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DanielleElmas [232]
4 years ago
12

Enter what that means? Help someone.

Mathematics
1 answer:
Luda [366]4 years ago
7 0

\frac{x^2}{3} is the same as \frac{1}{3}x^2

Dividing by 3 is the same as multiplying by the fraction 1/3

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There are a few boxes with 12 red pens in each and a few boxes with 10 blue pens in each. If 10 red pens and 5 blue pens are rem
jolli1 [7]

Answer with explanation:

x=the number of boxes containing red pens and

y = the number of boxes containing blue pens.

The expression which describes the number of red pen and blue pen in Boxes is

 = (12 x-10)+(10 y-5)

The equivalent expression are

1.By Adding like terms

=12 x +1 0 y -10 -5

=12 x +1 0 y -15

or Using Associative Property of addition of three numbers which is

⇒a+(b+c)=(a+b)+c=(a+c)+b

= 10 y +12 x -15

or

= -15 +10 y +12 x

or

= 12 x -15 +10 y

3 0
3 years ago
Read 2 more answers
Can somone explain to me how to do this and help me thansk!!!
Gnesinka [82]
=> x² = (2.4)² + (4.8)²

=> x² = 5.76 + 23.04

=> x² = 28.8

=> x = 5.366563146

=> x = 5.4

Ans: 2nd option (5.4)

Hope this helps!
6 0
3 years ago
Solve the radical equation x – 7 = square root of -4x+28. Which statement is true about the solutions to the radical equation? T
Julli [10]

Answer:

x = 3 and 7

There are two true solutions.

Step-by-step explanation:

To solve x-7 = \sqrt{-4x+28}, use inverse operations by squaring both sides of the equal sign.

(x-7)^2 = (\sqrt{-4x+28})^2\\x^2-14x+49 = -4x+28\\x^2-14x+4x+49-28 = 0\\x^2-10x+21=0

The quadratic expression can be factored into binomials and set equal to 0 by the zero product property to find x.

(x - 3) ( x - 7) = 0

x-3 = 0 so x=3

x-7 = 0 so x=7

Now check each solution into the original equation to be sure it solve the solution and is not extraneous.

7-7 = \sqrt{-4(7)+28}\\ 0=0

and

3-7 = \sqrt{-4(3)+28}\\[tex]-4 = \sqrt{16}\\-4 =-4[/tex]

3 0
3 years ago
Paul's dad made pot pie and the family ate 4/8 of it and Paul ate 2/8. How much remains?
Lera25 [3.4K]
2/8 because 4/8+2/8= 2/8
4 0
3 years ago
Read 2 more answers
Here you go amiragumbs2
Simora [160]
Umm , what is this for? do you need help on anything or? lol
5 0
3 years ago
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