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3 years ago

8

Aaron bought 2 pounds of cheese for $6. Assuming the situation is proportional, determine how much it will cost for Aaron to buy

15 pounds of cheese.

1 answer:

Inessa [10]3 years ago

3 0

The answer is $45 hope that helps

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What is the quotient? StartFraction 3 y 2 Over 3 y EndFraction divided by StartFraction 6 y squared 4 y Over 3 y 2 EndFraction S

Dima020 [189]

The quotient of the expression is $\frac{1}{2y}$

The expression is given as:

$\frac{3y^2}{3y} \div \frac{6y^4}{3y^2}$

Divide 3y^2 by 3y

$\frac{3y^2}{3y} \div \frac{6y^4}{3y^2} = y \div \frac{6y^4}{3y^2}$

Rewrite the equation as a product

$\frac{3y^2}{3y} \div \frac{6y^4}{3y^2} = y \times \frac{3y^2}{6y^4}$

Divide 6 by 3

$\frac{3y^2}{3y} \div \frac{6y^4}{3y^2} = y \times \frac{y^2}{2y^4}$

Multiply y and y^2

$\frac{3y^2}{3y} \div \frac{6y^4}{3y^2} = \frac{y^3}{2y^4}$

Divide y^3 by y^4

$\frac{3y^2}{3y} \div \frac{6y^4}{3y^2} = \frac{1}{2y}$

Hence, the quotient of the expression is $\frac{1}{2y}$

Read more about quotients at:

brainly.com/question/12217706

8 0

2 years ago

A theorem is a statement that can be easily prove using a corollary

Reika [66]

False. Although, I'm not 100% sure.

7 0

3 years ago

Read 2 more answers

What are the solutions to the equation 0 = x2 - X-6? Select two options

Ganezh [65]

Step-by-step explanation:

x2 - x - 6 = 0

x2 - 3x + 2x - 6 = 0

x(x - 3) + 2(x - 3) = 0

x - 3 = 0. x + 2 = 0

x = 3 x = - 2

Option no 2 and 5 are the correct answer

8 0

3 years ago

The kite shown has a width of 24 inches and a total height of 42 inches. What is the total area of the kite without the tail? Sh

joja [24]

Answer:

504 in²

Step-by-step explanation:

The area of a kite is given by :

(a * b) / 2

Where a and b are the width and height of the kite :

Width = 24 inches ; height = 42 inches

Hence,

Area of kite = (24 * 42) / 2

Area of kite = 24 * 21

Area of kite = 504 in²

3 0

3 years ago

What is the tenth term of the geometric sequence that has a common ratio of `1/3` and 36 as its fifth term?

katrin [286]

$\bf \begin{array}{llll} term&value\\ \cline{1-2} a_5&36\\ a_6&36\left( \frac{1}{3} \right)\\ a_7&36\left( \frac{1}{3} \right)\left( \frac{1}{3} \right)\\ a_8&36\left( \frac{1}{3} \right)\left( \frac{1}{3} \right)\left( \frac{1}{3} \right)\\ a_9&36\left( \frac{1}{3} \right)\left( \frac{1}{3} \right)\left( \frac{1}{3} \right)\left( \frac{1}{3} \right)\\ a_{10}&36\left( \frac{1}{3} \right)\left( \frac{1}{3} \right)\left( \frac{1}{3} \right)\left( \frac{1}{3} \right)\left( \frac{1}{3} \right) \end{array}$

$\bf a_{10}=36\left( \frac{1}{3} \right)^5\implies a_{10}=36\cdot \cfrac{1^5}{3^5}\implies a_{10}=\cfrac{36}{243}\implies a_{10}=\cfrac{4}{27}$

3 0

3 years ago