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2 years ago

10

The door to the clubhouse has an area of 1,792 sq cms If the length of the door is 56cms how wide is the door ?

1 answer:

LuckyWell [14K]2 years ago

4 0

Step-by-step explanation:

area = length x width

area = 1,792 sq cm

length = 56 sq cm

width = ?

equation = 1,792-56= 1,736

Answer:

1,736 sq cm

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What is the classification for this polynomial?<br> u^2 v^3 w

andrew11 [14]

Answer:

A...

Step-by-step explanation:

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2 years ago

Determine algebraically whether the function is even, odd, or neither even nor odd.

m_a_m_a [10]

For a function (fn) to be odd:
f(x) = - f(-x)
For a fn to be even:
f(x) = f(-x)
For a fn to be neither even nor odd
f(x) != f(-x) [No Relation]

(-x)^n = x^n for n -> even
(-x)^n = -x^n for n -> odd
In your example:

f(x) = -4x^3 + 4x
f(-x) = -4 (-x)^3 + 4 (-x)^1 ( 3 and 1 are odd powers )
f(-x) = 4x^3 - 4x (take -1 common to do the check)
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2 years ago

Find the distance between (2, 2) and (-4, 4). Round answer to the nearest tenth.

Naya [18.7K]

Answer:

6.3

Step-by-step explanation:

First off, we need the distance formula, which is:

$\sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$

If we plug in the points, we get:

$\sqrt{(-4 - 2)^2 + (4 - 2)^2}$

If we simplify everything under the square root, we get:

$\sqrt{(-6)^2 + (2)^2}$

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2 years ago

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

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Answer:

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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:

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$\%V = 54.34\,\%$

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

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2 years ago

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