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3 years ago

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An oxygen tank contained 212 12/3 liters of oxygen before 27 1/3 liters were used. If the tank can hold 240 3/8 liters, how much

space in the tank is unused

1 answer:

lukranit [14]3 years ago

5 0

So, initially 240 $\frac{3}{8}$ -212 $\frac{12}{3}$ were empty:

this is :

$240 \frac{3}{8} - 212 \frac{12}{3} = 240 \frac{3}{8} - 212 +4 = 240 \frac{3}{8} - 216= 24 \frac{3}{8}$

12/4 is 4 so that's why i substituted it with 4.

now, later 27 1/3 were used so we add this tho the original empty space

$24 \frac{3}{8}+ 27 \frac{1}{3}$ = $24 \frac{9}{24}+ 27 \frac{8}{24}$ = $51 \frac{17}{24}$

which is the result!

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Answer:

The answer is 64.

16-12=4.

4×16×1=64.

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A bag hold 1 blue block and 3 yellow block. How many green Blocks should be added to the bag to makr the probability of randomly

Natasha_Volkova [10]

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3 years ago

Select the correct answer.

sp2606 [1]

The perimeter of the pentagon with apothem of 5cm is 28.87cm

<h3>How to calculate the perimeter of a pentagon</h3>

The sum of all the external sides of the pentagon is its perimeter.

First we need to calculate the half of one of its sides using the SOH CAH TOA

tan 30 = x/5

x = 5 tan 30

x = 2.89

One of the sides = 2(2.89) = 5.77 cm

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Hence the perimeter of the pentagon with apothem of 5cm is 28.87cm

Learn more on perimeter of pentagon here: brainly.com/question/11484726

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2 years ago

An airplane is traveling at 625 miles per hour. At this rate, about how long will it take the plane to travel 3250 miles?

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determine the value of m and n for mx^3 +20x^2+ nx-36 when divided by (x+1) gives remainder 0 and when divided by (x-2) gives re

soldi70 [24.7K]

Given:

The given polynomial is

$mx^3+20x^2+nx-36$

When it divided by (x+1) gives remainder 0 and when divided by (x-2) gives remainder of 45.

To find:

The values of m and n.

Solution:

According to the remainder theorem, if a polynomial P(x) is divided by (x-c), then the remainder is P(c).

Let the given polynomial be P(x).

$P(x)=mx^3+20x^2+nx-36$

When it divided by (x+1) gives remainder 0. So, P(-1)=0.

$P(-1)=m(-1)^3+20(-1)^2+n(-1)-36$

$0=-m+20-n-36$

$0=-m-n-16$

$m+n=-16$ ...(i)

When P(x) is divided by (x-2) gives remainder of 45. So, P(2)=0

$m(2)^3+20(2)^2+n(2)-36=45$

$8m+80+2n-36=45$

$8m+2n+44=45$

$8m+2n=45-44$

$8m+2n=1$ ...(ii)

Multiply 2 on both sides of (i).

$2m+2n=-32$ ...(iii)

Subtract (iii) from (ii).

$6m=33$

$m=\dfrac{33}{6}$

$m=5.5$

Putting m=5.5 in (i), we get

$5.5+n=-16$

$n=-16-5.5$

$n=-21.5$

Therefore, the values of m and n are 5.5 and -21.5 respectively.

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3 years ago