The area of a trapezoid is found using the formula A=1/2h(b^1+b^2), where b^1 and b^2 are the parallel sides of the trapezoid an

d h is the height. The area of a trapezoid is 18km^2 , with bases 5 km and 7 km. What is the height, in kilometers? PLEASE HELP I NEED THIS SOOON

Mathematics

1 answer:

OverLord2011 [107]3 years ago

3 0

$\bf \textit{area of a trapezoid}\\\\ A=\cfrac{1}{2}h(b_1+b_2)~~ \begin{cases} A=18\\ b_1=5\\ b_2=7 \end{cases}\implies 18=\cfrac{1}{2}h(5+7) \\\\\\ 18=\cfrac{h(12)}{2}\implies 18=6h\implies \cfrac{18}{6}=h\implies 3=h$

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GalinKa [24]

Part A. What is the slope of a line that is perpendicular to a line whose equation is −2y=3x+7?

Rewrite the equation −2y=3x+7 in the form $y=-\dfrac{3}{2}x-\dfrac{7}{2}.$ Here the slope of the given line is $m_1=-\dfrac{3}{2}.$ If $m_2$ is the slope of perpendicular line, then

$m_1\cdot m_2=-1,\\ \\m_2=-\dfrac{1}{m_1}=\dfrac{2}{3}.$

Answer 1: $\dfrac{2}{3}$

Part B. The slope of the line y=−2x+3 is -2. Since $-\dfrac{3}{2}\neq -2\quad \text{and}\quad \dfrac{2}{3}\neq -2,$ then lines from part A are not parallel to line a.

Since $-2\cdot \left(-\dfrac{3}{2}\right)=3\neq -1\quad \text{and}\quad -2\cdot \dfrac{2}{3}=-\dfrac{4}{3}\neq -1,$ both lines are not perpendicular to line a.

Answer 2: Neither parallel nor perpendicular to line a

Part C. The line parallel to the line 2x+5y=10 has the equation 2x+5y=b. This line passes through the point (5,-4), then

2·5+5·(-4)=b,

10-20=b,

b=-10.

Answer 3: 2x+5y=-10.

Part D. The slope of the line $y=\dfrac{x}{4}+5$ is $\dfrac{1}{4}.$ Then the slope of perpendicular line is -4 and the equation of the perpendicular line is y=-4x+b. This line passes through the point (2,7), then