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kotegsom [21]
3 years ago
6

The dimensions of a triangle are shown below. If the height of the triangle is increased by the factor of 4, which statement wil

l be true about the area of the triangle?
A. the area will increase by a factor of 2.
B. the area will increase by a factor of 4.
C. the area will increase by a factor of 8
D. the area will not change
Mathematics
1 answer:
Varvara68 [4.7K]3 years ago
6 0

Answer:

Option B. The area will increase by a factor of 4

Step-by-step explanation:

we know that

The area of triangle is equal to

A=\frac{1}{2}bh

where

b is the base of triangle

h is the height of triangle

If the height of the triangle is increased by the factor of 4

then

The new area of the triangle will be

Anew=\frac{1}{2}b(4h)

Anew=4[\frac{1}{2}bh]

Anew=4[A]

The new area is 4 times the original area

therefore

The area will increase by a factor of 4

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Step-by-step explanation:

Probability of winning a game = 1/4

Probability of winning two games in a row = 1/4 * 1/4 = 1/16

Probability of winning three games in a row = 1/4 * 1/4 * 1/4 = 1/64

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1/64 = 0.015625

1/64 = 1.6% (rounding to the next tenth)

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7 0
2 years ago
An ideal gas is confined within a closed cylinder at a pressure of 2.026 × 105 Pa by a piston. The piston moves until the volume
Elis [28]

Answer:

The final pressure of the gas when its temperature returns to its initial value 1.8234\times 10^6 Pa.

Step-by-step explanation:

Given : An ideal gas is confined within a closed cylinder at a pressure of 2.026\times 10^5 Pa by a piston. The piston moves until the volume of the gas is reduced to one-ninth of the initial volume.

To find : What is the final pressure of the gas when its temperature returns to its initial value?

Solution :

Since the temperature is constant .

The relation between P and V is given by,

P_1\times V_1 = P_2\times V_2

\frac{P_1}{P_2}=\frac{V_2}{V_1} ....(1)

The piston moves until the volume of the gas is reduced to one-ninth of the initial volume i.e. V_2=\frac{V_1}{9}

or \frac{V_2}{V_1}=\frac{1}{9}

P_1=2.026\times 10^5

Substitute in equation (1),

\frac{2.026\times 10^5}{P_2}=\frac{1}{9}

P_2=9\times 2.026\times 10^5

P_2=18.234\times 10^5

P_2=1.8234\times 10^6

The final pressure of the gas when its temperature returns to its initial value 1.8234\times 10^6 Pa.

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3 years ago
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