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Volgvan
3 years ago
8

Solve 960 divided by 4 using the area model Show your work

Mathematics
1 answer:
Alborosie3 years ago
6 0
The answer is 960 ÷ 4 = 240. I hope this helps
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Please help me I will give you the brain thing and extra points. (image below) 20/30
Dmitriy789 [7]

Answer: the correct answer is B

explanation:

The y - intercept shows starts at 4 and you can see that the y axis is the number of comic books you have. So you start at 4 comic books.

6 0
3 years ago
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the local DQ sells 4 sundaes for every 7 blizzards if they sell 539 blizzards in a month how many sundaes were sold? plsssss hel
kap26 [50]
I believe the answer is 77.

Explanation

539 divided by 7 is 77.

Hopefully this helps.
7 0
3 years ago
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Circle O has a circumference of 887 cm.<br> What is the length of the radius of the circle?<br> cm
Salsk061 [2.6K]

Answer:

44cm

Step-by-step explanation:

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where r is the radius

r = C / 2π

= \frac{88\pi }{2\pi }

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7 0
4 years ago
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Give an example of a function with both a removable and a non-removable discontinuity.
navik [9.2K]

Answer:

(x+5)(x-3) / (x+5)(x+1)

Step-by-step explanation:

A removeable discontinuity is always found in the denominator of a rational function and is one that can be reduced away with an identical term in the numerator.  It is still, however, a problem because it causes the denominator to equal 0 if filled in with the necessary value of x.  In my function above, the terms (x + 5) in the numerator and denominator can cancel each other out, leaving a hole in your graph at -5 since x doesn't exist at -5, but the x + 1 doesn't have anything to cancel out with, so this will present as a vertical asymptote in your graph at x = -1, a nonremoveable discontinuity.

8 0
3 years ago
How do you use the limit comparison test on this particular series?<br>Calculus series tests​
barxatty [35]

Compare \dfrac1{\sqrt{n^2+1}} to \dfrac1{\sqrt{n^2}}=\dfrac1n. Then in applying the LCT, we have

\displaystyle\lim_{n\to\infty}\left|\frac{\frac1{\sqrt{n^2+1}}}{\frac1n}\right|=\lim_{n\to\infty}\frac n{\sqrt{n^2+1}}=1

Because this limit is finite, both

\displaystyle\sum_{n=1}^\infty\frac1{\sqrt{n^2+1}}

and

\displaystyle\sum_{n=1}^\infty\frac1n

behave the same way. The second series diverges, so

\displaystyle\sum_{n=0}^\infty\frac1{\sqrt{n^2+1}}=1+\sum_{n=1}^\infty\frac1n

is divergent.

4 0
4 years ago
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