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11111nata11111 [884]

3 years ago

14

The volume of a gas in a container varies inversely with the pressure on the gas. A container of helium has a volume of 370in3 u

nder a pressure of 15psi (pounds per square inch). Write the equation that relates the volume, V, to the pressure, P. What would be the volume of this gas if the pressure was increased to 25psi?

1 answer:

Elan Coil [88]3 years ago

5 0

Answer:

Step-by-step explanation:

When two variables vary inversely, it means that an increase in one would lead to a decrease in the other and vice versa. Since the volume of a gas, V in a container varies inversely with the pressure on the gas, P, if we introduce a constant of proportionality, k, the expression would be

V = k/P

If V = 370 in³ and P = 15psi, then

370 = k/15

k = 370 × 15 = 5550

The equation that relates the volume, V, to the pressure, P would be

V = 5550/P

if the pressure was increased to 25psi, the volume would be

V = 5550/25 = 222 in³

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TIMED PLEASE HURRY HELP WITH 2 EASY QUESTIONS

Alex_Xolod [135]

<h2> Question # 1</h2><h2>Which statements are true?</h2><h2 /><h3><u>Analyzing and solving the first statement:</u></h3>

$4g^2-g=g^2\left(4-g\right)$

Solving the expression

$4g^2-g$

$\mathrm{Apply\:exponent\:rule}:\quad \:a^{b+c}=a^ba^c$

$g^2=gg$

So,

$4gg-g$

$\mathrm{Factor\:out\:common\:term\:}g$

$g\left(4g-1\right)$

So,

$4g^2-g:\quad g\left(4g-1\right)$

Therefore, the statement $4g^2-g=g^2\left(4-g\right)$ is NOT CORRECT.

<h3><u>Analyzing and solving the second statement:</u></h3>

$35g^5-25g^2=\:5g^2\left(7g^3-5\right)$

Solving the expression

$35g^5-25g^2$

$\mathrm{Apply\:exponent\:rule}:\quad \:a^{b+c}=a^ba^c$

$g^5=g^3g^2$

So,

$35g^3g^2-25g^2$

$\mathrm{Rewrite\:}25\mathrm{\:as\:}5\cdot \:5$

$\mathrm{Rewrite\:}35\mathrm{\:as\:}5\cdot \:7$

$5\cdot \:7g^3g^2-5\cdot \:5g^2$

$\mathrm{Factor\:out\:common\:term\:}5g^2$

$5g^2\left(7g^3-5\right)$

So,

$35g^5-25g^2=\:5g^2\left(7g^3-5\right)$

Therefore, the statement $35g^5-25g^2=\:5g^2\left(7g^3-5\right)$ is CORRECT.

<h3><u>Analyzing and solving the third statement:</u></h3>

$24g^4+18g^2=\:6g^2\left(4g^2+3g\right)$

<h3 />

Solving the expression

<h3> $24g^4+18g^2$ </h3><h3> $24g^2g^2+18g^2$ </h3><h3> $\mathrm{Rewrite\:}18\mathrm{\:as\:}6\cdot \:3$ </h3><h3> $\mathrm{Rewrite\:}24\mathrm{\:as\:}6\cdot \:4$ </h3><h3> $6\cdot \:4g^2g^2+6\cdot \:3g^2$ </h3><h3> $\mathrm{Factor\:out\:common\:term\:}6g^2$ </h3><h3> $6g^2\left(4g^2+3\right)$ </h3>

So,

<h3> $24g^4+18g^2=6g^2\left(4g^2+3\right)$ </h3>

Therefore, the statement $24g^4+18g^2=\:6g^2\left(4g^2+3g\right)$ is CORRECT.

<h3><u>Analyzing and solving the fourth statement:</u></h3>

$9g^3+12=\:3\left(3g^3+4\right)$

Solving the expression

$9g^3+12$

$\mathrm{Rewrite\:}12\mathrm{\:as\:}3\cdot \:4$

$\mathrm{Rewrite\:}9\mathrm{\:as\:}3\cdot \:3$

$3\cdot \:3g^3+3\cdot \:4$

$\mathrm{Factor\:out\:common\:term\:}3$

$3\left(3g^3+4\right)$

So,

$9g^3+12=\:3\left(3g^3+4\right)$

Therefore, the statement $9g^3+12=\:3\left(3g^3+4\right)$ is CORRECT.

<h2> Question # 2</h2><h2>Which expressions are completely factored?</h2>

<u>Solving first expression</u>

Considering the expression

$30a^6-24a^2$

$30a^6-24a^2$

$30a^4a^2-24a^2$

$\mathrm{Rewrite\:}24\mathrm{\:as\:}6\cdot \:4$

$\mathrm{Rewrite\:}30\mathrm{\:as\:}6\cdot \:5$

$6\cdot \:5a^4a^2-6\cdot \:4a^2$

$\mathrm{Factor\:out\:common\:term\:}3a^2$

$3a^2\left(10a^4-8\right)$

Thus, the expression $30a^6-24a^2=3a^2\left(10a^4-8\right)\:$ is completely factored.

<u>Solving second expression</u>

Considering the expression

$12a^3-8a$

$12a^3-8a$

$\mathrm{Apply\:exponent\:rule}:\quad \:a^{b+c}=a^ba^c$

$a^3=a^2a$

So,

$12a^2a-8a$

$\mathrm{Rewrite\:}8\mathrm{\:as\:}4\cdot \:2$

$\mathrm{Rewrite\:}12\mathrm{\:as\:}4\cdot \:3$

$4\cdot \:3a^2a-4\cdot \:2a$

$\mathrm{Factor\:out\:common\:term\:}4$

$4\left(3a^3-2a\right)$

Thus, the expression $12a^3-8a=\:4\left(3a^3-2a\right)$ is completely factored.

<u>Solving third expression</u>

$16a^5-20a^3\:\:\:\:\:\:\:\:\:\:\:\:$

$16a^5-20a^3$

$\mathrm{Apply\:exponent\:rule}:\quad \:a^{b+c}=a^ba^c$

$a^5=a^2a^3$

So,

$16a^2a^3-20a^3$

$\mathrm{Rewrite\:}20\mathrm{\:as\:}4\cdot \:5$

$\mathrm{Rewrite\:}16\mathrm{\:as\:}4\cdot \:4$

$4\cdot \:4a^2a^3-4\cdot \:5a^3$

$\mathrm{Factor\:out\:common\:term\:}4a^3$

$4a^3\left(4a^2-5\right)$

Thus, the expression $16a^5-20a^3\:=4a^3\left(4a^2-5\right)$ is completely factored.

<u>Solving fourth expression</u>

$24a^4+18$

$24a^4+18$

$\mathrm{Rewrite\:}18\mathrm{\:as\:}6\cdot \:3$

$\mathrm{Rewrite\:}24\mathrm{\:as\:}6\cdot \:4$

$6\cdot \:4a^4+6\cdot \:3$

$\mathrm{Factor\:out\:common\:term\:}6$

$6\left(4a^4+3\right)$

Thus, the expression $24a^4+18=6\left(4a^4+3\right)$ is completely factored.

Keywords: expression, factoring

Learn more about expression factoring from brainly.com/question/14051207

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