I don’t understand this question what so ever

Mathematics

1 answer:

Mama L [17]3 years ago

3 0

bearing in mind that perpendicular lines have negative reciprocal slopes, let's find firstly the slope of AC.

$\bf (\stackrel{x_1}{2}~,~\stackrel{y_1}{1})\qquad (\stackrel{x_2}{1}~,~\stackrel{y_2}{6}) \\\\\\ slope = m\implies \cfrac{\stackrel{rise}{ y_2- y_1}}{\stackrel{run}{ x_2- x_1}}\implies \cfrac{6-1}{1-2}\implies \cfrac{5}{-1}\implies -5 \\\\[-0.35em] ~\dotfill\\\\ \stackrel{\textit{perpendicular lines have \underline{negative reciprocal} slopes}} {\stackrel{slope}{\cfrac{-5}{1}}\qquad \qquad \qquad \stackrel{reciprocal}{\cfrac{1}{-5}}\qquad \stackrel{negative~reciprocal}{\cfrac{1}{5}}}$

so, we're really looking for the equation of a line whose slope is 1/5 and that passes through (3,3)

$\bf (\stackrel{x_1}{3}~,~\stackrel{y_1}{3}) ~\hspace{10em}slope = m\implies \cfrac{1}{5} \\\\\\ \begin{array}{|c|ll} \cline{1-1} \textit{point-slope form}\\ \cline{1-1} \\ y-y_1=m(x-x_1) \\\\ \cline{1-1} \end{array}\implies y-3=\cfrac{1}{5}(x-3) \implies y-3=\cfrac{1}{5}x-\cfrac{3}{5} \\\\\\ y=\cfrac{1}{5}x-\cfrac{3}{5}+3\implies y=\cfrac{1}{5}x+\cfrac{12}{5}$

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Find the length of the third side. If necessary, write in simplest radical form.

Mazyrski [523]

Answer:

$\boxed {\boxed {\sf 8}}$

Step-by-step explanation:

This triangle has a small square, which represents a right angle. Therefore, we can use the Pythagorean Theorem.

$a^2+b^2=c^2$

Where a and b are the legs of the triangle and c is the hypotenuse.

In this triangle, 7 and √15 are the legs, because these sides make up the right angle. The unknown side is the hypotenuse, because it is opposite the right angle. So, we know two values:

$a= 7 \\b= \sqrt{15}$

Substitute these values into the formula.

$(7)^2+(\sqrt{15})^2=c^2$

Solve the exponents.

(7)²= 7*7=49

$49+ (\sqrt{15})^2=c^2$

(√15)²=√15*√15=15

$49+15=c^2$

Add.

$64=c^2$

Since we are solving for c, we must isolate the variable. It is being squared and the inverse of a square is the square root. Take the square root of both sides.

$\sqrt{64}=\sqrt{c^2} \\\sqrt{64}= c\\8=c$