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Leokris [45]
3 years ago
7

Y=-5/3x + 28/3 in standard form

Mathematics
2 answers:
Temka [501]3 years ago
7 0
3y + 5x - 28 = 0

anything please ask me
Katena32 [7]3 years ago
7 0
Formula is ax+by=c.multiply every side by 3 to get rid of 3
3y=-5x+28 then take x to the other side 
3y+5x=28


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four students have to give a speech in class today. In how many different orders can they give their speeches?
bulgar [2K]

Answer:

There are 24 possible outcomes.

Step-by-step explanation:

1. 1-2-3-4

2. 1-4-3-2

3. 1-4-2-3

4. 1-2-4-3

5. 1-3-2-4

6. 1-4-3-2

7. 2-4-3-1

8. 2-1-3-4

9. 2-1-4-3

10. 2-4-1-3

11. 2-3-1-4

12. 2-3-4-1

13. 3-2-1-4

14. 3-2-4-1

15. 3-1-2-4

16. 3-1-4-2

17. 3-4-1-2

18. 3-4-2-1

19. 4-3-2-1

20. 4-3-1-2

21. 4-2-3-1

22. 4-2-1-3

23. 4-1-2-3

24. 4-1-3-2

Hope this helps!!


5 0
3 years ago
HELP ME ASAP.
slavikrds [6]

Answer:

The first number lets say is x

the second is y so

y=1/2x+8

x+1/2x+8=58

1 1/2x=50

x= 33 1/3

Hope This Helps!!!

6 0
3 years ago
Read 2 more answers
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
Solve the equation.
Sergeeva-Olga [200]
The answer to your question is 4
3 0
4 years ago
Five snails are having a race. The snail's speeds are given in the table. Which two snails are going the same rate of speed?
skad [1K]

Answer:

2 1/5 for alice

4 3/14 for Bert

5 for Carl

5 for Eva

5 1/2 for Dennis

Carl and Eva

Step-by-step explanation:

brainliest please

7 0
3 years ago
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