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Oksanka [162]
4 years ago
10

Solve 1/5x+3=10. Show all work and check it

Mathematics
2 answers:
goblinko [34]4 years ago
8 0


1/5x+3=10

1/5×-3=10-3

1/5×=7

5*(1/5x)=7*5

x=35
iren2701 [21]4 years ago
5 0
1/5x + 3 = 10

Subtract by 3 on both sides. 

1/5x = 7.

Now to isolate the variable, flip the coefficient in front of the variable and multiply both sides by it. 

5(1/5x) = 7(5)

x = 7(5)
x = 35

Now to check:

1/5(35) + 3 = 10

35/5 + 3 = 10

35/5 is 7.

7 + 3 = 10

10 = 10
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Amy purchased a brand-new mountain bike, which she's riding through
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Answer:

The Acceleration of bike after 3 minutes of ride is  14 meter² per minutes

Step-by-step explanation:

Given as :

The displacement of bike is the function of time

i.e s(t) = t³ - 2 t² + 3  meters

The Time of bike acceleration = t = 3 minutes

Let The Acceleration of bike = a meter² per minutes

Now, According to question

∵ <u>Acceleration is define as the rate of change of velocity</u>

i.e acceleration = \dfrac{\textrm velocity}{\textrm time}

or, a =  \dfrac{\textrm v}{\textrm t}

And

<u>Velocity is define as rate of change of displacement</u>

i.e velocity =  \dfrac{\textrm displacement}{\textrm time}

Or, v =  \dfrac{\textrm s}{\textrm t}

Since here displacement is function of time

So , v = \frac{\partial s(t)}{\partial t}

Or, v =   \frac{\partial (t³ - 2 t² + 3)}{\partial t}

Or, v = 3 t² - 4 t + 3

So, velocity is the function of time = v = 3 t² - 4 t + 3   meter per minutes

Now, Again

Acceleration = a = \frac{\partial v}{\partial t}

Or, a = \frac{\partial (3 t² - 4 t + 3)}{\partial t}

Or, a = 6 t - 4

∴  Acceleration is the function of time = a = 6 t - 4

Now, Acceleration of bike after 3 minutes

So, at t = 3 min

i.e a = 6 t - 4

Or, a = 6 × 3 - 4

Or, a = 18 - 4

∴  a = 14 meter² per minutes

Hence, The Acceleration of bike after 3 minutes of ride is  14 meter² per minutes . Answer

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6 0
3 years ago
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1. Main Show Tank Calculation:
natulia [17]

Step-by-step explanation:

Volume of a sphere is

\frac{4}{3} \pi {r}^{3}

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\pi {r}^{2} h

from the second prompt we know the tanks are identical since they are said to be congruent which means the hight and radius are the same. Plug in 120 f for h and 35 ft for r to get the volume of a cylinder and then divide that by 2 to get the volume of a single tank. the volume of both tanks would be

\pi {35}^{2} (120) = 461,814

To show tank model is a sixth of the original so you divide the radius by 6 and perform the same calculation as in the first portion of the problem ro get the volume a quarter sphere

\frac{4}{3} \pi {( \frac{70}{6} )}^{3}  = 6,652 \:  {ft}^{3}

To get the percentage of the model relative to the original you divide the models volume by the full scale volume

\frac{6652}{1436755} \times 100  = 0.46

which is less than 1% of the full scale volume.

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4 years ago
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