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klio [65]
3 years ago
14

Harlon wrote the equation uppercase A = 18 (x squared + 1) to find the area of a rectangle that has a length of x squared + 1 an

d a width of 18. He solved for x and got x = plus-or-minus StartRoot StartFraction uppercase A Over 18 EndFraction minus 1 EndRoot Which of the following is an equivalent equation for x?
Mathematics
1 answer:
Arturiano [62]3 years ago
3 0

Answer:

its A

Step-by-step explanation:

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3 years ago
What is the perimeter of the trapezoid with vertices Q(8, 8), R(14, 16), S(20, 16), and T(22, 8)? Round to the nearest hundredth
EleoNora [17]
Check the picture below.

so... you can pretty much see how long RS and QT are, you can just count the units off the grid.

now, let's find QR's length

\bf ~~~~~~~~~~~~\textit{distance between 2 points}\\\\
\begin{array}{ccccccccc}
&&x_1&&y_1&&x_2&&y_2\\
%  (a,b)
&Q&(~ 8 &,& 8~) 
%  (c,d)
&R&(~ 14 &,& 16~)
\end{array}~~ 
%  distance value
d = \sqrt{( x_2- x_1)^2 + ( y_2- y_1)^2}
\\\\\\
QR=\sqrt{(14-8)^2+(16-8)^2}\implies QR=\sqrt{6^2+8^2}
\\\\\\
QR=\sqrt{36+64}\implies QR=\sqrt{100}\implies QR=10

and let's also find the length for ST

\bf ~~~~~~~~~~~~\textit{distance between 2 points}\\\\
\begin{array}{ccccccccc}
&&x_1&&y_1&&x_2&&y_2\\
%  (a,b)
&S&(~ 20 &,& 16~) 
%  (c,d)
&T&(~ 22 &,& 8~)
\end{array}~ 
d = \sqrt{( x_2- x_1)^2 + ( y_2- y_1)^2}
\\\\\\
ST=\sqrt{(22-20)^2+(8-16)^2}\implies ST=\sqrt{2^2+(-8)^2}
\\\\\\
ST=\sqrt{4+64}\implies ST=\sqrt{68}\implies ST=\sqrt{4\cdot 17}
\\\\\\
ST=\sqrt{2^2\cdot 17}\implies ST=2\sqrt{17}

so, add the lengths of all sides, and that's the perimeter of the trapezoid.

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