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melomori [17]
2 years ago
15

stumpy the skeleton goes for a bike ride every night in the cemetery. He rides 1/2 way around the cemetery and stops. Stumpy the

n bikes 3/4 of the way around before stopping again. he finishes his ride by biking another 3/8. how far around the cemetery does stumpy bike?
Mathematics
1 answer:
marin [14]2 years ago
4 0
He bikes 13/8 of the cemetery
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Explain how to resolve this problem. there are 13 books.some books cost $9 each and the rest cost $8 each. A total of $108 spent
Katen [24]
X=one type that cost 9
y=one type that cost 8

total books=type1+type2
total books=13
x+y=13

the total cost=cost of each added together
cost of each=number of books times cost per book
total cost=108
9x+8y=108


we have
x+y=13
9x+8y=108
solve for x and y

multiply first equation by -8 and add to first equation
-8x-8y=-104
<u>9x+8y=108 +</u>
x+0y=4

x=4
subsitute
x+y=13
4+y=13
minus 4
y=9

4 of the $9
9 of the $8

thats how to solve
6 0
2 years ago
Read 2 more answers
Which answer choice shows 12.444 rounded to the nearest half?
vesna_86 [32]

Answer:

C. 12

Step-by-step explanation:

Remember 0 to 4 turn it back

5 or above give it a shove.

Since 12.444 is around 0 to 4.

We can round that to 12.

3 0
2 years ago
Read 2 more answers
The following formula for the sum of the cubes of the first n integers is proved in Appendix E. Use it to evaluate the limit in
Marina86 [1]

Answer:

\lim_{n\to\infty} (1+ \frac{2}{n} +\frac{1}{n^2})

And when we apply the limit we got that:

\lim_{n\to\infty} (1+ \frac{2}{n} +\frac{1}{n^2}) =1

Step-by-step explanation:

Assuming this complete problem: "The following formula for the sum of the cubes of the first n integers is proved in Appendix E. Use it to evaluate the limit . 1^3+2^3+3^3+...+n^3=[n(n+1)/2]^2"

We have the following formula in order to find the sum of cubes:

\lim_{n\to\infty} \sum_{n=1}^{\infty} i^3

We can express this formula like this:

\lim_{n\to\infty} \sum_{n=1}^{\infty}i^3 =\lim_{n\to\infty} [\frac{n(n+1)}{2}]^2

And using this property we need to proof that: 1^3+2^3+3^3+...+n^3=[n(n+1)/2]^2

\lim_{n\to\infty} [\frac{n(n+1)}{2}]^2

If we operate and we take out the 1/4 as a factor we got this:

\lim_{n\to\infty} \frac{n^2(n+1)^2}{n^4}

We can cancel n^2 and we got

\lim_{n\to\infty} \frac{(n+1)^2}{n^2}

We can reorder the terms like this:

\lim_{n\to\infty} (\frac{n+1}{n})^2

We can do some algebra and we got:

\lim_{n\to\infty} (1+\frac{1}{n})^2

We can solve the square and we got:

\lim_{n\to\infty} (1+ \frac{2}{n} +\frac{1}{n^2})

And when we apply the limit we got that:

\lim_{n\to\infty} (1+ \frac{2}{n} +\frac{1}{n^2}) =1

3 0
2 years ago
For each object , choose an appropriate scale for drawing that fits on a regular sheet of paper . Not all scales on the list wil
olganol [36]

Answer:

Step-by-step explanation:

First you will have to put the formula of the rectangular rectangle L x W

Making you the answer you will do this

120 x 53 and then you divide the answer with 360 and then multiply 6 x2 and then add it and you get the answer

5 0
2 years ago
I'm having trouble on problem 36
denpristay [2]
Jason has x dollars, and Ryan has 5 more dollars than Jason. How many dollars does Ryan have? How many dollars do both boys have?

Jason has x
Ryan has x + 5
both boys have x + x + 5 = 2x + 5


5 0
2 years ago
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