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Nataliya [291]
2 years ago
7

Determine whether AB is a median,altitude,or neither

Mathematics
1 answer:
nasty-shy [4]2 years ago
7 0

Answer:

median

Step by step:

the line segment from a vertex to the midpoint of the opposite side. It is also an angle bisector when the vertex is an angle in an equilateral triangle or the non-congruent angle of an isoceles triangle.

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Step-by-step explanation:

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Find the value of x
marin [14]

Answer:

x=14

Step-by-step explanation:

The given triangle is a right triangle, meaning it is a triangle with a right angle, this is indicated by the box around one of the angles. When given a right triangle, one can use the right triangle trigonometric ratios. These ratios describe the relationship between an angle in a right triangle, and the sides in a right triangle. Such ratios are as follows,

sin(\theta)=\frac{opposite}{hypotenuse}\\\\cos(\theta)=\frac{adjacent}{hypotenuse}\\\\tan(\theta)=\frac{opposite}{adjacent}

Bear in mind that the way one names the sides, in the sense of (opposite or adjacent) will change based on the angle one uses to describe the triangle. However, the hypotenuse is the same no matter the angle, as the hypotenuse is the side opposite the right angle.

In this case, one is given one of the angles, and the side opposite the angle. One is asked to find the hypotenuse of the triangle. One can use the sine (sin) ratio to achieve this.

sin(\theta)=\frac{opposite}{hypotenuse}

Substitute,

sin(30)=\frac{7}{x}

Inverse operations,

x=\frac{7}{sin(30)}

Simplify,

x=14

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3 years ago
2.1 &amp; 2.2 Assessment: Vertex &amp; Standard Form of Quadratic Functions
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~~~~~~\textit{vertical parabola vertex form} \\\\ y=a(x- h)^2+ k\qquad \begin{cases} \stackrel{vertex}{(h,k)}\\\\ \stackrel{"a"~is~negative}{op ens~\cap}\qquad \stackrel{"a"~is~positive}{op ens~\cup} \end{cases} \\\\[-0.35em] ~\dotfill

\begin{cases} h=1\\ k=-9 \end{cases}\implies y=a(x-1)^2 - 9\qquad \textit{we also know that} \begin{cases} x=2\\ y=-7 \end{cases} \\\\\\ -7=a(2-1)^2-9\implies 2=a(1)^2\implies 2=a \\\\[-0.35em] ~\dotfill\\\\ ~\hfill {\Large \begin{array}{llll} y=2(x-1)^2 -9 \end{array}} ~\hfill

6 0
1 year ago
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