3 years ago

Can u solve this puzzle?

Mathematics

1 answer:

finlep [7]3 years ago

4 0

Answer:

i dont think so sorry

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A cylindrical oil tank 8 ft deep holds 620 gallons when filled to capacity. How many gallons remain in the tank when the depth o

vekshin1

Answer:

271.25 gal

Step-by-step explanation:

Assuming the tank is oriented so that the quantity of oil is proportional to depth, the amount remaining is ...

(3 1/2 ft)/(8 ft) × (620 gal) = 271 1/4 gal

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3 years ago

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Write an algebraic equation for each question below and then solve the equation.

Aneli [31]

Answer:

$number = x \\ number \: decreaed \: by \: 8 = > x - 8 = 26 \\ x - 8 = 26 \\ x = 8 + 26 \\ x = 34 \\ thank \: you$

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3 years ago

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Solve for W in A = 2(L + W)

Roman55 [17]

$\Large\begin{aligned}\\\\&A = 2(L + W)\\&A = 2L+2W\\&2W=A-2L\\&W=\dfrac{A-2L}{2}\end$

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2 years ago

There are currently 17 frogs in a (large) pond. The frog population grows exponentially, tripling every 6 days. How long will it

Tema [17]

Answer:

t=11,11 days

Step-by-step explanation:

F=frogs poblation, t=time, be the variables dF/dt = KF, dF/F=Kdt, integrating $\int\limits^ {} \, dF/F =K\int\limits^ {} \, dt$ ⇒ LnF=Kt+c, $F=ce^{Kt}$ ; Knowing t=0, F=17 and t=6 F=51 (tripling every 6 days (17*3)), $F=ce^{K0} = F=c=17; F=17e^{6K} =51$ ⇒ $e^{6K} =51/17; K6=ln\frac{51}{17} ; K=ln\frac{51}{17}/6=0.183$ , so $F=17e^{0,183t}$ , now if F=130, t=? we have:

$130=17e^{0.183t} =e^{0.183t} =130/17; 0.183t=ln(130/17); t=ln(130/17)/0.183 = 11,11$

8 0

3 years ago

What factors do 16 and 24 have in common?

ycow [4]

Common factor of 16 and 24 is 8

5 0

3 years ago