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3 years ago

Please help!! will give brainliest ASAP

Mathematics

1 answer:

soldier1979 [14.2K]3 years ago

8 0

Answer:

c = 36.06

Step-by-step explanation:

a² + b² = c²

20² + 30² = c²

400 + 900 = c²

c² = 1300

$\sqrt{c^2} =\sqrt{1300}$

c = 36.06

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jabu left town A to travel to town B at 6:00 a.m. he drove at an average speed of 80km/h. At 8.30a.m. Sipho left town A to trave

7nadin3 [17]

Answer:

Sipho catches up Jabu at 6:30 pm

Step-by-step explanation:

Jabu left town A to travel to town B at 6:00 a.m. he drove at an average speed of 80km/h.

At 8:30 am, when Sipho leaves, we will have traveled 2.5 hours at 80 km/h, so he will be 80*2.5 = 200 km ahead.

From then, his position will be given by:

$J(t) = 200 + 80t$

Sipho left town A to travel to town B. He drove at an average speed of 100 km/h.

His position will be given by:

$S(t) = 100t$

At what time will Sipho catch up jabu?

He will catch up in t hours, and t is found when:

$S(t) = J(t)$

$100t = 200 + 80t$

$20t = 200$

$t = \frac{200}{20} = 10$

He will catch up 10 hours after 8:30 am, that is, 6:30 pm.

Sipho catches up Jabu at 6:30 pm

4 0

2 years ago

This is a "water tank" calculus problem that I've been working on and I would really appreciate it if someone could look at my w

Sedaia [141]

Part A

Everything looks good but line 4. You need to put all of the "2h" in parenthesis so the teacher will know you are squaring all of 2h. As you have it right now, you are saying "only square the h, not the 2". Be careful as silly mistakes like this will often cost you points.

============================================================

Part B

It looks like you have the right answer. Though you'll need to use parenthesis to ensure that all of "75t/(2pi)" is under the cube root. I'm assuming you made a typo or forgot to put the parenthesis.

dh/dt = (25)/(2pi*h^2)
2pi*h^2*dh = 25*dt
int[ 2pi*h^2*dh ] = int[ 25*dt ] ... applying integral to both sides
(2/3)pi*h^3 = 25t + C
2pi*h^3 = 3(25t + C)
h^3 = (3(25t + C))/(2pi)
h^3 = (75t + 3C)/(2pi)
h^3 = (75t + C)/(2pi)
h = [ (75t + C)/(2pi) ]^(1/3)

Plug in the initial conditions. If the volume is V = 0 then the height is h = 0 at time t = 0
0 = [ (75(0) + C)/(2pi) ]^(1/3)
0 = [ (0 + C)/(2pi) ]^(1/3)
0 = [ (C)/(2pi) ]^(1/3)
0^3 = (C)/(2pi)
0 = C/(2pi)
C/(2pi) = 0
C = 0*2pi
C = 0

Therefore the h(t) function is...
h(t) = [ (75t + C)/(2pi) ]^(1/3)
h(t) = [ (75t + 0)/(2pi) ]^(1/3)
h(t) = [ (75t)/(2pi) ]^(1/3)

Answer:
h(t) = [ (75t)/(2pi) ]^(1/3)

============================================================

Part C

Your answer is correct.
Below is an alternative way to find the same answer

--------------------------------------

Plug in the given height; solve for t
h(t) = [ (75t)/(2pi) ]^(1/3)
8 = [ (75t)/(2pi) ]^(1/3)
8^3 = (75t)/(2pi)
512 = (75t)/(2pi)
(75t)/(2pi) = 512
75t = 512*2pi
75t = 1024pi
t = 1024pi/75
At this time value, the height of the water is 8 feet

Set up the radius r(t) function
r = 2*h
r = 2*h(t)
r = 2*[ (75t)/(2pi) ]^(1/3) .... using the answer from part B

Differentiate that r(t) function with respect to t
r = 2*[ (75t)/(2pi) ]^(1/3)
dr/dt = 2*(1/3)*[ (75t)/(2pi) ]^(1/3-1)*d/dt[(75t)/(2pi)]
dr/dt = (2/3)*[ (75t)/(2pi) ]^(-2/3)*(75/(2pi))
dr/dt = (2/3)*(75/(2pi))*[ (75t)/(2pi) ]^(-2/3)
dr/dt = (25/pi)*[ (75t)/(2pi) ]^(-2/3)

Plug in t = 1024pi/75 found earlier above
dr/dt = (25/pi)*[ (75t)/(2pi) ]^(-2/3)
dr/dt = (25/pi)*[ (75(1024pi/75))/(2pi) ]^(-2/3)
dr/dt = (25/pi)*[ (1024pi)/(2pi) ]^(-2/3)
dr/dt = (25/pi)*(1/64)
dr/dt = 25/(64pi)
getting the same answer as before

----------------------------

Thinking back as I finish up, your method is definitely shorter and more efficient. So I prefer your method, which is effectively this:
r = 2h, dr/dh = 2
dh/dt = (25)/(2pi*h^2) ... from part A
dr/dt = dr/dh*dh/dt ... chain rule
dr/dt = 2*((25)/(2pi*h^2))
dr/dt = ((25)/(pi*h^2))
dr/dt = ((25)/(pi*8^2)) ... plugging in h = 8
dr/dt = (25)/(64pi)
which is what you stated in your screenshot (though I added on the line dr/dt = dr/dh*dh/dt to show the chain rule in action)

8 0

3 years ago

The game piece has 12 sides labeled 1 to 12 what is the probability of rolling on 11

maks197457 [2]

Answer:

The probability would be 1/12 because there are 12 sides and there is only 1 11.

Hope this helped

*Befri.stends

Step-by-step explanation:

6 0

3 years ago

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