$\bf \textit{area of a trapezoid}\\\\ A=\cfrac{h(a+b)}{2}~~ \begin{cases} a,b=&\stackrel{bases}{parallel}\\ &sides\\ h=&height\\ \cline{1-2} a=&2\\ b=&6\\ A=&24 \end{cases}\implies 24=\cfrac{h(2+6)}{2}\implies 24=\cfrac{h(8)}{2} \\\\\\ 24=4h\implies \cfrac{24}{4}=h\implies 6=h$

8 0

3 years ago

Here's a question ~

Gre4nikov [31]

Answer:

(0, 5√3/2 + 2)

Step-by-step explanation:

Given is the absolute value function.

Observations:

It has a slope of ±√3 and the y- intercept of 2.
There is no horizontal shift, so the the y-axis is the line of symmetry.
The y-axis is also an angle bisector of the two lines.
The foot P₁P₂ is parallel to the x-axis since it's perpendicular to the y- axis.

We need to find the coordinates of intersection of the line P₁P₂ with the y- axis (the point Y in the picture).

Consider the triangle AYP₂.

We know AP₂ = 5.

The angle YAP₂ is:

arctan (1/√3) = 30°

The distance AY is:

AY = AP₂ cos 30° = 5*√3/2

The distance from the x-axis to the point Y is:

5√3/2 + 2, added the y- intercept of the graphed lines

The coordinates of the point Y:

(0, 5√3/2 + 2)

3 0

3 years ago