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2 years ago

If A = lw, find the width of a rectangle with a length of 30 miles and an area of 210 square miles.

Mathematics

1 answer:

Lera25 [3.4K]2 years ago

6 0

Answer:

Yes and no nor

Step-by-step explanation:

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What is the following product

dangina [55]

Option B is correct.

Step-by-step explanation:

We need to solve: $\sqrt[3]{x^2}\sqrt[4]{x^3}$

We know that: $\sqrt[n]{x}\sqrt[b]{x} =\sqrt[n*b]{x.x}= \sqrt[n*b]{x^2}$

Applying the above rule:

$\sqrt[3]{x^2}\sqrt[4]{x^3}\\=\sqrt[3*4]{x^2.x^3}\\=\sqrt[12]{x^5}$

So, Option B is correct.

Keywords: Solving with Exponents

Learn more about Solving with Exponents at:

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2 years ago

What is the

olya-2409 [2.1K]

Answer:

joe mamma!!!

Step-by-step explanation:

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3 years ago

How many days are in a month

ser-zykov [4K]

It could be 30 to 31 but February has 28 expected from a leap year with 29 days

7 0

2 years ago

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Sophia‘s favorite homemade cookie recipe requires one cup of chocolate chips for 10 servings if the number of cups required for

weqwewe [10]

Answer:

3 cups

Step-by-step explanation:

We can use a proportion to find how many cups of chocolate chips she needs for 30 servings. Assuming c = cups of chocolate chips and b = batches

$\frac{c}{b}$

$\frac{1}{10} = \frac{c}{30}$

We can now multiply the diagonal values that don't include the missing variable (30 and 1) and then divide it by the value that is diagonal to the variable (10)

$30 \cdot 1 = 30\\30 \div 10 = 3$

Therefore, she needs 3 cups of chocolate chips to make 30 servings.

8 0

2 years ago

Read 2 more answers

Find the critical numbers of the function. (Enter your answers as a comma-separated list. If an answer does not exist, enter DNE

EleoNora [17]

Answer:

0, 10

Step-by-step explanation:

The given function is:

$g(y) = \frac{y-5}{y^2-3y+15}$

According to the quotient rule:

$d(\frac{f(y)}{h(y)}) = \frac{f(y)*h'(y)-h(y)*f'(y)}{h^2(y)}$

Applying the quotient rule:

$g(y) = \frac{y-5}{y^2-3y+15}\\g'(y)=\frac{(y-5)*(2y-3)-(y^2-3y+15)*(1)}{(y^2-3y+15)^2}$

The values for which g'(y) are zero are the critical points:

$g'(y)=0=\frac{(y-5)*(2y-3)-(y^2-3y+15)*(1)}{(y^2-3y+15)^2}\\(y-5)*(2y-3)-(y^2-3y+15)=0\\2y^2-3y-10y+15-y^2+3y-15\\y^2-10y=0\\y=\frac{10\pm \sqrt 100}{2}\\y_1=\frac{10-10}{2}= 0\\y_2=\frac{10+10}{2}=10$

The critical values are y = 0 and y = 10.

5 0

2 years ago