MATHEMATICS COLLEGE

A cylindrical can without a top is made to contain 25 3 cm of liquid. What are the dimensions of the can that will minimize the cost to make the can if the metal for the sides will cost $1.25 per 2 cm and the metal for the bottom will cost $2.00 per 2 cm ?

Answers

Answer 1

Answer:

Answer:

Therefore the radius of the can is 1.71 cm and height of the can is 2.72 cm.

Step-by-step explanation:

Given that, the volume of cylindrical can with out top is 25 cm³.

Consider the height of the can be h and radius be r.

The volume of the can is V= $\pi r^2h$

According to the problem,

$\pi r^2 h=25$

$\Rightarrow h=(25)/(\pi r^2)$

The surface area of the base of the can is = $\pi r^2$

The metal for the bottom will cost $2.00 per cm²

The metal cost for the base is =$(2.00× $\pi r^2$ )

The lateral surface area of the can is = $2\pi rh$

The metal for the side will cost $1.25 per cm²

The metal cost for the base is =$(1.25× $2\pi rh$ )

$=\$2.5 \pi r h$

Total cost of metal is C= 2.00 $\pi r^2$ + $2.5 \pi r h$

Putting $h=(25)/(\pi r^2)$

$\therefore C=2\pi r^2+2.5 \pi r * (25)/(\pi r^2)$

$\Rightarrow C=2\pi r^2+ (62.5)/( r)$

Differentiating with respect to r

$C'=4\pi r- (62.5)/( r^2)$

Again differentiating with respect to r

$C''=4\pi + (125)/( r^3)$

To find the minimize cost, we set C'=0

$4\pi r- (62.5)/( r^2)=0$

$\Rightarrow 4\pi r=(62.5)/( r^2)$

$\Rightarrow r^3=(62.5)/( 4\pi)$

⇒r=1.71

Now,

$\left C''\right|_(x=1.71)=4\pi +(125)/(1.71^3)>0$

When r=1.71 cm, the metal cost will be minimum.

Therefore,

$h=(25)/(\pi* 1.71^2)$

⇒h=2.72 cm

Therefore the radius of the can is 1.71 cm and height of the can is 2.72 cm.

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Answers

Answer:

$y=mx+(y_1-mx_1), b=y_1-mx_1$

Step-by-step explanation:

The point slope form of the equation of a line is given as:

$y-y_1=m(x-x_1)$

The slope-intercept form of the equation of a line is given as:

$y=mx+b$

where: m=slope, b=y-intercept.

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Step 1: Distribute the right hand side

$y-y_1=m(x-x_1)\ny-y_1=mx-mx_1$

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$y=mx-mx_1+y_1\ny=mx+(y_1-mx_1)$

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A point is on a circle of the distance form the center from the center of the circle to the the point is equal to the

Answers

Answer:

The Radius?

.......... ....

Answer:

C. Radius

Step-by-step explanation:

Have a good day!

If b^2=a then b is what of a?

Answers

Answer:

$\huge\boxed{b = √(a)}$

Step-by-step explanation:

If we have an equation $b^2 = a$ and we want to find what b is in relation to a, we can change the equation so that we have b on one side and whatever is on the other side is what b is.

$b^2 = a$

To isolate b, we can take the square root of both sides as taking the square root of something squared results in the base.

$√(b^2) = √(a)$

$b = √(a)$

So b is the square root of a.

Hope this helped!

1:: If
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The time that it takes a randomly selected job applicant to perform a certain task has a distribution that can be approximated by a normal distribution with a mean value of 145 sec and a standard deviation of 25 sec. The fastest 10% are to be given advanced training. What task times qualify individuals for such training? (Round the answer to one decimal place.)

Answers

Answer:

A task time of 177.125s qualify individuals for such training.

Step-by-step explanation:

Problems of normally distributed samples can be solved using the z-score formula.

In a set with mean $\mu$ and standard deviation $\sigma$ , the zscore of a measure X is given by

$Z = (X - \mu)/(\sigma)$

After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X. Subtracting 1 by the pvalue, we This p-value is the probability that the value of the measure is greater than X.

In this problem, we have that:

A distribution that can be approximated by a normal distribution with a mean value of 145 sec and a standard deviation of 25 sec, so $\mu = 145, \sigma = 25$ .

The fastest 10% are to be given advanced training. What task times qualify individuals for such training?

This is the value of X when Z has a pvalue of 0.90.

Z has a pvalue of 0.90 when it is between 1.28 and 1.29. So we want to find X when $Z = 1.285$ .

$Z = (X - \mu)/(\sigma)$

$1.285 = (X - 145)/(25)$

$X - 145 = 32.125$

$X = 177.125$

A task time of 177.125s qualify individuals for such training.

A line and the point (3,4) are graphed in the coordinate plane. what is the equation of the line that passes through the point (3,4) and is parallel to the line shownA. y= 1/4x+3
B. y=1/4x+13/4
C. y=4x+3
D.y=4x+13/4

Answers

Answer:

B. y=1/4x+13/4

Step-by-step explanation:

Find the surface area of a cylinder with a radius of 5 cm and a height of 30 cm. Use 3.14

Answers

Whole formula 2*pi*radius^2 + 2*pi*radius*height:

157+942= 1099

SA= 1099

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