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podryga [215]
1 year ago
7

Rewrite one tenth times x cubed times y plus two tenths times x times y squared using a common factor.

Mathematics
1 answer:
zalisa [80]1 year ago
6 0

The expression which can be rewritten using a common factor is; xy/10 (x² + 2y).

<h3>What is an expression?</h3>

Expression in maths is defined as the collection of numbers variables and functions by using signs like addition, subtraction, multiplication, and division.

We have the following information given as;

One tenth times x cubed times y plus two tenths times x times y squared using a common factor.

We know that 'times' has to be replaced with multiplication.

Now, we will change this into mathematical expression as;

x³y/ 10 + 2xy²/ 10

Now, the common factor can be ;

xy/10

The Common factor refers to the number and the variables taken as a common factor should be able to completely divide the terms given any expression.

Now, taking it as a common factor, we have the following expression:

xy/10 (x² + 2y)

Hence, the required expression is; xy/10 (x² + 2y).

To know more about factor here;

brainly.com/question/24182713

#SPJ1

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Simplify this expression 10b+7-3b-2
iren2701 [21]

Answer:

7b+5

Step-by-step explanation:

10b + 7 − 3b −2

=10b + 7 + −3b + −2

Combine Like Terms:

=10b + <u>7 </u>+ −3b + <u>−2 </u>

=(10b + −3b) + ( <u>7 </u>+ <u>−2</u>)

=7b + <u>5</u>

Answer:

7b + 5

Hope this helps : )

3 0
3 years ago
Read 2 more answers
I JUST WANNA PASS MY SOPHOMORE YEAR BUT I SUCK AT MATH!!!!! PLEASE HELP!!!!!!
Natasha2012 [34]

Mr. Sanchez's class: 1.65p

Mr. Kelly's class: 1.36j

where p = fruit pies and j = fruit juice

____________________________________________________________

(a) Creating a system of equations

p + j = 79

1.65p + 1.36j = 118.17

are your system of equations

____________________________________________________________

(a) Solving the system of equations for p and j

Solve for p in the first equation. Subtract j from both sides.

p = 79 - j

____________________________________________________________

Plug p into the second equation.

1.65(79 - j) + 1.36j = 118.17

Distribute 1.65 inside the parentheses.

130.35 - 1.65j + 1.36j = 118.17

Combine like terms.

130.35 - 0.29j = 118.17

Subtract 130.35 from both sides.

-0.29j = -12.18

Divide both sides by -0.29.

j = 42

____________________________________________________________

Plug j into the first equation.

p + (42) = 79

Subtract 42 from both sides.

p = 37

____________________________________________________________

(b) To find out how much each class earned

Now plug in p and j into Mr. Sanchez's class and Mr. Kelly's class equations (at the top).

Mr. Sanchez's class: 1.65(37) = 61.05

Mr. Kelly's class: 1.36(42) = 57.12

____________________________________________________________

(c) To see how much more Mr. S's class earned

Subtract 57.12 from 61.05.

61.05 - 57.12 = 3.93

____________________________________________________________

(a)

p + j = 79

1.65p + 1.36j = 118.17

(b)

Mr. Sanchez's class earned more money (61.05 > 57.12).

(c)

Mr. Sanchez's class earned $3.93 more than Mr. Kelly's class.

3 0
3 years ago
Drag each equation to show if it could be a correct first step to solving the equation 2(x+7)=36
sdas [7]

2(x + 7) = 36 |use distributive property

(2⋅x)+(2⋅7)=36

2·x+2·7=36

2x + 14 = 36

2(x + 7) = 36 |:2

x + 7 = 18

Answer (bold expressions).

7 0
3 years ago
Read 2 more answers
Which is the graph of g(x)?
hichkok12 [17]

Answer:

option d

Step-by-step explanation:

We are given with three function

Plug in x=-2 in first function

f(x)= 3 so f(-2) = 3

Plug in x=-2 in second function

f(x) = -\frac{x}{2} +2

f(-2) = -\frac{-2}{2} +2=3

Plug in x=2 in second function

f(x) = -\frac{x}{2} +2

f(-2) = -\frac{2}{2} +2=1

Plug in x=2 in third function

f(x)= 2x-3

f(2)= 2(2) -3 = 1

We can see that f(-2)=3 is same for first and second function

also f(2) = 1 is same for second and third functions

So the graph is continuous

option D satisfies the points


8 0
3 years ago
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<span>The answers to this are:
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</span>
6 0
3 years ago
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