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3 years ago

Twenty-seven is the product of three and a number

Mathematics

2 answers:

n200080 [17]3 years ago

7 0

It’s the product of three and nine.

Ksivusya [100]3 years ago

3 0

This situation can be modeled by an equal.

The word is means equal.

So 27=

The product means multiplication
27 = 3*

Then "a number" is an unknown (we'll call it x)

So the equation is
21=3*x
9=x

The number is 9

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Help I'll give you 100 points

Dvinal [7]

✏ Answer:

$y=-9$

Step-by-step explanation:

$2-\frac{1}{8}y=11+\frac{7}{8}y$

➠ Subtract 2 from both sides:-

 $2-\frac{1}{8}y-2=11+\frac{7}{8}y-2$

$-\frac{1}{8}y=\frac{7}{8}y+9$

➠ Subtract 7/8y from both sides:-

 $-\frac{1}{8}y-\frac{7}{8}y=\frac{7}{8}y+9-\frac{7}{8}y$

$-y=9$

➠ Divide both sides by -1:-

 $\frac{-y}{-1}=\frac{9}{-1}$

$y=-9$

OAmalOHopeO

✎--------------------------

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777dan777 [17]

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3 years ago

To play basketball with her friends, Evangeline needs to pump air in her ball, which is completely deflated. Before inflating it

Nadya [2.5K]

Answer:

ρ_air = 0.15544 kg/m^3

Step-by-step explanation:

Solution:-

- The deflated ball ( no air ) initially weighs:

m1 = 0.615 kg

- The air is pumped into the ball and weight again. The new reading of the ball's weight is:

m2 = 0.624 kg

- The amount of air ( mass of air ) pumped into the ball can be determined from simple arithmetic between inflated and deflated weights of the ball.

m_air = Δm = m2 - m1

m_air = 0.624 - 0.615

m_air = 0.009 kg

- We are to assume that the inflated ball takes a shape of a perfect sphere with radius r = 0.24 m. The volume of the inflated ( air filled ) ball can be determined using the volume of sphere formula:

V_air = 4*π*r^3 / 3

V_air = 4*π*0.24^3 / 3

V_air = 0.05790 m^3

- The density of air ( ρ_air ) is the ratio of mass of air and the volume occupied by air. Expressed as follows:

ρ_air = m_air / V_air

ρ_air = 0.009 / 0.05790

Answer: ρ_air = 0.15544 kg/m^3

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Lemur [1.5K]

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Deffense [45]

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