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Veronika [31]
3 years ago
13

Solve for all values of x:

Mathematics
1 answer:
kati45 [8]3 years ago
3 0

Answer:

X= 6/D

Step-by-step explanation:

i did the math

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How many solutions does 2x+ 2x+ 2=4+2x have
yuradex [85]
Let's see!
2x + 2x + 2 = 4 + 2x\\ 4x + 2 = 4 + 2x\\ 4x - 2x + 2 = 4 + 2x - 2x\\ 2x + 2 = 4\\ 2x + 2 - 2 = 4 - 2\\ 2x + 0 = 2\\ 2x = 2\\ 2x/2 = 2/2\\ x = 1\\
The answer is B
3 0
3 years ago
B + 4 = 2b - 5 <br> B= ??
Sergio039 [100]

Answer:

B=9

Step-by-step explanation:

b+4=2b-5

b-2b+4=-5

b-2b=-5-4

-b=-5-4

-b=-9

b=9

6 0
3 years ago
Read 2 more answers
Can someone please help me with the question in the attachment? I will mark you as brainliest!​
san4es73 [151]

Answer:

a=327 m=416

Step-by-step explanation:

subtract the numbers and add to make sure ur answer is correct

8 0
2 years ago
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Prepare a perpetual inventory​ record, using the FIFO inventory costing​ method, and determine the​ company's cost of goods​ sol
krek1111 [17]

Answer:

The First-In, First-Out (FIFO) inventory costing method assumes that the inventory items ordered first are the first ones sold.

Step-by-step explanation:

The First-In, First-Out inventory costing method assumes that the inventory items ordered first are the first sold. This is ideal for goods that are highly perishable, for example fresh milk. Since no figures or dates are given, we will assume that the month is March 2019 and use any figures to make the example.

Date Item      Quantity of stock Cost Price

01  Opening stock bought on Feb 28  10   100

05  Sale of 5 goods (cost is $10 each)  (5)   50

15 Purchase of stock (20 goods at $20 each) 20   400

25 Sale of 15 goods                     (15)   250  

(5 at $10 each & 10 at $20 each)

31 Closing Stock               10   200

       (20 goods bought on 15th - 10 goods sold on 25th)

The quantity on hand at the end of the month is 10 units.  

Total cost of goods on hand at end of the month = 10 units * $20 = 200.

Total cost of goods purchased during the month = $20 * 20 units = $400

Total cost of goods sold during the month = [($10 *5) + ($10 * 5)+ ($20 * 10)] = $200

6 0
4 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
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