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2 years ago

Find the surface area of the composite figure.

Mathematics

1 answer:

Neko [114]2 years ago

8 0

Answer:

3cm I think that's the answer

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The rainfall amount last year was 6 inches below normal. The rainfall this year was 13 inches more than last year. What is the d

Lostsunrise [7]

Answer:

Step-by-step explanation:

The term difference means to subtract. Therefore, 13in - 6in, is equal to 7 inches.

13-6=7

Hope this Helps!!!

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3 years ago

What is the density of a sample with a mass of 5250g and a volume of 4.5mL?

just olya [345]

Step-by-step explanation:

Density=5250/4.5=1166.66 g/ml

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3 years ago

Need help on khan NOW! question is on picture. will report link and mark brainliest if right

Genrish500 [490]

Answer: R= 1

Explanation: no matter how you do it you still get 1 bc its x divided by y which is 1

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3 years ago

Dan bought one package if blackberries for 3 dollars/euros. How many packages can Darryl buy if he has 15 dollars/euros?

myrzilka [38]

Answer:

Dan can buy 5 bags for 15 dollars

Step-by-step explanation:

divide 15 and 3 it will equal 5

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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