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Anettt [7]
3 years ago
8

Whats the answer? I will mark brainliest! please help!

Mathematics
2 answers:
vivado [14]3 years ago
7 0

Answer:

D) 12

Step-by-step explanation:

Since PR is a midsegment of ABC, you need 2 of them.

(x + 3)2 = x + 9                   Distribute 2 to (x + 3)

2x + 6 = x + 9

- x          - x                         Subtract x from both sides

x + 6 = 9

  - 6   - 6                            Subtract 6 from both sides

x = 3

Plug this into the equation of segment AC

3 + 9

12 is the length of segment AC

If this answer is correct, please make me Brainliest!

Umnica [9.8K]3 years ago
6 0

Answer:

D) 12

Step-by-step explanation:

Given: Segment PR is a midsegment of triangle ABC.

So if PR is a midsegment the triangle, then we expect the base segment, AC, to be double the length of segment PR;

I.e 2(x + 3) = x + 9

2x + 6 = x + 9

2x - x = 9 - 6

x = 3

So the length of segment AC is 3 + 9 = 12

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Read 2 more answers
Please can someone help me ill mark brainliest
kramer
#5 is 4, #6 is graph A, and #7 is graph B. I hope this helped, Please mark as brainliest because I am 3 more away from the next level.
7 0
3 years ago
11. A city engineer determines that 5,500 cubic meters of
PolarNik [594]

The city do not have up to 5,500 cubic meters of sand that will be needed to combat erosion.

<h3>Volume</h3>

Volume is the amount of space occupied by a three dimensional object or figure. The volume (V) of a cone is given by:

V = (1/3)πr²h

Where h is the height, and r is the radius

Given that h = 35 m, the radius using Pythagoras theorem:

37² = r² + 35²

r = 12

V = (1/3)π(12)² * 35 = 5280 m³

The city do not have up to 5,500 cubic meters of sand that will be needed to combat erosion.

Find out more on Volume at: brainly.com/question/1972490

3 0
2 years ago
Let O be an angle in quadrant III such that cos 0 = -2/5 Find the exact values of csco and tan 0.​
vivado [14]

well, we know that θ is in the III Quadrant, where the sine is negative and the cosine is negative as well, or if you wish, where "x" as well as "y" are both negative, now, the hypotenuse or radius of the circle is just a distance amount, so is never negative, so in the equation of cos(θ) = - (2/5), the negative must be the adjacent side, thus

cos(\theta)=\cfrac{\stackrel{adjacent}{-2}}{\underset{hypotenuse}{5}}\qquad \textit{let's find the \underline{opposite side}} \\\\\\ \textit{using the pythagorean theorem} \\\\ c^2=a^2+b^2\implies \sqrt{c^2-a^2}=b \qquad \begin{cases} c=hypotenuse\\ a=adjacent\\ b=opposite\\ \end{cases} \\\\\\ \pm\sqrt{5^2 - (-2)^2}=b\implies \pm\sqrt{25-4}\implies \pm\sqrt{21}=b\implies \stackrel{III~Quadrant}{-\sqrt{21}=b}

\dotfill\\\\ csc(\theta)\implies \cfrac{\stackrel{hypotenuse}{5}}{\underset{opposite}{-\sqrt{21}}}\implies \stackrel{\textit{rationalizing the denominator}}{-\cfrac{5}{\sqrt{21}}\cdot \cfrac{\sqrt{21}}{\sqrt{21}}\implies -\cfrac{5\sqrt{21}}{21}} \\\\\\ tan(\theta)=\cfrac{\stackrel{opposite}{-\sqrt{21}}}{\underset{adjacent}{-2}}\implies tan(\theta)=\cfrac{\sqrt{21}}{2}

4 0
2 years ago
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