Just need help finding X and Y

Mathematics

1 answer:

Lapatulllka [165]3 years ago

5 0

Answer:

Step-by-step explanation:

For x use the Pythagorean theorem:

$x^2+6^2=10^2$

$x^2+36=100$ <em>subtract 36 from both sides</em>

$x^2=64\to x=\sqrt{64}\\\\x=8$

We know: x + y = 20. Substitute:

$8+y=20$ <em>subtract 8 from both sides</em>

$y=12$

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Kam had bought 10 packets of chocolates where one packet contains 15 small packs. He wants to share equally among his four child

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3 years ago

the line joining A(a, 3) to B(2 -3) is perpendicular to the line joining C(10,1) to B. The value of a is?

RideAnS [48]

Well, first off, let's find what is the slope of BC

$\bf \begin{array}{lllll}
&x_1&y_1&x_2&y_2\\
% (a,b)
B&({{ 2}}\quad ,&{{ -3}})\quad 
% (c,d)
C&({{ 10}}\quad ,&{{ 1}})
\end{array}
\\\\\\
% slope = m
slope = {{ m}}= \cfrac{rise}{run} \implies 
\cfrac{{{ y_2}}-{{ y_1}}}{{{ x_2}}-{{ x_1}}}\implies \cfrac{1-(-3)}{10-2}\implies \cfrac{1+3}{10-2}
\\\\\\
\cfrac{4}{8}\implies \cfrac{1}{2}$

now, a line perpendicular to that one, will have a negative reciprocal slope, thus

$\bf \textit{perpendicular, negative-reciprocal slope for slope}\quad \cfrac{1}{2}\\\\
slope=\cfrac{1}{{{ 2}}}\qquad negative\implies -\cfrac{1}{{{ 2}}}\qquad reciprocal\implies - \cfrac{{{ 2}}}{1}\implies \boxed{-2}$

now, we know the slope "m" of AB is -2 then, thus

$\bf \begin{array}{lllll}
&x_1&y_1&x_2&y_2\\
% (a,b)
A&({{ a}}\quad ,&{{ 3}})\quad 
% (c,d)
B&({{ 2}}\quad ,&{{ -3}})
\end{array}
\\\\\\
% slope = m
slope = {{ m}}= \cfrac{rise}{run} \implies 
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\\\\\\
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\\\\\\
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3 years ago

The figure below shows a straight line AB intersected by another straight line t:

BaLLatris [955]

When two straight lines intersect, the pairs of nonadjacent angles in opposite posi-tions are known as vertical angles.

If a segment AB is intersected by a transversal labeled t, then ∠1 and ∠3 and ∠2 and ∠4 are vertically angles formed by the transversal t on the segment AB.

Angles ∠1 and ∠2 can be described as adjacent and supplementary angles, so

$m\angle 1+m\angle 2=180^{\circ}$ .

Angles ∠3 and ∠2 can be also described as adjacent and supplementary angles, so

$m\angle 3+m\angle 2=180^{\circ}$ .

Subtract from the first equation the second equation:

$m\angle 1+m\angle 2-(m\angle 3+m\angle 2)=180^{\circ}-180^{\circ},\\ m\angle 1-m\angle 3=0,\\ m\angle 1=m\angle 3$ .

Similarly you can prove that $m\angle 2=m\angle 4$ .