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3 years ago

10

4 1/3ft 12 ft 9 ft what is the volume

1 answer:

mina [271]3 years ago

8 0

Just multiply all them together for volume, 468 ft ^3

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WORTH 30 POINTS!!!! HELP ASAP!!!!!<br><br> Find 121/2% of 720. Show work please.

DedPeter [7]

121/2 = 720/p
720*2= 121p
1440=121p
divide by 121
=12%

8 0

3 years ago

Solve the system of equations algebraically using substitution and elimination

GrogVix [38]

Answer:

my answer is 2x+ 3y=22 because that the answer

3 0

2 years ago

Please help me answer for 35 and 36

dexar [7]

Answer:

35. $f(2)=16$

36. $f(2)=77$

Step-by-step explanation:

Question 35:

Given:

$f(1)=32$

$f(n)=f(n-1)\times \frac{1}{2}$

Plug in 2 for $n$ and simplify. This gives,

$f(2)=f(2-1)\times \frac{1}{2}\\f(2)=f(1)\times \frac{1}{2}\\f(2)=32\times \frac{1}{2}\\f(2)=16$

Therefore, term 2 is 16.

Question 36:

Given:

$f(1)=11$

$f(n)=f(n-1)\times 7$

Plug in 2 for $n$ and simplify. This gives,

$f(2)=f(2-1)\times 7\\f(2)=f(1)\times 7\\f(2)=11\times 7\\f(2)=77$

Therefore, term 2 is 77.

8 0

3 years ago

One container is filled with mixture that is 25% acid. A second container is filled with a mixture that is 55% acid. The second

algol13

Answer:

Therefore the third container contains $\frac{735}{17}\%$ = 43.23 % acid.

Step-by-step explanation:

Given that, One container is filled with the mixture that 25% acid. 55% acid is contain by second container.

Let the volume of first container be x cubic unit.

Since the volume of second container is 55% larger than the first.

Then the volume of the second container is

$= (V\times \frac{100+55}{100})$ cubic unit.

$=\frac{155}{100}V$ cubic unit.

The amount of acid in first container is

$=V\times \frac{25}{100}$ cubic unit.

$=\frac{25}{100}V$ cubic unit.

The amount of acid in second container is

$=(\frac{155}{100}V \times \frac{55}{100})$ cubic unit.

$=\frac{8525}{10000}V$ cubic unit.

Total amount of acid $=(\frac{25}{100}V+\frac{8525}{10000}V)$ cubic unit.

$=(\frac{2500+8525}{10000}V)$ cubic unit.

$=(\frac{11025}{10000}V)$ cubic unit.

Total volume of mixture $=(V+\frac{155}{100}V)$ cubic unit.

$=\frac{100+155}{100}V$ cubic unit.

$=\frac{255}{100}V$ cubic unit.

The amount of acid in the mixture is

$=\frac{\textrm{The volume of acid}}{\textrm{The amount of mixture}}\times 100 \%$

$=\frac {\frac{11025}{10000}V}{\frac{255}{100}V}\times 100\%$

$=\frac{735}{17}\%$

Therefore the third container contains $\frac{735}{17}\%$ acid.

5 0

3 years ago

Can someone help me with this problem?

Diano4ka-milaya [45]

Can you show the answer choices but I believe is 18

3 0

2 years ago

Read 2 more answers