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3 years ago

What is the surface area of the right cylinder below?

Mathematics

1 answer:

attashe74 [19]3 years ago

5 0

Answer: C

Step by step explanation:

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A certain project can be completed by 28 men in 90 days. the numbers of men needed varies inversely to the time needed to comple

leva [86]

ANSWER

C. 35 men

EXPLANATION

Let m represent the number of men needed and d represent the number of days.

From the question,the number of men needed varies inversely to the time needed to complete the project.

We can write the inverse variation equation.

$m = \frac{k}{d}$

where k is the constant of variation.

When m=28, d =90.

We substitute these values into the variation equation to determine the value of k.

$28 = \frac{k}{90}$

$k = 90 \times 28$

$k = 252$

Now the equation becomes:

$m= \frac{2520}{d}$

When d=72,

$m = \frac{2520}{72}$

$m = 35$

Therefore he needs to have 35 men working.

3 0

4 years ago

Solve the equation. +8+3x=X-6 Ox=-18 Ox=-14 Ox=2 Ox=4

professor190 [17]

Answer:

x=-7

Step-by-step explanation:

$8+3x=x-6\\\\8+3x+6=x-6+6\\\\8+3x+6=x+0\\\\(8+6)+3x=x\\\\14+3x=x\\\\14+3x-3x=x-3x\\\\14+0=x-3x\\\\14=(1-3)x\\\\14=-2x\\\\\frac{1}{2}(14)=\frac{1}{2}(-2x)\\\\7=-x\\\\-(7)=-(-x)\\\\-7=x\\\\x=-7$

7 0

3 years ago

The toll on a highway is based on the number of miles driven. If the toll on a 15-mile stretch of highway is $0.60, find the cos

Crank

Answer:

$3.20

Step-by-step explanation:

We have

15-mile stretch = $0.60

80-mile stretch = x

Cross Multiply

15 × x = 80 × $0.60

x = 80 × $0.60/15

x = $3.20

Hence, the cost to drive a 80-mile stretch is $3.20

8 0

3 years ago

*need this answered ASAP please. Write the solution to the given inequality in interval notation.

belka [17]

Step-by-step explanation:

this is the right answer..hope it helped you .

5 0

3 years ago

Read 2 more answers

Solve for g. what is g equal to?

sweet [91]

Answer:

$\displaystyle g=\frac{4\pi^2l}{T^2}$

Step-by-step explanation:

Equation Solving

Given the equation

$\displaystyle l=\frac{gT^2}{4\pi^2}$

It's required to solve it for g. Let's swap sides of the equation:

$\displaystyle \frac{gT^2}{4\pi^2}=l$

Multiply by $4\pi^2$ to remove the denominator:

$\displaystyle gT^2=4\pi^2l$

Divide by $T^2$

$\boxed{\displaystyle g=\frac{4\pi^2l}{T^2}}$

4 0

3 years ago