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1 year ago

Factor the given expression by determining any common monomial factorsthat may exist 6k2m - 3k + 9km2 - 12k2m2

Mathematics

1 answer:

Varvara68 [4.7K]1 year ago

3 0

ANSWER

$=3k\left(3m^2+2km-4km^2-1\right)$

EXPLANATION

Given;

$6k^2m-3k+9km^2-12k^2m^2$

The common factor is 3k;

Hence, factor out the common term;

$\begin{gathered} =3k\left(2mk-1+3m^2-4m^2k\right) \\ =3k\left(3m^2+2km-4km^2-1\right) \end{gathered}$

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ANSWER

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Chloe and Libby want to tie Ryan to a bunch of approximately spherical helium balloons of diameter 0.3m. The volume of each ball

dem82 [27]

Answer: Aproximately 2,525 balloons

Step-by-step explanation:

1. Find the volume of a balloon with the formula given in the problem, where $r$ is the radius ( $r=\frac{0.3m}{2}=0.15m$ ), then:

$V=4(0.15m)^{3}=0.0135m^{3}$

2. Convert the volume from m³ to lliters by multiplying it by 1,000:

$V=0.0135m^{3}*1000=13.5L$

3. You know that that 1 liter of helium can lift 1 gram and that Ryan weighs 5 stone and 5 pounds. So you must make the following conversions:

1 g=0.0022 lb

From 5 stones to pounds

$(5stone)(\frac{14pounds}{1stone})=70pounds=70lbs$

4. Then Ryan's weigh is:

5lb+70lb=75lb

5. Then, if 1 liter of helium can lift 0.0022 lb, to lift 75 lb (which is the weight of Ryan) they need:

$\frac{75lb*1L}{0.0022lb}=34,090.90L$

6. Then, to calculate the aproximated number of balloons they need to make him float (which can call $n$ ), you must divide the liters of helium needed to lift the weight of Ryan by the volume of a balloon, then the result is:

$n=\frac{34,090.90L}{13.5L}$

$n=2,525.25$ ≈2,525 balloons.

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4 years ago