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tino4ka555 [31]
2 years ago
9

Can someone help me on this 6th grade problem? Its worth 15 points for anyone to answer

Mathematics
2 answers:
Zolol [24]2 years ago
8 0

\bold{Heya!}

<h2>→ <u>Answer :-</u></h2>

<h3>B. Put the numbers in the order from least to greatest and then find the middle number.</h3>

<h2>→ <u>EXPLANATION :</u></h2>

<u></u>

Because you will always have to put the numbers from the smallest, and to the Iargest/biggest Then you need to find the middle number and that's how you will get your answer from that point of view.

Hopefully This Helps ! ~

#LearnWithBrainly

\underline{Answer :}

<em>Jaceysan ~</em>

mafiozo [28]2 years ago
7 0

Answer:

<u>I Believe your answer is B, put all the numbers in order then find the middle number</u>

Step-by-step explanation:

it may be thought of as the "middle" value of a data set. For example, in the data set {1, 3, 3, 6, 7, 8, 9}, the median is 6,

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OleMash [197]
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2 years ago
Anne-marienhas saved 9 dollars for a new coat that is 1/6 as much money as she needs how much does the coat cost
Tatiana [17]

Answer:

Cost of the coat = $54

Step-by-step explanation:

Given:

She saved = $ 9

To Find:

Cost of the coat = ?

Solution:

Let the price of coat be x

She saved $9

and according to given condition it is 1/6 of the total money she needs for buying coat

by this we have

\frac{1}{6}*x = 9

Multiplying both sides by 6

\frac{1}{6}*x*6 = 9*6

it becomes

x = 9*6

x = $54

As x is the amount she needs

so

Cost of the coat = $54


7 0
3 years ago
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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.
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