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Gnesinka [82]
2 years ago
8

The area of a second soup can’s base is 5 pi inches squared. The can has a height of 10 inches. How would you use that informati

on to find how much soup the can will hold? Explain.
Mathematics
2 answers:
Ilia_Sergeevich [38]2 years ago
8 0

Answer:

The amount of soup the can will hold is;

50π inches cubed = 157.08 inches cubed

Step-by-step explanation:

The amount of soup the can will hold is equal to the volume of the can.

Volume of the can = base area × height

Given;

Base Area = 5π inches squared

height = 10 inches

Volume of can = 5π × 10 = 50π inches cubed

The amount of soup the can will hold is;

50π inches cubed = 157.08 inches cubed

kenny6666 [7]2 years ago
8 0

Answer: 157.10 cubic inches

Step-by-step explanation:

Hi, to answer this question we have to calculate the volume of the can,.

Since the can is a cylinder we can apply the next formula:

Volume: π x radius ²x height  

Since = π x radius ² = area of the base of the cylinder

Replacing with the values given:

V = are of the base x height

V = 5 pi inch2 x 10 inch

V = 5πx10 = 157.10 cubic inches

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Answer:

3.5 cm

Step-by-step explanation:

6x-1=4x+6

6x-4x=1+6

2x=7

x=7÷2

x=3.5

I hope my answer will help you and i guess it's correct

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2 years ago
What does 5x evaluate to if x is equal to 2?
stiks02 [169]

Answer: Hi!

Since 5 is being multiplied by x, and x is equal to 2, be would multiply 5 and 2. 5 * 2 is equal to 10.

Hope this helps!

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Express 3/4 in sixty-fourths
vesna_86 [32]
It would be 48/64 this is done by finding how many times 4 fits in 64 (64/4) then multiplying 3 by that answer
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Julie rides her bike from the sports complex to the school. Then she rides from the school to the mall, and then on to the libra
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Between the two persons presented above, Julie had ridden farther.  

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Jane wishes to bake an apple pie for dessert. The baking instructions say that she should bake the pie in an oven at a constant
Viktor [21]

Answer:

Therefore k= \frac{ln2 }{18}, A=184

Step-by-step explanation:

Given function is

T(t)=230 -e^{-kt}

where T(t) is the temperature in °C and t is time in minute and A and k are constants.

She noticed that after 18 minutes the temperature of the pie is 138°C

Putting T(t) =138°C and t= 18 minutes

138=230 -Ae^{-k\times 18}

\Rightarrow  -Ae^{-18k}=138-230

\Rightarrow  Ae^{-18k}=92 .....(1)

Again after 36 minutes it is 184°C

Putting T(t) =184°C and t= 36 minutes

184=230-Ae^{-k\times 36}

\Rightarrow Ae^{-36k}=230-184

\Rightarrow Ae^{-36k}=46.......(2)

Dividing (2) by (1)

\frac{Ae^{-36k}}{Ae^{-18k}}=\frac{46}{92}

\Rightarrow e^{-18k}=\frac{46}{92}

Taking ln both sides

ln e^{-18k}=ln\frac{46}{92}

\Rightarrow -18k =ln (\frac12)

\Rightarrow -18k= ln1-ln2

\Rightarrow k= \frac{ln2 }{18}

Putting the value k in equation (1)

Ae^{-18\frac{ln2}{18}}=92

\Rightarrow A e^{ln2^{-1}}=92

\Rightarrow A.2^{-1}=92

\Rightarrow \frac{A}{2}=92

\Rightarrow A= 92 \times 2

⇒A= 184.

Therefore k= \frac{ln2 }{18}, A=184

7 0
2 years ago
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