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11111nata11111 [884]
4 years ago
6

Could someone please help me solve these?

Mathematics
1 answer:
ivanzaharov [21]4 years ago
5 0
In advance, I hope nothing I say sounds condescending; I just don’t know how much you know lol. Also geometry sucks

They’re vertical angles (they’re opposite angles when 2 lines intersect). Since vertical angles are congruent, we can set up an equation to solve for the value of x.
If they’re both equal, we’ll say exactly that in our equation!

2x + 90 = 4x + 30

Combine like terms

2x = 60

Simplify

x = 30

Now, just plug in 30 for one of the x values in either of the equations

2(30) + 90 = 150

Angle = 150 °

Hope this helped!
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A trapezoid has an area of 20 cm2 and a height z cm. The lengths of the parallel sides are (2z + 3) cm and (6z – 1) cm. Find the
andriy [413]
Check the picture below.

\bf A=\cfrac{h(a+b)}{2}\quad 
\begin{cases}
A=20\\
a=6z-1\\
b=2z+3\\
h=z
\end{cases}\implies 20=\cfrac{z[(6z-1)~+~(2z+3)]}{2}
\\\\\\
20=\cfrac{z(8z+2)}{2}\implies 20=\cfrac{2z(4z+1)}{2}\implies 20=z(4z+1)
\\\\\\
20=4z^2+z\implies 0=4z^2+z-20

\bf \qquad \qquad \textit{quadratic formula}\\\\
\begin{array}{llccll}
0=&{{ 4}}z^2&{{ +1}}x&{{ -20}}\\
&\uparrow &\uparrow &\uparrow \\
&a&b&c
\end{array} 
\qquad \qquad 
z= \cfrac{ - {{ b}} \pm \sqrt { {{ b}}^2 -4{{ a}}{{ c}}}}{2{{ a}}}
\\\\\\
z=\cfrac{-1\pm\sqrt{1+320}}{8}\implies z=\cfrac{-1\pm\sqrt{321}}{8}\implies z\approx 
\begin{cases}
\boxed{2.1146}\\
-2.3646
\end{cases}

since the height is just a length unit, it can't be -2.3646.

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