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3 years ago

−5+(−5r)+10 answer plz

Mathematics

2 answers:

nata0808 [166]3 years ago

4 0

Answer:

-5r + 5

Step-by-step explanation:

We simply simplify by adding like terms. Only thing we can combine are the constants -5 and 10, which we get 5. So then our answer is -5r + 5.

Drupady [299]3 years ago

3 0

Answer:

5(1-r)

Simplified expression 5-5r

Step-by-step explanation:

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if an area can be washed at a rate of 4,900 cm2/ minute, how many square inches can be washed per hour?

Olenka [21]

Answer : 4.5 × 10⁴ square inches can be washed per hour.

Step-by-step explanation :

As we are given that an area can be washed at a rate of 4,900 cm²/min. Now we have to determine the square inches can be washed per hour.

Given conversions are:

1 cm = 0.39 in and 1 hr = 60 min

As, 1 cm² = (0.39)² in² and 1 min = 1/60 hr

So, $1cm^2/min=(0.39)^2\times 60in^2/hr$

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Now we have to determine the square inches can be washed per hour.

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Therefore, 4.5 × 10⁴ square inches can be washed per hour.

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3 years ago

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

Basile [38]

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× $2\pi rh$ )

$=\$2.5 \pi r h$

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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