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Sveta_85 [38]
2 years ago
12

What is DC to the nearest tenth

Mathematics
1 answer:
Goryan [66]2 years ago
3 0

Answer:

DC= 32.3 units

Step-by-step explanation:

Use Pythag theorem to find BC   first

15 ^2 +  BC ^2 = 22^2      then BC = sqrt 259

Then use the Pythagorean Theorem again to find DC

DC^2 = 28^2 + sqrt(259)^2

       DC=32.3

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8 0
4 years ago
Find the value of x in the isosceles triangle shown below.
guajiro [1.7K]

Answer:

B) x = \sqrt{60}

Step-by-step explanation:

1. As you can see from the image, there are two right triangles in the isosceles triangle. The base of each right triangle is 2 (because 4 ÷ 2 = 2), and the hypotenuse is 8.

2. Let's use Pythagorean Theorem to find the height of each right triangle, which will also give us the height of the isosceles triangle: a^2 + b^2 = c^2, where a = height, b = base, and c= hypotenuse.

  • a^2 + 2^2 = 8^2
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  • a^2 + 4 - 4 = 64 - 4 (Subtract 4 from both sides)
  • a^2 = 60
  • \sqrt{a^2} = \sqrt{60} (Take square root of both sides
  • a = \sqrt{60}

3. The√(60) is the height of each right triangle, and it's also the height of the isosceles triangle. Therefore, the answer is B) x = √60.

8 0
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Let $abcdefgh$ be a cube of side length 5, as shown. let $p$ and $q$ be points on $\overline{ab}$ and $\overline{ae}$, respectiv
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Let's buils the intersection plane:
Point P is on AB and AP=2, then PB=3; point Q is on AE and AQ=1, then QE=4. Let P' be a point on CD such that CP'=2 and Q' be a point on the plane CDHG such that P'Q'=1 and P'Q' is perpendicular to CD. The line CQ' intersects HD at point R and the plane CPQR is intersection plane.
Consider triangles ΔCDR and ΔCP'Q', they are similar. So,
\frac{CP'}{CD}=  \frac{P'Q'}{RD}  \\  \frac{2}{5}  =\frac{1}{RD}  \\ RD=2.5,
so R is a midlepoint of the side HD (for details see picture).



5 0
4 years ago
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