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2 years ago

Please just help me simplify these NOT SOLVE THEM IF AT ALL POSSIBLE! THANKS!!!

Mathematics

1 answer:

alexandr402 [8]2 years ago

3 0

5th question: $\frac{-8-4a}{-4} =2 + a$

6th question: $11(-4q +3d + 3) = -44q +33d+33$

7th question: $\frac{15w +5-35w}{-5} =-3w-1+7w=4w-1$

8th question: $4(y-1)+9(y+2) = 4y-4 + 9y+18= 13y+14$

Hope it helps!

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2 years ago

The average density of the material in intergalactic space is approximately 2.5 × 10–27 kg/m3. what is the volume of a lead samp

Lesechka [4]

Answer:

e. $1.8\times 10^{-6}m^3$

Step-by-step explanation:

It is given that,

The density of intergalactic space material is $2.5 \times 10^{-27}$ kg per cubic meter.

And the volume of intergalactic space material is $8.0\times 10^{24} m^3$

So the mass of that much intergalactic space material is,

$m= \rho\times v$ , where 'm' is the mass, and 'v' is the volume.

Putting the values we get,

$m=2.5 \times 10^{-27}\times 8.0 \times 10^{24}$

$m=20\times 10^{(-27+24)}=20\times 10^{-3} kg$

It is also given that the mass of lead is the same as the mass of the intergalactic space material. Therefore, the mass of lead is $20 \times 10^{-3}=2 \times 10^{-2}kg$

So the volume of lead is,

$v=\frac{m}{\rho} =\frac{2 \times 10^{-2}}{11300} = 0.00000177 m^3$

$v=1.77 \times 10^{-6}=1.8\times 10^{-6}m^3$

So the volume of lead is $v=1.8\times 10^{-6}m^3$ .

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2 years ago

Please just help me simplify these NOT SOLVE THEM IF AT ALL POSSIBLE! THANKS!!!​

Please just help me simplify these NOT SOLVE THEM IF AT ALL POSSIBLE! THANKS!!!