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ollegr [7]
2 years ago
7

A triangle has lengths of 3 cm, 5 cm, and 9 cm, can it form 1 or more triangles?

Mathematics
1 answer:
AlladinOne [14]2 years ago
4 0

Answer:Yes

Step-by-step explanation:

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solniwko [45]

Answer:

3000

Step-by-step explanation:

6 0
2 years ago
Need help solving for x
yawa3891 [41]

Answer:

9.2

Step-by-step explanation:

The given triangle is a right angled triangle. To solve for any of the side length of such triangle, apply the trigonometry ratio formula which can easily be remembered as SOHCAHTOA.

SOH is Sin θ = opposite/hypothenuse,

CAH is Cos θ = Adjacent/hypotenuse

TOA is Tan θ = Opposite/adjacent

Thus, in the right triangle given, we have:

θ = 38°

Opposite side to the given angle = x

Hypotenuse = 15

We're going to use, sin θ = opposite/hypotenuse

Sin(38) = x/15

Multiply both sides by 15 to solve for x

15*sin(38) = x

15*0.616 = x

9.24 = x

<em>x ≈ 9.2 (to nearest tenth)</em>

3 0
2 years ago
The points A(1, 4), B(5,1) lie on a circle. The line segment AB is a chord. Find the equation of a diameter of the circle.
tangare [24]

Check the picture below.

well, we want only the equation of the diametrical line, now, the diameter can touch the chord at any several angles, as well at a right-angle.

bearing in mind that <u>perpendicular lines have negative reciprocal</u> slopes, hmm let's find firstly the slope of AB, and the negative reciprocal of that will be the slope of the diameter, that is passing through the midpoint of AB.

\bf A(\stackrel{x_1}{1}~,~\stackrel{y_1}{4})\qquad B(\stackrel{x_2}{5}~,~\stackrel{y_2}{1}) ~\hfill \stackrel{slope}{m}\implies \cfrac{\stackrel{rise} {\stackrel{y_2}{1}-\stackrel{y1}{4}}}{\underset{run} {\underset{x_2}{5}-\underset{x_1}{1}}}\implies \cfrac{-3}{4} \\\\[-0.35em] ~\dotfill\\\\ \stackrel{\textit{slope of AB}}{-\cfrac{3}{4}}\qquad \qquad \qquad \stackrel{\textit{\underline{negative reciprocal} and slope of the diameter}}{\cfrac{4}{3}}

so, it passes through the midpoint of AB,

\bf ~~~~~~~~~~~~\textit{middle point of 2 points } \\\\ A(\stackrel{x_1}{1}~,~\stackrel{y_1}{4})\qquad B(\stackrel{x_2}{5}~,~\stackrel{y_2}{1}) \qquad \left(\cfrac{ x_2 + x_1}{2}~~~ ,~~~ \cfrac{ y_2 + y_1}{2} \right) \\\\\\ \left( \cfrac{5+1}{2}~~,~~\cfrac{1+4}{2} \right)\implies \left(3~~,~~\cfrac{5}{2} \right)

so, we're really looking for the equation of a line whose slope is 4/3 and runs through (3 , 5/2)

\bf (\stackrel{x_1}{3}~,~\stackrel{y_1}{\frac{5}{2}}) \stackrel{slope}{m}\implies \cfrac{4}{3} \\\\\\ \begin{array}{|c|ll} \cline{1-1} \textit{point-slope form}\\ \cline{1-1} \\ y-y_1=m(x-x_1) \\\\ \cline{1-1} \end{array}\implies y-\stackrel{y_1}{\cfrac{5}{2}}=\stackrel{m}{\cfrac{4}{3}}(x-\stackrel{x_1}{3})\implies y-\cfrac{5}{2}=\cfrac{4}{3}x-4 \\\\\\ y=\cfrac{4}{3}x-4+\cfrac{5}{2}\implies y=\cfrac{4}{3}x-\cfrac{3}{2}

4 0
3 years ago
CAC SOMEONE IN THIS APP help
anyanavicka [17]

Answer:

The tree is 8 foot long

hope this helps

4 0
2 years ago
Hey guys i need help finding m&lt;1 PLEASE HELPPP
valentinak56 [21]

Step-by-step explanation:

You see the 90 degree angle marking at the top left so that angle is 90 degrees. However, it is bisected into two so both sections of that 90 degree angle equal 45. One triangle is equal to 180 degrees, so the triangle at the tops' missing angle is 52. I don't know if you heard of vertical angles but that angle opposite 83, because it is a vertical angle, is also 83 degrees. This is as far as I've gotten. Not sure how to solve the rest without more information

8 0
2 years ago
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