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swat32
3 years ago
14

A rectangular gasoline tank can hold 49.0 kg of gasoline when full. what is the depth of the tank (in m) if it is 0.400 m wide b

y 0.900 m long?
Mathematics
1 answer:
Arturiano [62]3 years ago
5 0
We know that

<span>ρ = density of gasoline = 737 kg/m³ (at T = 60°F = 15.6°C)
</span>ρ = m/V
ρV = m
V = m/ρ
V = 49.0 kg / 737 kg/m³
<span>V = 0.066 m³ 

[volume of the tank]=L*W*H-----> H=volume/[L*W]----> H=0.066/(0.9*0.4)
H=0.1833 m

the answer is
t</span><span>he depth of the tank is 0.18 m</span>


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Step-by-step explanation:

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Triangle ABC is being dilated with the center of dilation at the origin. The image of C, point C' has coordinates of (7.2, 3.6).
MAXImum [283]

General Idea:

When we are given a point P(x, y) centered at origin with a scale factor of k, then the dilated point will be given by P' (kx, ky)

Applying the concept:

In the diagram given, the coordinate of C is (6, 3). Triangle ABC is being dilated with the center of dilation at the origin. The image of C, point C' has coordinates of (7.2, 3.6).

If C (x, y) centered at origin with a scale factor of k, then the dilated point will be given by C' (kx, ky)

We know x = 6 & kx = 7.2

Substituting 6 for x in the equation kx = 7.2, we get 6k = 7.2.

Dividing 6 on both sides\frac{6k}{6}=\frac{7.2}{6}

Simplifying fractions on both sidesk=1.2

Point A from the diagram is (-3, 3)

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y-coordinate of A' = ky =1.2*3=3.6

Point A' is given by (-3.6, 3.6)

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yaroslaw [1]

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Step-by-step explanation:

Given the geometric sequence

7, 14, 28, ...

We know that a geometric sequence has a constant ratio 'r' and is defined by

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Computing the ratios of all the adjacent terms

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now substituting r = 2 and a₁ = 7 in the nth term

a_n=a_1\cdot r^{n-1}

a_n=7\cdot \:2^{n-1}

Therefore, the nth term of the geometric sequence 7, 14, 28, ... is:

a_n=7\cdot \:2^{n-1}

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