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2 years ago

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Mathematics

1 answer:

PIT_PIT [208]2 years ago

5 0

The solution to the equation is p = 1/3 and q = undefined

<h3>How to solve the equation?</h3>

The equation is given as:

p^2 - 2qp + 1/q = (p - 1/3)

The best way to solve the above equation is by the use of a graphing calculator i.e. graphically

However, it can be solved algebraically too (to some extent)

Recall that the equation is given as:

p^2 - 2qp + 1/q = (p - 1/3)

Split the equation

So, we have

p^2 - 2qp + 1/q = 0

p - 1/3 = 0

Solve for p in p - 1/3 = 0

p = 1/3

Substitute p = 1/3 in p^2 - 2qp + 1/q = 0

So, we have

(1/3)^2 - 2q(1/3) + 1/q = 0

This gives

1/9 - 2/3q + 1/q = 0

This gives

2/3q + 1/q = -1/9

Multiply though by q

So, we have

2/3q^2 + 1 = -1/9q

Multiply through by 9

6q^2 + 9 = -q

So, we have

6q^2 + q + 9 = 0

Using the graphing calculator, we have

q = undefined

Hence. the solution to the equation is p = 1/3 and q = undefined

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A box has a width of 10 cm and a length of 17 cm. The volume of the box is decreasing at a rate of 527 cubic cm per minute, with

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Answer:

$-3.1$ cm per minute.

Step-by-step explanation:

We have been given that a box has a width of 10 cm and a length of 17 cm. The volume of the box is decreasing at a rate of 527 cubic cm per minute, with the width and length being held constant.

We know that volume of a cuboid is length times width times height.

$V=lwh$

Upon substituting our given width and length, we will get:

$V=17\cdot 10\cdot h$

$V=170\cdot h$

Now, we will find derivative of volume with respect to time as:

$\frac{dV}{dt}=170\cdot \frac{dh}{dt}$

Since the volume of the box is decreasing at a rate of 527 cubic cm per minute, so we will substitute $\frac{dV}{dt}=-527$ as:

$-527=170\cdot \frac{dh}{dt}$

$\frac{-527}{170}=\frac{170\cdot \frac{dh}{dt}}{170}$

$\frac{dh}{dt}=-3.1$

Therefore, the rate of change in height is $-3.1$ cm per minute.

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