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klemol [59]
3 years ago
12

What is the value of x?

Mathematics
2 answers:
OLga [1]3 years ago
5 0

Answer:

x = 24

Step-by-step explanation:

Pythagorean Theorem: a² + b² = c²

<em>a</em> = a leg

<em>b</em> = another leg

<em>c</em> = hypotenuse

Step 1: Plug in known variables

x² + 10² = 26²

Step 2: Evaluate

x² + 100 = 676

Step 3: Isolate <em>x </em>term

x² = 576

Step 4: Isolate <em>x</em>

√x² = √576

x = 24

tatiyna3 years ago
4 0

Answer:

x=24

Step-by-step explanation:

Since this is a right triangle, we can use the Pythagorean Theorem.

The Pythagorean Theorem is:

a^2+b^2=c^2

The hypotenuse is c, and the legs are a and b.

Plug in 26 for c since 26 is the hypotenuse, and let 10 for a and x be b. Therefore:

x^2+10^2=26^2

Square:

x^2+100=676

Subtract 100 from both sides:

x^2=576

Take the square root of both sides:

\sqrt{x^2}=\sqrt{576}\\x=\sqrt{576}=24

The answer is 24.

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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}

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Answer:

Step-by-step explanation:

With reference < H

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