2 years ago

Could someone please help me with this problem? I don't understand.

Mathematics

1 answer:

sasho [114]2 years ago

4 0

Try dividing 150 by 2 1/2 and your answer would be 6.25 you may want to double check that math though

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Greg has 21 new magazines to read. Let M be the number of magazines he would have left to read after reading R of them. Write an

Rina8888 [55]

M = 21 - R is our equation. Plug your value of 12 into the equation for R.

M = 21 - 12.

M = 9.

Greg would have 9 magazines left after reading 12 of them

7 0

2 years ago

M + 5n = p solve for m

atroni [7]

-5n on both sides
M= -5n + p

8 0

3 years ago

Read 2 more answers

4(3x^2y^4)^3 / (2x^3y^5)^4

tatyana61 [14]

Solving the expressions $\frac{4(3x^2y^4)^3}{(2x^3y^5)^4}$ we get $\frac{27}{4x^{6}y^{8}}$

Step-by-step explanation:

We need to Solve the expression: $\frac{4(3x^2y^4)^3}{(2x^3y^5)^4}$

Solving:

$\frac{4(3x^2y^4)^3}{(2x^3y^5)^4}$

$=\frac{4(3^3x^6y^{12})}{(2^4x^{12}y^{20})}\\=\frac{4(27x^6y^{12})}{(16x^{12}y^{20})}\\=\frac{108x^6y^{12}}{16x^{12}y^{20}}$

Applying exponent rule: $\frac{x^a}{x^b}=x^{a-b}$

$=\frac{108x^{6-12}y^{12-20}}{16}\\=\frac{27x^{-6}y^{-8}}{4}$

Another exponent rule says: $x^{-a}=\frac{1}{x^a}=$

$=\frac{27}{4x^{6}y^{8}}$

So, solving the expressions $\frac{4(3x^2y^4)^3}{(2x^3y^5)^4}$ we get $\frac{27}{4x^{6}y^{8}}$

Keywords: Solving Exponents

Learn more about Solving Exponents at:

#learnwithBrainly

8 0

2 years ago

How to solve this problem and the answer.

weqwewe [10]

I’m not sure sorry cau

8 0

2 years ago

Read 2 more answers

Kevin is 3 times as old as Daniel. 4 years ago, Kevin was 5 times as old as Daniel.

DochEvi [55]

X is Kevin; Y is Dan

X = 3Y

X - 4 = 5(Y - 4)

3Y - 4 = 5(Y - 4)
3Y - 4 = 5Y - 20
   + 4 +4
4 cancels each other out.
3Y = 5Y - 16
   - 5 -5
5 cancels each other out.
-2Y = -16
/2 /2
   Y = 8

X = 3Y;
X = 3 x 8;

Thus,
X = 24;
Kevin(x) is 24

8 0

3 years ago