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erma4kov [3.2K]
3 years ago
12

Toni recipe also calls for seasonings. She needs to stir in 215 milligrams of pepper and 437 milligrams of salt. What is the dif

ference between the total amount of seasonings and 1 gram.
Mathematics
1 answer:
marishachu [46]3 years ago
6 0
215g + 437g = 652g

You need to find the difference between the amount of seasoning and 1 gram.
652 - 1 = 651g

The answer is:651g
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--2m + 4n2 for m = -6 and n=-5<br> Evaluate the expression
Tamiku [17]

Answer: 112

Step-by-step explanation:

Since we were given m and n, we plug them into the expression and solve.

-2(-6)+4(-5)²            [exponent]

-2(-6)+4(25)            [multiply]

12+100                     [add]

112

Now, we know that the value of the expression is 112.

7 0
3 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
A machine cuts 25 circuit boards every 5 minutes. What is the unit rate of production for the machine?
Sladkaya [172]

Answer:

  • B- 5/1 boards/minutes

Step-by-step explanation:

<u>Unit rate is the number of circuit boards a minute:</u>

  • 25 boards every 5 minutes >>
  • 25/5 boards every 5/5 minutes >>
  • 5/1 boards/minutes

Correct choice is B

3 0
2 years ago
Sub to Via's Creations channel on you tube!
Elden [556K]

Answer:

You got it! ill do it now

Step-by-step explanation:

3 0
3 years ago
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Which of the following numbers is a COUNTEREXAMPLE to the statement "All odd numbers greater than 2 are prime numbers" ? 2 3 5 7
GuDViN [60]
9 would be ur answer....because 9 is an odd number greater then 2, but it is not a prime number
5 0
3 years ago
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