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vichka [17]
3 years ago
15

Find the radius of a cylinder with a volume 686mm2 and a height of 14mm​

Mathematics
1 answer:
shtirl [24]3 years ago
8 0

Answer:

r = 3.95 mm (nearest hundredth)

Step-by-step explanation:

\textsf{volume of a cylinder} = \pi r^2h

(where r = radius and h = height)

Given, volume = 686 mm² and height = 14 mm:

686=\pi r^2\times14

\implies\pi r^2=686 \div14=49

\implies r^2=49 \div \pi

\implies r=\sqrt{49 \div \pi }=3.949327085...

\implies r=3.95 \ \textsf{mm}

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3 years ago
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How to convert 2500mcg to mg
Volgvan

Answer:2500mcg would be 2.5mg

Step-by-step explanation:

1000mcg = 1mg so 2500mcg = 2.5mg

6 0
3 years ago
How many liters each of a 35% acid solution and a 80% acid solution must be used to produce 60 liters of a 65% acid solution? (R
serg [7]

Solving a system of equations we will see that we need to use <u>40 liters of the 80% acid solution</u>, and the other <u>20 liters are of the 35% acid solution</u>.

<h3>How many liters of each solution do we need to use?</h3>

First, we need to define the variables:

  • x = liters of the 35% acid used.
  • y = liters of the 80% acid used.

We know that we want to produce 60 liters of 65% acid, then we have the system of equations:

x + y = 60

x*0.35 + y*0.80 = 60*0.65

(in the second equation we wrote the percentages in decimal form).

To solve this we need to isolate one of the variables in one equation and then replace it in other one, isolating x we get:

x = 60 - y

Replacing that in the other equation:

(60 - y)*0.35 + y*0.80 = 60*0.65

y*(0.80 - 0.35) = 60*(0.65 - 0.35)

y*0.45 = 60*0.30

y = 60*0.30/0.45 = 40

So we need to use <u>40 liters of the 80% acid solution</u>, and the other <u>20 liters are of the 35% acid solution</u>.

If you want to learn more about systems of equations:

brainly.com/question/13729904

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7 0
2 years ago
In Exercises 26 and 27 write an equation of the line with the given slope and y-intercept.
Neko [114]
26. y= -7/3+ 0 or just y= -7/3

27. y= -10
7 0
3 years ago
The population of Henderson City was 3,381,000 in 1994, and is growing at an annual rate 1.8%
liq [111]
<h2>In the year 2000, population will be 3,762,979 approximately. Population will double by the year 2033.</h2>

Step-by-step explanation:

   Given that the population grows every year at the same rate( 1.8% ), we can model the population similar to a compound Interest problem.

   From 1994, every subsequent year the new population is obtained by multiplying the previous years' population by \frac{100+1.8}{100} = \frac{101.8}{100}.

   So, the population in the year t can be given by P(t)=3,381,000\textrm{x}(\frac{101.8}{100})^{(t-1994)}

   Population in the year 2000 = 3,381,000\textrm{x}(\frac{101.8}{100})^{6}=3,762,979.38

Population in year 2000 = 3,762,979

   Let us assume population doubles by year y.

2\textrm{x}(3,381,000)=(3,381,000)\textrm{x}(\frac{101.8}{100})^{(y-1994)}

log_{10}2=(y-1994)log_{10}(\frac{101.8}{100})

y-1994=\frac{log_{10}2}{log_{10}1.018}=38.8537

y≈2033

∴ By 2033, the population doubles.

4 0
3 years ago
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