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3 years ago

Quick:Three different factorizations of 32m+56mp.

Mathematics

1 answer:

Nat2105 [25]3 years ago

8 0

Answer:

8m(4+7p)

Step-by-step explanation:

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Artemon [7]

Answer:

so yea what was the answer again

Step-by-step explanation:

6 0

3 years ago

Solve for h. 15.3 + 1 h =1.3- 1 h

Iteru [2.4K]

Answer:

-7 = h

Step-by-step explanation:

15.3 + 1h = 1.3 - 1h

+1h +1h

15.3 + 2h = 1.3

Remember what you do to one side you MUST do to the other.

15.3 + 2h = 1.3

-15.3 -15.3

2h = -14

Now to isolate the variable you divide both sides by 2 to get, h = -7

Hope that helps and have a great day!

6 0

2 years ago

Read 2 more answers

Is (6, 0) a coordinate, a slope, or neither

MrRissso [65]

Answer:

Coordinate

Step-by-step explanation:

It is a coordinate because with slope you would need at least two ordered pairs

6 0

3 years ago

Read 2 more answers

Can someone please help me with my maths question

DIA [1.3K]

Answer:

$a. \ \dfrac{625 \cdot m}{27 \cdot n^{11}}$

$b. \ \dfrac{x^{3 \cdot m - 2}}{y^{ 3 + n}}$

Step-by-step explanation:

The question relates with rules of indices

(a) The give expression is presented as follows;

$\dfrac{m^3 \times \left (n^{-2} \right )^4 \times (5 \cdot m)^4}{\left (3 \cdot m^2 \cdot n \right )^3}$

By expanding the expression, we get;

$\dfrac{m^3 \times n^{-8} \times 5^4 \times m^4}{\left 3^3 \times m^6 \times n^3}$

Collecting like terms gives;

$\dfrac{m^{(3 + 4 - 6)} \times 5^4}{ 3^3 \times n^{3 + 8}} = \dfrac{625 \cdot m}{27 \cdot n^{11}}$

$\dfrac{m^3 \times \left (n^{-2} \right )^4 \times (5 \cdot m)^4}{\left (3 \cdot m^2 \cdot n \right )^3}= \dfrac{625 \cdot m}{27 \cdot n^{11}}$

(b) The given expression is presented as follows;

$x^{3 \cdot m + 2} \times \left (y^{n - 1} \right )^3 \div (x \cdot y^n)^4$

Therefore, we get;

$x^{3 \cdot m + 2} \times \left (y^{n - 1} \right )^3 \times x^{-4} \times y^{-4 \cdot n}$

Collecting like terms gives;

$x^{3 \cdot m + 2 - 4} \times \left (y^{3 \cdot n - 3 -4 \cdot n}} \right ) = x^{3 \cdot m - 2} \times \left (y^{ - 3 -n}} \right ) = x^{3 \cdot m - 2} \div \left (y^{ 3 + n}} \right )$

$x^{3 \cdot m - 2} \div \left (y^{ 3 + n}} \right ) = \dfrac{x^{3 \cdot m - 2}}{y^{ 3 + n}}$

$x^{3 \cdot m + 2} \times \left (y^{n - 1} \right )^3 \times x^{-4} \times y^{-4 \cdot n} =\dfrac{x^{3 \cdot m - 2}}{y^{ 3 + n}}$

4 0

3 years ago

3х – 6+1 = - 2x – 5+5х

marysya [2.9K]

Answer:0=12

Step-by-step explanation:

Rearrange the equation by subtracting what is to the right of the equal sign from both sides of the equation:

3x-7=3x-5

Combine like terms:

3x-3x=7+5

0=12

4 0

3 years ago

Read 2 more answers