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aev [14]
3 years ago
15

The equation r=3c +5 represents the values shown in the table below C- 6, 8, 12, 18 R- 23, 29, ?, 59 What is the missing value i

n the table?

Mathematics
2 answers:
Kipish [7]3 years ago
4 0

C: 41 is the missing value in the table

almond37 [142]3 years ago
3 0
The missing value is 53.

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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
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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=\frac{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)

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

Putting h=\frac{25}{\pi r^2}

\therefore C=2\pi r^2+2.5 \pi r \times \frac{25}{\pi r^2}

\Rightarrow C=2\pi r^2+ \frac{62.5}{ r}

Differentiating with respect to r

C'=4\pi r- \frac{62.5}{ r^2}

Again differentiating with respect to r

C''=4\pi + \frac{125}{ r^3}

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

4\pi r- \frac{62.5}{ r^2}=0

\Rightarrow 4\pi r=\frac{62.5}{ r^2}

\Rightarrow  r^3=\frac{62.5}{ 4\pi}

⇒r=1.71

Now,

\left C''\right|_{x=1.71}=4\pi +\frac{125}{1.71^3}>0

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

Therefore,

h=\frac{25}{\pi\times 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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Given information

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When translated 4 units down, we got (5, -4)

Step 2: Reflect over the y-axis

The general rule for reflecting over the y-axis is (-x, y)

This means the value of x will be negated and the value of y will remain the same

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Step 3: the graph the point

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