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DENIUS [597]
4 years ago
11

Anyone can help me with this question? I would so much appreciate for helping me out!

Mathematics
1 answer:
AlexFokin [52]4 years ago
6 0
I strongly recommend that you draw this situation.  The vertices of the right triangle are (2,0), (p,0) and (p, -p^2+8p-12).

We are to maximize the area of the triangle.  To do this, we find the equation for the area (which is based upon the usual b*h/2), differentiate this formula, set the derivative = to 0 and solve for p.  I obtained p = 14/3, or 4 2/3 units.

Keep in mind that the correct expression for the base here is p-2.  Can you see why?  The height is simply the function itself:  p(x) = -x^2+8x-12.
Thus, the area of the triangle, in terms of p, is

            (p-2)(-p^2+8p-12)
A(p) = ---------------------------
                           2

Let me know if you need further help with this problem.  Good luck!

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bogdanovich [222]
It is -5.1 then -4 then -14/3 
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a designer needs to create perfectly circular necklace. the necklace each need to have a radius of 10cm. what is the largest num
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Ok so find circumference of 1 neclace
C=2pir
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how many 20pi's make 1000?
20pi times x=1000
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xnecklaces=1000/20pi
xnecklaces=50/pi
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xnecklaces=50/3.14
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round down since you can't sell 0.9 necklasce

15 is answer

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If a sphere’s volume is doubled, what is the corresponding change in its radius?
velikii [3]
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The sphere's volume is proportional to the cube of its radius,
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The only way to double a sphere's volume is to increase
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What is the value of the expression below when y=8y and z=8z? 10y-3z
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Answer:

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Step-by-step explanation:

6 0
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This 2 pls i need help with explanation sum
Nataliya [291]

9514 1404 393

Explanation:

<h3>8.</h3>

An exterior angle is equal to the sum of the remote interior angles. Define ∠PQR = 2q, and ∠QPR = 2p. The purpose of this is to let us use a single character to represent the angle, instead of 4 characters.

The above relation tells us ...

  ∠PRS = ∠PQR +∠QPR = 2q +2p

Then ...

  ∠TRS = (1/2)∠PRS = (1/2)(2q +2p) = q +p

and

  ∠TRS = ∠TQR +∠QTR . . . . . exterior is sum of remote interior

  q +p = (1/2)(2q) +∠QTR . . . . substitute for ∠TRS and ∠TQR

  p = ∠QTR = 1/2(∠QPR) . . . . . subtract q

__

<h3>9.</h3>

For triangle ABC, draw line DE parallel to BC through point A. Put point D on the same side of point A that point B is on the side of the median from vertex A. Then we have congruent alternate interior angles DAB and ABC, as well as EAC and ACB. The angle sum theorem tells you that ...

  ∠DAB +∠BAC +∠CAE = ∠DAE . . . . a straight angle = 180°

Substituting the congruent angles, this gives ...

  ∠ABC +∠BAC +∠ACB = 180° . . . . . the desired relation

4 0
3 years ago
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