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Marina CMI [18]
2 years ago
8

Sale! 60% OFF

Mathematics
1 answer:
ra1l [238]2 years ago
8 0
$12 hope this helps hehehehehe
You might be interested in
A carpenter needs 9.3 feet of rope for a project. How much will the carpenter spend if the rope costs $3.82 per foot?
den301095 [7]

The total amount the carpenter will spend if the rope costs $3.82 per foot is $35.526

<h3>How to determine how much will the carpenter spend if the rope costs $3.82 per foot?</h3>

The length of rope is given as:

Length = 9.3 feet

The price per rope is given as:

Price per rope = $3.82 per foot

The total amount spent by the carpenter is calculated as:

Total amount = Length * Price per rope

Substitute the known values in the above equation

Total amount = 9.3 feet * $3.82 per foot

Evaluate the product

Total amount = $35.526

Hence, the total amount the carpenter will spend if the rope costs $3.82 per foot is $35.526

Read more about products at:

brainly.com/question/10873737

#SPJ1

4 0
1 year ago
Read 2 more answers
Dilate by a scale<br> factor of 1/2<br> AL<br> D
Firdavs [7]

Answer:

The image triangle ΔD'L'W' after the dilation by a scale factor of 1/2 will have the vertices:

  • D' (0, 1)
  • L' (1/2, 5/2) or L'(0.5, 2.5)
  • W' (5/2, 2) or W'(2.5, 2)

`The figure of the dilated image is attached below.

Step-by-step explanation:

Given the triangle ΔDLW withe the vertices

  • D(0, 2)
  • L(1, 5)
  • W(5, 4)

As the scale factor 1/2 < 0, thus the image triangle will be smaller than the original triangle.

The rule of dilation is given by

Dilation: (x, y)  →  (1/2x, 1/2y), centered at (0, 0)

D(0, 2)  → D' (1/2x, 1/2y) → D' (1/2(0), 1/2(2)) = D' (0, 1)

L(1, 5)  → L'(1/2x, 1/2y) →  L'(1/2(1), 1/2(5)) = L' (1/2, 5/2)

W(5, 4)  → W'(1/2x, 1/2y) →  W'(1/2(5), 1/2(4)) = W' (5/2, 2)

Therefore, the image triangle ΔD'L'W' after the dilation by a scale factor of 1/2 will have the vertices:

  • D' (0, 1)
  • L' (1/2, 5/2) or L'(0.5, 2.5)
  • W' (5/2, 2) or W'(2.5, 2)

`The figure of the dilated image is attached below.

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
I can't think of any right answer?​
madam [21]
I think the answer is true
Hope this helps!!
Please consider brainliest!<3
8 0
2 years ago
Read 2 more answers
Write the equation of the line in slope-intercept form using y=mx+b​
soldi70 [24.7K]
Equation Answer: y= -2x -7
4 0
2 years ago
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