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3 years ago

A student is 7 years old. If you triple the teachers age and add the student's age the answer is 163. How old is the teacher?

Mathematics

2 answers:

pogonyaev3 years ago

4 0

Answer:

Step-by-step explanation:

We can write this out as an equation. Let's say that the teacher's age is x. Triple the teachers age plus the students age is 163, which can be written out as:

$3x + 7 = 163$

We want to isolate the variable, so subtract 7 from both sides. This gives us:

$3x = 156$

Finally, we divide both sides by 3, giving:

$x=52$

So the teacher is 52 years old.

Hope this helps!

asambeis [7]3 years ago

3 0

Answer: The teacher is 52.

Equation: 7+3t=163

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viktelen [127]

Answer:

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Step-by-step explanation:

Given that:

the inverted conical water tank with a height of 20 ft and a radius of 8 ft is drained through a hole in the vertex (bottom) at a rate of 4 ft^3/sec.

We are meant to find the rate of change of the water depth when the water depth is 10 ft.

The diagrammatic expression below clearly interprets the question.

From the image below, assuming h = the depth of the tank at a time t and r = radius of the cone shaped at a time t

Then the similar triangles ΔOCD and ΔOAB is as follows:

$\dfrac{h}{r}= \dfrac{20}{8}$ ( similar triangle property)

$\dfrac{h}{r}= \dfrac{5}{2}$

$\dfrac{h}{r}= 2.5$

h = 2.5r

$r = \dfrac{h}{2.5}$

The volume of the water in the tank is represented by the equation:

$V = \dfrac{1}{3} \pi r^2 h$

$V = \dfrac{1}{3} \pi (\dfrac{h^2}{6.25}) h$

$V = \dfrac{1}{18.75} \pi \ h^3$

The rate of change of the water depth is :

$\dfrac{dv}{dt}= \dfrac{\pi r^2}{6.25}\ \dfrac{dh}{dt}$

Since the water is drained through a hole in the vertex (bottom) at a rate of 4 ft^3/sec

Then,

$\dfrac{dv}{dt}= - 4 \ ft^3/sec$

Therefore,

$-4 = \dfrac{\pi r^2}{6.25}\ \dfrac{dh}{dt}$

the rate of change of the water at depth h = 10 ft is:

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$100 \pi \dfrac{dh}{dt} = -4 \times 6.25$

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Answer:

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Step-by-step explanation:

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