What is the total surface area of the pyramid below?

Mathematics

2 answers:

mel-nik [20]3 years ago

8 0

Answer:I guess 6 ft as you gave same options

Step-by-step explanation:

tamaranim1 [39]3 years ago

8 0

Answer:

17 ft that's the ans

I don't know if it is correct

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SIMPLIFY<br> −6u2v−uv2−22u2v2.

nignag [31]

Answer:

-uv • (22uv + 6u + v)

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8 0

3 years ago

What is the value of s?

KonstantinChe [14]

Answer:

35°

Step-by-step explanation:

Inscribed angle s is half the measure of the arc it subtends. That arc is the supplement to the 110° arc shown. The arc is 70°, so angle s is 35°.

6 0

3 years ago

Find the area of the triangle ABC with the coordinates of A(10, 15) B(15, 15) C(30, 9).

lions [1.4K]

Check the picture below. so, that'd be the triangle's sides hmmm so let's use Heron's Area formula for it.

$~\hfill \stackrel{\textit{\large distance between 2 points}}{d = \sqrt{( x_2- x_1)^2 + ( y_2- y_1)^2}}~\hfill~ \\\\[-0.35em] ~\dotfill\\\\ (\stackrel{x_1}{10}~,~\stackrel{y_1}{5})\qquad (\stackrel{x_2}{15}~,~\stackrel{y_2}{15}) ~\hfill a=\sqrt{[ 15- 10]^2 + [ 15- 5]^2} \\\\\\ ~\hfill \boxed{a=\sqrt{125}} \\\\\\ (\stackrel{x_1}{15}~,~\stackrel{y_1}{15})\qquad (\stackrel{x_2}{30}~,~\stackrel{y_2}{9}) ~\hfill b=\sqrt{[ 30- 15]^2 + [ 9- 15]^2} \\\\\\ ~\hfill \boxed{b=\sqrt{261}}$

$(\stackrel{x_1}{30}~,~\stackrel{y_1}{9})\qquad (\stackrel{x_2}{10}~,~\stackrel{y_2}{5}) ~\hfill c=\sqrt{[ 10- 30]^2 + [ 5- 9]^2} \\\\\\ ~\hfill \boxed{c=\sqrt{416}} \\\\[-0.35em] ~\dotfill$

$\qquad \textit{Heron's area formula} \\\\ A=\sqrt{s(s-a)(s-b)(s-c)}\qquad \begin{cases} s=\frac{a+b+c}{2}\\[-0.5em] \hrulefill\\ a=\sqrt{125}\\ b=\sqrt{261}\\ c=\sqrt{416}\\ s\approx 23.87 \end{cases} \\\\\\ A\approx\sqrt{23.87(23.87-\sqrt{125})(23.87-\sqrt{261})(23.87-\sqrt{416})}\implies \boxed{A\approx 90}$