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fredd [130]
3 years ago
6

What is the simplified expression for 5ab+9ab+ab

Mathematics
2 answers:
OLEGan [10]3 years ago
8 0
It is basically 5+9+1ab
Margaret [11]3 years ago
4 0
The answer would be 15ab
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Interpret the coefficient of variation according to the context, determining the most homogeneous set
lord [1]

Answer:

2009 data set is most homogeneous

Step-by-step explanation:

As per coefficient of variation, which is around 5% for the year 2009 and around 20% for the year 2010 we can state that the data set for the year 2009 is most homogeneous

Homogeneous means similar, it can be observed on the graph as well that the data for the year 2009 is less variant

Hope it answers your question.

4 0
3 years ago
Solve this equation<br> 3x^2 − 16x = x^2 − 30
ValentinkaMS [17]

3x^2-16x=x^2-30

-x^2        -x^2

2x^2-16x=-30

+30          +30

2x^2-16x+30=0

2x-6=0 or x-5=0

x=3    or x=5

A. 3,5

If you need an explanation, I can provide one. Hope it helped! If it didn't tell me what went wrong or what confused you.

3 0
3 years ago
Read 2 more answers
Simplify the expression.
kotegsom [21]

Answer:

\large\boxed{D.\ 2x^2y\sqrt[5]{7xy^3}}

Step-by-step explanation:

\sqrt[5]{224x^{11}y^8}\\\\=\sqrt[5]{(32)(7)x^{5+5+1}y^{5+3}}\\\\\text{use}\ a^na^m=a^{n+m}\\\\=\sqrt[5]{(2^5)(7)x^5x^5x^1y^5y^3}\\\\\text{use}\ \sqrt[n]{ab}=\sqrt[n]{a}\cdot\sqrt[n]{b}\\\\=\sqrt[5]{2^5}\cdot\sqrt[5]{x^5}\cdot\sqrt[5]{x^5}\cdot\sqrt[5]{y^5}\cdot\sqrt[5]{7xy^3}\\\\\text{use}\ \sqrt[n]{a^n}=a\\\\=2\cdot x\cdot x\cdot y\cdot\sqrt[5]{7xy^3}\\\\=2x^2y\sqrt[5]{7xy^3}

6 0
3 years ago
Convert standard to slope-intercept forms. 1. Standard form: 10x − 7y = −8
garri49 [273]

Answer:

Step-by-step explanation:

-7y = -10x - 8

y = 10/7x + 8/7

8 0
3 years ago
Suppose lines EF and GH are reflected over the line Y equals X to form the lines JK and LM which statement would be true about t
katrin [286]

Answer:

A) JK and LM will be parallel to each other.

Step-by-step explanation:

On reflection on y=x line the x co-ordinate changes with y co-ordinate and y co-ordinate changes with x co-ordinate

(x,y)\rightarrow (y,x)

Points on line EF

(0,6) , (-5,-2)

On reflection of this line on y=x the new points we get for line JK are

(6,0),(-2,-5)

Points on line GH

(-4,9),(-9,1)

On reflection on y=x line the new points we get for line LM are

(9,-4),(1,-9)

Slope of line JK

m=\frac{y_2-y_1}{x2-x1}\\m=\frac{(-5)-0}{(-2)-6} \\m=\frac{-5}{-8}=\frac{5}{8}

Slope of line LM

m=\frac{y_2-y_1}{x2-x1}\\m=\frac{(-9)-(-4)}{1-9} \\m=\frac{-9+4}{-8}=\frac{-5}{-8}\\m=\frac{5}{8}

For two line to be parallel, their slopes will be same.

m_{JK} =\frac{5}{8} , m_{LM}=\frac{5}{8}

Since slopes of lines JK and LM are same therefore we can say that these are parallel to each other.

5 0
3 years ago
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