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IceJOKER [234]
3 years ago
6

NEED SOME ASSISTANCE ASAP

Mathematics
1 answer:
Tasya [4]3 years ago
5 0

Answer:

hi ube8ejeissb susbsmsns sjebsizg sushi underbrush jsueushd

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Item 7 you and a friend are running on treadmills. you run 0.5 mile every 3 minutes, and your friend runs 2 miles every 14 minut
sweet [91]
<span>You run 0.5 miles every 3 minutes => 1 mile every 6 minutes. Your friend runs 2 miles every 14 minutes => 1 mile every 7 minutes. You run a whole number of miles every number of minutes that is a multiple of 6. Your freind runs a whole number of miles every number of minutes that is a multiple of 7 Then the least possible number of miles that you both run to end at the same time is the least common factor of 7 and 6 minutes. This is 7 * 6 = 42 minutes. You will have run 42 min / (6 miles/min) = 7 miles, and your friend will have run 42 min / (7 miles/min) = 6 miles</span>
5 0
3 years ago
Read 2 more answers
How do you do 69% as a fraction? Please show work.​
MakcuM [25]
69% is the same as .69, which is the same as 69/100
7 0
3 years ago
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What's -8.2 x 10 to the fifth power? Please explain your answer. Thank you!
Stolb23 [73]

First lets find -8.2 x 10 = -82 know we find 82^5 which is 82*82*82*82*82

-82*-82=6724 and then 6724*-82= -551368 and then -551368*-82 =45212176 and then 45212176*-82= -3707398432 and finally -3707398432 is our answer i checked the calc to after i did the hard math that took me forever i had some help from the calc and i got this hopefully this helps

8 0
3 years ago
How do I solve this question?
Yuri [45]
It's 18/100. The reduced / simplest form is 9/50 because you divide it by 2 since it's the GCS.
4 0
3 years ago
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Find the limit, if it exists, or type dne if it does not exist.
Phantasy [73]
\displaystyle\lim_{(x,y)\to(0,0)}\frac{\left(x+23y)^2}{x^2+529y^2}

Suppose we choose a path along the x-axis, so that y=0:

\displaystyle\lim_{x\to0}\frac{x^2}{x^2}=\lim_{x\to0}1=1

On the other hand, let's consider an arbitrary line through the origin, y=kx:

\displaystyle\lim_{x\to0}\frac{(x+23kx)^2}{x^2+529(kx)^2}=\lim_{x\to0}\frac{(23k+1)^2x^2}{(529k^2+1)x^2}=\lim_{x\to0}\frac{(23k+1)^2}{529k^2+1}=\dfrac{(23k+1)^2}{529k^2+1}

The value of the limit then depends on k, which means the limit is not the same across all possible paths toward the origin, and so the limit does not exist.
8 0
4 years ago
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