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3 years ago

5

13. I have 22 nickels, dimes and quarters worth $3.45. There are twice as many quarters as nickels. Find the number of each coin

.

1 answer:

damaskus [11]3 years ago

6 0

Answer:

12 nickels 4 dimes 3 quaters

Step-by-step explanation:

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You have a wire that is 20 cm long. You wish to cut it into two pieces. One piece will be bent into the shape of a square. The o

Aleksandr [31]

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Therefore the circumference of the circle is $=\frac{20\pi}{4+\pi}$

Step-by-step explanation:

Let the side of the square be s

and the radius of the circle be r

The perimeter of the square is = 4s

The circumference of the circle is =2πr

Given that the length of the wire is 20 cm.

According to the problem,

4s + 2πr =20

⇒2s+πr =10

$\Rightarrow s=\frac{10-\pi r}{2}$

The area of the circle is = πr²

The area of the square is = s²

A represent the total area of the square and circle.

A=πr²+s²

Putting the value of s

$A=\pi r^2+ (\frac{10-\pi r}{2})^2$

$\Rightarrow A= \pi r^2+(\frac{10}{2})^2-2.\frac{10}{2}.\frac{\pi r}{2}+ (\frac{\pi r}{2})^2$

$\Rightarrow A=\pi r^2 +25-5 \pi r +\frac{\pi^2r^2}{4}$

$\Rightarrow A=\pi r^2\frac{4+\pi}{4} -5\pi r +25$

For maximum or minimum $\frac{dA}{dr}=0$

Differentiating with respect to r

$\frac{dA}{dr}= \frac{2\pi r(4+\pi)}{4} -5\pi$

Again differentiating with respect to r

$\frac{d^2A}{dr^2}=\frac{2\pi (4+\pi)}{4}$ > 0

For maximum or minimum

$\frac{dA}{dr}=0$

$\Rightarrow \frac{2\pi r(4+\pi)}{4} -5\pi=0$

$\Rightarrow r = \frac{10\pi }{\pi(4+\pi)}$

$\Rightarrow r=\frac{10}{4+\pi}$

$\frac{d^2A}{dr^2}|_{ r=\frac{10}{4+\pi}}=\frac{2\pi (4+\pi)}{4}>0$

Therefore at $r=\frac{10}{4+\pi}$ , A is minimum.

Therefore the circumference of the circle is

$=2 \pi \frac{10}{4+\pi}$

$=\frac{20\pi}{4+\pi}$

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3 years ago