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MariettaO [177]
3 years ago
13

Box A has a volurme of 12 cublic meters. Box B is similar to box A. To create Box B, Box A's dimensions were multiplied by five.

What is the volume of Box B?
A) 60 m^3
B)300 m^3
C) 1,500 m^3
D) 7,500 m^3
Mathematics
1 answer:
CaHeK987 [17]3 years ago
8 0

Answer:

<h2>C) 1,500 m³</h2>

Step-by-step explanation:

k - \text{scale of similar}

\text{If}\ AB\sim CD\ \text{in scale}\ k,\ \text{then}\ \dfrac{length_{AB}}{length_{CD}}=k

\text{If}\ A\sim B\ \text{in scale}\ k,\ \text{then}\ \dfrac{area_A}{area_B}=k^2

\text{If}\ A\sim B\ \text{in scale}\ k,\ \text{then}\ \dfrac{volume_A}{volume_B}=k^3

\text{We have}\ B\sim A\ \text{in scale}\ k=5.\ V_A=12\ m^3,\ \text{therefore}\\\\\dfrac{V_B}{V_A}=k^3\to\dfrac{V_B}{12}=5^3

\dfrac{V_B}{12}=125            <em>multiply both sides by 12</em>

V_B=1500\ m^3

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69.808

Step-by-step explanation:

69.808

4 0
3 years ago
One week Beth bought 3 apples and 8 pears for $14.50. The next week she bought 6 apples and 4 pears and paid $14. Find the cost
zalisa [80]
Let us assume the cost of 1 apple = x dollars
Let us also assume the cost of 1 pear = y dollars
Then we can form two equations from the details given in the question. Based on those details the required answer to the question can be easily deduced.
3x + 8y = 14.50
And
6x + 4y = 14
Dividing both sides of the equation by 2 we get
3x + 2y = 7
2y = 7 - 3x
y = (7 - 3x)/2
Putting the value of y from the second equation in the first equation we get
3x + 8y = 14.50
3x + 8[(7 - 3x)/2] = 14.50
3x + 4 (7 - 3x) = 14.50
3x + 28 - 12x = 14.50
- 9x = 14.50 - 28
- 9x = - 13.5
9x = 13.5
x = 13.5/9
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Putting the value of x in the second equation we get
6x + 4y = 14
(6 * 1.5) + 4y = 14
9 + 4y = 14
4y = 14 - 9
4y = 5
y = 5/4
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So we can find from the above deduction that the cost of 1 apple is 1.5 dollars and the cost of 1 pear is 1.25 dollars
Then
Cost of 2 apples = 2 * 1.5 dollars
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5 0
4 years ago
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Step-by-step explanation:

Given

\frac{\sqrt{2} }{2} + \frac{2}{\sqrt{2} } ← rationalise the denominator by multiplying by \frac{\sqrt{2} }{\sqrt{2} }

= \frac{\sqrt{2} }{2} + ( \frac{2}{\sqrt{2} } × \frac{\sqrt{2} }{\sqrt{2} } )

= \frac{\sqrt{2} }{2} + \frac{2\sqrt{2} }{2}

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5 0
3 years ago
The doubling period of a bacterial population is 20 20 minutes. At time t = 100 t=100 minutes, the bacterial population was 9000
Nuetrik [128]

Answer:

The initial population was 2810

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Step-by-step explanation:

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P = P_0 \times e^{rt}

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Data: The doubling period of a bacterial population is 20 minutes (1/3 hour). Replacing this information in the formula we get:

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\frac{9000}{e^{2.08 \; 5/3}} = P_0

2810 = P_0

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P = 2810 \times e^{2.08 \; 5}

P = 92335548

3 0
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8 0
4 years ago
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