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andrew11 [14]
3 years ago
8

8. Expand (2a – 3b + 4c)2 Helllpppp plz Nowww it’s imp

Mathematics
1 answer:
Elis [28]3 years ago
7 0

Step-by-step explanation:

(2a–3b+4c)²=

4a²+9b²+16c²-12ab-24bc+8ac

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Brandon is selling gift cards. He earns a one-time pay out of $159 and then earns $30 for each gift card he sells. Write a equat
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Answer:

g * $30 = p

Step-by-step explanation:

He earns $30 per gift card.

You multiply the amount of gift cards sold by the money you earn per gift card sold. g, being the gift cards sold. p, being the total pay.

4 0
3 years ago
The table shows data from a survey about the number of times families eat at restaurants during a week. The families are either
kifflom [539]

Hey there,

To solve this problem, let us first define what is mean and median. Mean is the average of all the numbers in the data set while the median is the number in the middle of the data set in ascending order. If we create a box plot for the data of Rome and New York, we can see that there is an outlier in the data for New York. Since New York has an outlier, so the mean is not a good representation on the central tendency of the data. The mean is skewed (distorted) by the outlier. So in this case it is better to use the median. While the Rome data is nice and symmetrical, it does not seem to have an outlier,  so we can use the mean for this data set.

Therefore the answer is:

The Rome data center is best described by the mean. The New York data center is best described by the median

Hoped I Helped

6 0
4 years ago
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Can someone please check this for me? distance formula
seropon [69]
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Solve the quadratic equation<br> give your answer to 2 decimal places<br> : 3x^2+x-5=0
Dima020 [189]

Given:

The quadratic equation is:

3x^2+x-5=0

To find:

The solution for the given equation rounded to 2 decimal places.

Solution:

Quadratic formula: If a quadratic equation is ax^2+bx+c=0, then:

x=\dfrac{-b\pm \sqrt{b^2-4ac}}{2a}

We have,

3x^2+x-5=0

Here, a=3,b=1,c=-5. Using the quadratic formula, we get

x=\dfrac{-1\pm \sqrt{1^2-4(3)(-5)}}{2(3)}

x=\dfrac{-1\pm \sqrt{1+60}}{6}

x=\dfrac{-1\pm \sqrt{61}}{6}

x=\dfrac{-1\pm 7.81025}{6}

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x=\dfrac{-1+7.81025}{6}

x=1.13504167

x\approx 1.14

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x=\dfrac{-1-7.81025}{6}

x=-1.468375

x\approx -1.47

Therefore, the required solutions are 1.14 and -1.47.

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