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Soloha48 [4]
3 years ago
9

Using a protractor and a ruler, reflect points A and B across line PQ?

Mathematics
1 answer:
MA_775_DIABLO [31]3 years ago
8 0

Answer:

shiii

Step-by-step explanation:

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The answer is c because S=Ph+2b
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babunello [35]

Answer: y = 90   x = 29

Step-by-step explanation:

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A number is greater than 8. The same number is less than 10. The inequalities x > 8 and x < 10 represent the situation
wlad13 [49]

Answer:

Option 4

Step-by-step explanation:

Let any two real number a and b (no matter +ve, -ve or 0). a ≥ b

The average of them will always lie in between them or be equal(if 0).

Let's prove : According to the statement,

a ≥ (a + b)/2 ≥ b

2a ≥ a + b ≥ 2b

2a ≥ a + b and a + b ≥ 2b

a ≥ b and a ≥ b, as we assumed.

Moreover, as the average exists in between a and b, we have the average (a + b)/2. Similarly, there exists one more average of (a + b)/2 and a or b, which definitely lie between a and b as (a + b)/2 lies there and smaller than a and b.

In the same order, we can have many average and the process would stop. This leads to infinite number between a and b.

Notice that we talked about all the numbers moreover there are many irrational(non-terminating like 9.898989.... etc numbers as well.

Option (4), infinite solutions.

Note: we solved for all the number (not specifically odd, even, natural, whole, integer, etc).

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yawa3891 [41]

Answer:

\frac{50}{7}

Step-by-step explanation:

Swap the fraction,

5 *\frac{10}{7} = \frac{50}{7}

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3 0
3 years ago
A rectangular box has a square base. The combined length of a side of the square base, and the height is 20 in. Let x be the len
aniked [119]

Answer:

a. V = (20-x) x^{2} in^{3}  

b . 1185.185 in^{3}

Step-by-step explanation:

Given that:

  • The height:  20  - x (in )
  • Let x be the length of a side of the base of the box (x>0)

a. Write a polynomial function in factored form modeling the volume V of the box.

As we know that, this is a rectangular box has a square base so the Volume of it is:

V = h *x^{2} in^{3}

<=> V = (20-x) x^{2}  in^{3}

b. What is the maximum possible volume of the box?

To  maximum the volume of it, we need to use first derivative of the volume.

<=> dV / Dx = -3x^{2} + 40x

Let dV / Dx = 0, we have:

-3x^{2} + 40x  = 0

<=> x = 40/3

=>the height h = 20/3

So  the maximum possible volume of the box is:

V = 20/3 * 40/3 *40/3

= 1185.185 in^{3}

7 0
3 years ago
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