All categories

2 years ago

Solve -r/4 < 8 A.r<-32 B. r. >-32 C.r < -2D.r>-2

Mathematics

1 answer:

Gnom [1K]2 years ago

4 0

$\frac{-r}{4}$

$Ans.\ B$

You might be interested in

Classify each triangle by its angles and sides. ABE BEC DEF CDF

lapo4ka [179]

Answer:

scalene right

isosceles obtuse

equilateral equilateral

scalene acute

7 0

2 years ago

Noah needs to order some new supplies for the restaurant where he works. The restaurant needs at least 503 knives. There are cur

anastassius [24]

Answer:

10x+391≥503

Step-by-step explanation:

let x = # of sets

If you need at least 503 knives, both the sets of knives and the initial amount we already have (391) must add to have a number of knives greater than or equal to the minimum required amount.

4 0

1 year ago

Estimate then record the product 293*4

jolli1 [7]

I believe that the equations answer is around 1170. The answer to the equation is 1172

3 0

2 years ago

A dozen plain hats cost 3.60,what is the unit rate of a hat?

deff fn [24]

I think the answer should be 12.

7 0

2 years ago

Read 2 more answers

After being treated with chemotherapy, the radius of the tumor decreased by 23%. What is the corresponding percentage decrease i

Crazy boy [7]

Answer:

The volume of the tumor experimented a decrease of 54.34 percent.

Step-by-step explanation:

Let suppose that tumor has an spherical geometry, whose volume ( $V$ ) is calculated by:

$V = \frac{4\pi}{3}\cdot R^{3}$

Where $R$ is the radius of the tumor.

The percentage decrease in the volume of the tumor ( $\%V$ ) is expressed by:

$\%V = \frac{\Delta V}{V_{o}} \times 100\,\%$

Where:

$\Delta V$ - Absolute decrease in the volume of the tumor.

$V_{o}$ - Initial volume of the tumor.

The absolute decrease in the volume of the tumor is:

$\Delta V = V_{o}-V_{f}$

$\Delta V = \frac{4\pi}{3}\cdot (R_{f}^{3}-R_{o}^{3})$

The percentage decrease is finally simplified:

$\%V = \left[1-\left(\frac{R_{f}}{R_{o}}\right)^{3} \right]\times 100\,\%$

Given that $R_{o} = R$ and $R_{f} = 0.77\cdot R$ , the percentage decrease in the volume of tumor is:

$\%V = \left[1-\left(\frac{0.77\cdot R}{R}\right)^{3} \right]\times 100\,\%$

$\%V = 54.34\,\%$

The volume of the tumor experimented a decrease of 54.34 percent.

5 0

2 years ago