Can u guys please help me

Mathematics

1 answer:

elena-s [515]3 years ago

5 0

1a. The equation is x-35+x+5+x

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By driving 8 mph faster than Bob, John can make a 230 mile trip in one half hour less.How fast does Bob drive on the trip? Round

aivan3 [116]

Answer:

57 mph

Step-by-step explanation:

Let b represent Bob's speed on the trip. Then his time is ...

time = distance/speed

Bob's time = 230/b

Then John's speed is b+8, and his time is ...

John's time = 230/(b+8)

The difference of their times is 1/2 hour:

Bob's time - John's time = 1/2

230/b -230/(b+8) = 1/2 . . . substitute the expressions for time

460(b+8 -b) = b(b+8) . . . . . multiply by 2b(b+8)

b^2 +8b -3680 = 0 . . . . . . .put in standard form

(b +4)^2 -3696 = 0 . . . . . . . add 16-16 to complete the square

b = -4 +√3696 ≈ 56.795 . . . solve for positive b

Bob's drove about 57 mph on the trip.

3 0

3 years ago

Gas is escaping from a spherical balloon at the rate of 12 ft3/hr. At what rate (in feet per hour) is the radius of the balloon

bija089 [108]

Answer:

This is the rate at which the radius of the balloon is changing when the volume is 300 $ft^3$ $\frac{dr}{dt}=-\frac{3}{225^{\frac{2}{3}}\pi ^{\frac{1}{3}}} \:\frac{ft}{h} \approx -0.05537 \:\frac{ft}{h}$

Step-by-step explanation:

Let $r$ be the radius and $V$ the volume.

We know that the gas is escaping from a spherical balloon at the rate of $\frac{dV}{dt}=-12\:\frac{ft^3}{h}$ because the volume is decreasing, and we want to find $\frac{dr}{dt}$

The two variables are related by the equation

$V=\frac{4}{3}\pi r^3$

taking the derivative of the equation, we get

$\frac{d}{dt}V=\frac{d}{dt}(\frac{4}{3}\pi r^3)\\\\\frac{dV}{dt}=\frac{4}{3}\pi (3r^2)\frac{dr}{dt} \\\\\frac{dV}{dt}=4\pi r^2 \frac{dr}{dt}$

With the help of the formula for the volume of a sphere and the information given, we find $r$

$V=\frac{4}{3}\pi r^3\\\\300=\frac{4}{3}\pi r^3\\\\r^3=\frac{225}{\pi }\\\\r=\sqrt[3]{\frac{225}{\pi }}$

Substitute the values we know and solve for $\frac{dr}{dt}$

$\frac{dV}{dt}=4\pi r^2 \frac{dr}{dt}\\\\\frac{dr}{dt}=\frac{\frac{dV}{dt}}{4\pi r^2} \\\\\frac{dr}{dt}=-\frac{12}{4\pi (\sqrt[3]{\frac{225}{\pi }})^2} \\\\\frac{dr}{dt}=-\frac{3}{\pi \left(\sqrt[3]{\frac{225}{\pi }}\right)^2}\\\\\frac{dr}{dt}=-\frac{3}{\pi \frac{225^{\frac{2}{3}}}{\pi ^{\frac{2}{3}}}}\\\\\frac{dr}{dt}=-\frac{3}{225^{\frac{2}{3}}\pi ^{\frac{1}{3}}} \approx -0.05537 \:\frac{ft}{h}$