8 times 8 so 64 is your answer :) hope this helps
Answer:
(a) $3800
(b) (a) -- April to May
Step-by-step explanation:
<h3>(a)</h3>
The least amount is found at the lowest point on the graph. That point is in May. It is on the line between 3700 and 3900, so the amount is $3800.
The least donation amount is a month is $3800.
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<h3>(b)</h3>
The greatest month-to-month decrease is found where the line on the graph has the steepest negative slope. There are two segments with negative slope:
- April - May (decrease of $500)
- June - July (decrease of $100)
The decrease from April to May is by far the largest of these two decreases.
The greatest decrease occurred April to May.
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.
Answer:
50 dollars
Step-by-step explanation:
40*25%=10
40+10= 50 dollars
Add it to find the original sales price which would end up being 50 dollars! Hope this helps, have a good day. c;