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3 years ago

What is the surface area of a square pyramid that has a base length of 5 inches and triangular faces that are 9 inches tall

Mathematics

1 answer:

scZoUnD [109]3 years ago

7 0

Answer:

115 in²

Step-by-step explanation:

SA = Area of triangular surfaces + area of the square base

SA= 2bh + b²

SA = 2(5)(9)+5²

SA = 90+25

SA = 115

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75 : 60

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2 years ago

The price for 7 lb of potatoes is $5 what is the price per pound.

Thepotemich [5.8K]

Answer:

The answer is $1.4.

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2 years ago

What is the MAD of the data set?

const2013 [10]

The answer to your question is 2.4

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3 years ago

1 2 3 4 5 6 7 8 9 10

daser333 [38]

<h3>Answer:</h3>

2.25

<h3>Explanation:</h3>

Consider the square ...

... (x+a)² = x² +2ax +a²

The constant term (a²) is the square of half the x-coefficient: a² = (2a/2)².

The x-coefficient in your expression is 3. The square of half that is ...

... (3/2)² = 9/4 = 2.25

Adding 2.25 to both sides gives ...

... x² +3x + 2.25 = 6 + 2.25

... (x +1.5)² = 8.25 . . . . completed square

4 0

3 years ago

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