Stephen deposited $1800 into a savings account for which interest is compounded weekly at a rate of 3.2%. How much interest will
1 answer:
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Step-by-step explanation:
This seems to be calculus 1.
<u>Question a</u>
We have
m = slope = derivative
Find the derivative / slope of
We do this by differentiating the polynomials. There are a few methods to do this but I am going to use the power rule, which we multiply the constant by the exponent on the variable and subtract one from the exponent.
when x = a
<em>Now that we have this information, we can answer question b</em>
<u>Question b</u>
<u>The tangent line for Point (1, 12)</u>
First find the slope by using our derivative.
Now that we have our slope, use point slope form to find our tangent line
<u>Now lets do the same for the Point (2, 13)</u>
Find the slope at the point.
Now find the tangent line using point slope form of a line.
Now graph the lines, which I have done and you can see by viewing the image I have attached.
Answer:
15 seconds
Step-by-step explanation:
If you make a table of values for the dog and the squirrel using d = rt, then the rates are easy: the dog's rate is 150 and the squirrel's is 100. The t is what we are looking for, so that's our unknown, and the distance is a bit tricky, but let's look at what we know: the dog is 200 feet behind the squirrel, so when the dog catches up to the squirrel, he has run some distance d plus the 200 feet to catch up. Since we don't know what d is, we will just call it d! Now it seems as though we have 2 unknowns which is a problem. However, if we solve both equations (the one for the dog and the one for the squirrel) for t, we can set them equal to each other. Here's the dog's equation:
d = rt
d+200 = 150t
And the squirrel's:
d = 100t
If we solve both for t and set them equal to each other we have:
Now we can cross multiply to solve for d:
150d = 100d + 20,000 and
50d = 20,000
d = 400
But we're not looking for the distance the squirrel traveled before the dog caught it, we are looking for how long it took. So sub that d value back into one of the equations we have solved for t and do the math:
That's 1/4 of a minute which is 15 seconds.