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ella [17]
2 years ago
15

Plzzzz help me

Mathematics
1 answer:
AVprozaik [17]2 years ago
4 0

y=kx

5=10k

k=10-5

k=5 is the correct answer

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Help I tried to do the page but I couldn't
ahrayia [7]
Sarah wanted to share her apple with 3 of her friends , but she only had 1/2 of an apple left. What math problem should she write to figure out how to split her apple. Explain your answer.
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2 years ago
What operation is implied between the whole number and fraction parts of the mixed number 4 2/3
agasfer [191]

Answer:

0

Step-by-step explanation:

7 0
3 years ago
The 22 members of a baseball team are trying to raise at least $1968.00 to cover the traveling cost for a holiday tournament. If
Rus_ich [418]

About $70.31.

You would take $1968.00 and subtract $421.00 from it. Once you've subtracted, you would have gotten $1547.00. You would divide $1547.00 by the 22 people and get about $70.31 per person. :)

8 0
2 years ago
I really really really need help!!!!
Yakvenalex [24]
For f to be continuous at x=1, you need to have the limit from either side as x\to1 to be the same.

\displaystyle\lim_{x\to1^-}f(x)=\lim_{x\to1^-}(|x-1|+2)=2
\displaystyle\lim_{x\to1^+}f(x)=\lim_{x\to1^+}(ax^2+bx)=a+b

If a=2 and b=3, then the limit from the right would be 2+3=5\neq2, so the answer to part (1) is no, the function would not be continuous under those conditions.

This basically answers part (2). For the function to be continuous, you need to satisfy the relation a+b=2.

Part (c) is done similarly to part (1). This time, you need to limits from either side as x\to2 to match. You have

\displaystyle\lim_{x\to2^-}f(x)=\lim_{x\to2^-}(ax^2+bx)=4a+2b
\displaystyle\lim_{x\to2^+}f(x)=\lim_{x\to2^+}(5x-10)=0

So, a and b have to satisfy the relation 4a+2b=0, or 2a+b=0.

Part (4) is done by solving the system of equations above for a and b. I'll leave that to you, as well as part (5) since that's just drawing your findings.
8 0
2 years ago
What is 72/15 simplified
fiasKO [112]
4 4/5 is the answer. :)

or.. 4.8!
7 0
3 years ago
Read 2 more answers
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