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klasskru [66]
3 years ago
6

Monique made several batches of soup for a potluck supper. Each batch required 3\4 0f a pound of potatoes, and she used a total

of 6 1/2 pounds of potatoes. Solve the division problem.
What is 6 1/2 ÷ 3/4
Mathematics
1 answer:
eimsori [14]3 years ago
6 0

Answer:

The\ value\ of\ \frac{6 \frac{1}{2}}{\frac{3}{4}} \ is\ \frac{26}{3}.

The\ value\ of\ \frac{6 \frac{1}{2}}{\frac{3}{4}} \ is\ 8.67 (Approx).

Step-by-step explanation:

As given

Monique\ made\ several\ batches\ of\ soup\ for\ a\ potluck\ supper.

Each\ batch\ required\ \frac{3}{4}\ of\ a\ pound\ of\ potatoes.

she\ used\ a\ total\ of\ 6 \frac{1}{2}\ pounds\ of\ potatoes.

As given the expression.

= \frac{6 \frac{1}{2}}{\frac{3}{4}}

= \frac{\frac{13}{2}}{\frac{3}{4}}

Solving the above

= \frac{13\times 4}{2\times 3}

= \frac{26}{3}

In the decimal form.

= 8.67 (Approx)

Therefore\ the\ value\ of\ \frac{6 \frac{1}{2}}{\frac{3}{4}} \ is\ \frac{26}{3}.

Therefore\ the\ value\ of\ \frac{6 \frac{1}{2}}{\frac{3}{4}} \ is\ 8.67 (Approx).

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g(x) = (x - 3)² - 4

given f(x) is shifted a units horizontally to the right → f(x) → f(x - a)

If f(x) is shifted by a units horizontally left → f(x) → f(x + a)

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If f(x) is shifted by b units vertically down → f(x) → f(x) - b

The vertex (0, 0) of f(x) has shifted to (3, -4)

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4 0
4 years ago
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wariber [46]
Hello!

The least common denominator is like the least common multiple. As you can see, these numbers already have the same denominators, and cannot be divided any further as they are prime. 

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I hope this helps!
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3 years ago
A storekeeper has two kinds of flour, one selling for 65 cents per pound, and the other selling for 95 cents per pound. How many
Murrr4er [49]

Answer:

The amount of 65 cents per pound flour to use is 40 pounds and the amount of 95 cents per pound flour to use is 60 pounds

Step-by-step explanation:

Let

x ----> amount of 65 cents per pound flour to use

y----> amount of 95 cents per pound flour to use

we know that

x+y=100 ----> y=100-x ----> equation A

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substitute equation A in equation B

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Solve for x

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Find the value of y

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therefore

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