint of a line segment with endpoints at (6, -4) and (15,8)?
1 answer:
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A) Let x stand for time, y stand for velocity.
We are given the points (2,50), (6, 54). We can make a line using the slope intercept form
y = mx + b.
slope is (54 - 50)/(6-2) = 4/4 = 1
y = 1x + b
plug in point (2,50) to find b
50 = 1(2) + b
50-2 = b
48 = b
the equation is y = 1x + 48
Make standard form.
<span>x - y = -48
</span><span>b) make a table and plot points for the first 7 hours
Table Is attached.
Good Luck!
If you like the answer please give me brainliest answer!
</span>
We are given the points (2,50), (6, 54). We can make a line using the slope intercept form
y = mx + b.
slope is (54 - 50)/(6-2) = 4/4 = 1
y = 1x + b
plug in point (2,50) to find b
50 = 1(2) + b
50-2 = b
48 = b
the equation is y = 1x + 48
Make standard form.
<span>x - y = -48
</span><span>b) make a table and plot points for the first 7 hours
Table Is attached.
Good Luck!
If you like the answer please give me brainliest answer!
</span>
<h3>Answer: 
</h3>
The -3 is not in the exponent
Explanation:
The parent function is . Plugging in x = 0 leads to y = 1. So the point (0,1) is on the f(x) curve. Going from (0,1) to (0,-2) is a vertical shift of 3 units downward. To represent this shift, we tack on a "-3" at the end of the f(x) function.
You could look at other points as well, but I find working with x = 0 is easiest.
As a check, plugging x = 0 into g(x) leads to...
This confirms our answer.