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Aleksandr-060686 [28]
3 years ago
6

You are working for a company that creates tubes that measure 10 cm long and have a diameter of 6 cm. The tubes need to be cut j

ust right. The company fired all of its math people so they do not know how to calculate the size of the tube needed. Please calculate the surface area needed for the tube. Use 3.14 for pi. *
Mathematics
1 answer:
Nataly_w [17]3 years ago
4 0

Answer:

A= 2\pi (3cm)^2 +1 2\pi (3cm)(10 cm)= 18 \pi + 60\pi = 78 \pi cm^2

And replacing the value of \pi =3.14 we got:

A = 78*3.14 cm^2= 244.92 cm^2

Step-by-step explanation:

For this case we have the long who represent the height 10 cm and the diameter is 6cm. So then the radius is given by:

r = \frac{D}{2}= \frac{6cm}{2}= 3cm

And the surface area for a cylinder is given by this formula:

A= 2\pi r^2 +2 \pi rh

And replacing the info given we got:

A= 2\pi (3cm)^2 +1 2\pi (3cm)(10 cm)= 18 \pi + 60\pi = 78 \pi cm^2

And replacing the value of \pi =3.14 we got:

A = 78*3.14 cm^2= 244.92 cm^2

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Answer:

A

Step-by-step explanation:

6*13+6*7

because 6 in both side

6(13+6)

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3 years ago
Which pair of triangles can be used to show that the slope of line a is the same anywhere along the line?
aalyn [17]

Answer:

C

Step-by-step explanation:

The pair of triangles in option C (image attached below) shows that the slope of line a is the same anywhere on the line.

Let's check it out:

Slope = rise/run

✔️For the first lower triangle:

Rise = 3.5

Run = 4.5

Slope = 3.5/4.5 ≈ 0.8

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Slope = 4/5 = 0.8

Therefore, it shows that the slope of the line is the same anywhere along the line of the graph.

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3 years ago
Please help me find RS, PQ, PS, RQ, and PRQ
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The arc equals the angle so
RS= 150°
PQ= 72°
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8 0
3 years ago
Read 2 more answers
Problem 3
Ilia_Sergeevich [38]

Problem 1 Answer: x=10

<u>Simplify both sides of the equation</u>

<u></u>(2)(x)+(2)(-3)=14(Distribute)<u></u>

<u></u>2x+-6=14<u></u>

<u></u>2x-6=14<u></u>

<u></u>

<u>Add 6 to both sides</u>

2x-6+6=14+6

2x=20

<u></u>

<u>Divide both sides by 2</u>

<u></u>2x/2=20/2<u></u>

<u></u>x=10<u></u>

----------------------------------------------------------------

Problem 2 Answer: x=-7

<u></u>

<u>Simplify both sides of the equation</u>

<u></u>(-5)(x)+(-5)(-1)=40(Distribute)<u></u>

<u></u>-5x+5=40<u></u>

<u></u>

<u>Subtract 5 from both sides</u>

<u></u>-5x+5-5=40-5<u></u>

<u></u>-5x=35<u></u>

<u></u>

<u>Divide both sides by -5</u>

<u></u>-5x/5=35/-5<u></u>

<u></u>x=-7<u></u>

----------------------------------------------------------------

Problem 3 Answer: x=-8

<u></u>

<u>Simplify both sides of the equation</u>

<u></u>(12)(x)+(12)(10)=24(Distribute)<u></u>

<u></u>12x+120=24<u></u>

<u></u>

<u>Subtract 120 from both sides</u>

<u></u>12x+120-120=24-120<u></u>

<u></u>12x=-96<u></u>

<u></u>

<u>Divide both sides by 12</u>

<u></u>12x/12=-96/12<u></u>

<u></u>x=-8<u></u>

----------------------------------------------------------------

Problem 4 Answer: x=-1/2

<u></u>

<u>Move all terms to the left:</u>

<u></u>2(x+6)-(11)=0<u></u>

<u></u>

<u>Multiply parentheses</u>

<u></u>2x+12-11=0<u></u>

<u></u>

<u>Add all the numbers together, and all the variables</u>

<u></u>2x+1=0<u></u>

<u></u>

<u>Move all terms containing x to the left, all other terms to the right</u>

<u></u>2x=-1\\x=-1/2<u></u>

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Problem 5 Answer: x = 44

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