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Temka [501]
3 years ago
7

i have a couple questions with my geometry homework which is due tomorrow, could someone try answering them? I've been trying to

solve them for the past hour.

Mathematics
1 answer:
sveticcg [70]3 years ago
7 0

Answer:

  2.  100°              22.  7

  18.  6                  23.  9

  19.  8                  24.  65°

  20.  55°             25.  AB = (1/2)DF

  21.  6                 26.  AB ║ DF

Step-by-step explanation:

Please be aware that the triangle measurements shown for problems 18–20 and 22–24 cannot exist. In the first case, the angle is closer to 48.6° (not 35°), and in the second case, the angle is closer to 51.1°, not 25°. So, you have to take the numbers at face value and not think too deeply about them. (This state of affairs is all too common in geometry problems these days.)

_____

2. You have done yourself no favors by marking the drawing the way you have. Look again at the given conditions. You will find that x+2x must total a right angle, so x=30°. Angle P is the complement of 40°, so is 50°. Then the sum of x and angle P is 30° +50° = 80°, and the angle of interest is the supplement of that, 100°.

__

18–20. Perpendicular bisector NO means  ∆NOL ≅ ΔNOM. Corresponding parts have the same measures, and angle L is the complement of the marked angle.

__

21. ∆ODA ≅ ∆ODB by hypotenuse-angle congruence (HA), so corresponding parts are the same measure. DB = DA = 6.

__

22–24. ∆VPT ≅ ∆VPR by LL congruence, so corresponding measures are the same. Once again, the angle in question is the complement of the given angle.

__

25–26. You observe that A is the midpoint of DE, and B is the midpoint of FE, so AB is what is called a "midsegment." The features of a midsegment are that it is ...

  • half the length of the base (DF)
  • parallel to the base

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A rectangular swimming pool is bordered by a concrete patio. the width of the patio is the same on every side. the area of the s
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Answer:

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where

l = length of the pool (w/o the patio)
w = width of the pool (w/o the patio)

Explanation: 

Let 

x = width of the patio
l = length of the pool (w/o the patio)
w = width of the pool (w/o the patio)

Since the pool is bordered by a complete patio, 

Length of the pool (with the patio) 
= (length of the pool (w/o the patio)) + 2*(width of the patio)
Length of the pool (with the patio) = l + 2x

Width of the pool (with the patio) 
= (width of the pool (w/o the patio)) + 2*(width of the patio)
Width of the pool (with the patio) = w + 2x

Note that

Area of the pool (w/o the patio)
=  (length of the pool (w/o the patio))(width of the pool (w/o the patio))
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Area of the pool (with the patio)
= (length of the pool (w/o the patio))(width of the pool (w/o the patio))
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= w(l + 2x) + 2x(l + 2x)
= lw + 2xw + 2xl + 4x²
Area of the pool (with the patio) = 4x² + 2x(l + w) + lw

Area of the patio
= (Area of the pool (with the patio)) - (Area of the pool (w/o the patio))
= (4x² + 2x(l + w) + lw) - lw
Area of the patio = 4x² + 2x(l + w)

Since the area of the patio is equal to the area of the surface of the pool, the area of the patio is equal to the area of the pool without the patio. In terms of the equation,

Area of the patio = Area of the pool (w/o the patio)
4x² + 2x(l + w) = lw
4x² + 2x(l + w) - lw = 0    (1)

Let 

a = numerical coefficient of x² = 4
b = numerical coefficient of x = 2(l + w)
c = constant term = -lw

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x = \frac{-b \pm  \sqrt{b^2 - 4ac}}{2a}&#10;\\ = \frac{-2(l + w) \pm  \sqrt{(2(l + w))^2 - 4(4)(-lw)}}{2(4)} &#10;\\ = \frac{-2(l + w) \pm  \sqrt{(4(l + w)^2) + 16lw}}{8} &#10;\\ = \frac{-2(l + w) \pm  \sqrt{(4(l^2 + 2lw + w^2) + 4(4lw)}}{8}&#10;\\ = \frac{-2(l + w) \pm  \sqrt{(4(l^2 + 2lw + w^2 + 4lw)}}{8}&#10;\\ = \frac{-2(l + w) \pm  \sqrt{(4(l^2 + 6lw + w^2)}}{8}
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Since (l + w) + \sqrt{l^2 + 6lw + w^2} \ \textgreater \  0, -\frac{1}{4}\left((l + w) + \sqrt{l^2 + 6lw + w^2}\right) is negative. Since x represents the patio width, x cannot be negative. Hence, the patio width is given by 

\boxed{x = \frac{1}{4}\left(-(l + w) + \sqrt{l^2 + 6lw + w^2} \right)}




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A container is leaking 0.42 liters of solution each hour. What is the overall change after 5 hours?
OLga [1]

Answer:

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Step-by-step explanation:

0.42 * 5

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