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dangina [55]
3 years ago
9

Value: 10

Mathematics
1 answer:
xxMikexx [17]3 years ago
6 0
Não vale 10 pontos só 5 ok letra b
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Find the point P on the line y=2x that is closest to the point (20,0) .what is the least distance between pans (20,0)?
Vsevolod [243]
The closest point on a line, to another point, will be a point that's on a normal of that line, or a line that is perpendicular to it, notice picture below

so, y = 2x, has a slope of 2, a perpendicular line to it, will have a slope of negative reciprocal that, or \bf 2\qquad negative\implies -2\qquad reciprocal\implies \cfrac{1}{-2}\implies -\cfrac{1}{2}

so, we know that line passes through the point 20,0, and has a slope of -1/2

if we plug that in the point-slope form, we get \bf y-0=-\cfrac{1}{2}(x-20)\implies y=-\cfrac{1}{2}x+10

now, the point that's on 2x and is also on that perpendicular line, is the closest to 20,0 from 2x, thus, is where both graphs intersect, as you can see in the graph

thus  \bf 2x=-\cfrac{1}{2}x+10  solve for "x'

------------------------------------------------------------

not sure on the 2nd part, but sounds like, what's the distance from that point to 20,0, well, if that's the case, just use the distance equation

\bf \textit{distance between 2 points}\\ \quad \\
\begin{array}{lllll}
&x_1&y_1&x_2&y_2\\
%  (a,b)
&({{ \square }}\quad ,&{{ \square }})\quad 
%  (c,d)
&({{ \square }}\quad ,&{{ \square }})
\end{array}\qquad 
%  distance value
d = \sqrt{({{ x_2}}-{{ x_1}})^2 + ({{ y_2}}-{{ y_1}})^2}


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3 years ago
A 12 foot ladder is leaning against the side of a building. The top of the ladder reaches 10 feet up the side of the building. A
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√135=3√15.....................

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The diagonals of a trapezoid are perpendicular <br> Always <br> Sometimes <br> Never
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Which represents r in terms of A and S?
ladessa [460]

Answer:

r=\frac{6\sqrt{10AS}}{S\sqrt{\pi }}

Step-by-step explanation:

Given the formula;

A=\frac{\pi r^2S}{360}

We want to solve the given formula for r.

Multiply both sides by \frac{360}{\pi S}

A\times \frac{360}{\pi S}=\frac{\pi r^2S}{360} \times \frac{360}{\pi S}

A\times \frac{360}{\pi S}=r^2

Take square root of both sides

r=\sqrt{\frac{360A}{\pi S}}

r=\frac{\sqrt{360A}}{\sqrt{\pi S}}

r=\frac{\sqrt{360A}}{\sqrt{\pi }\sqrt{S}}

r=\frac{\sqrt{360A}\times \sqrt{S}}{\sqrt{\pi }\sqrt{S} \times \sqrt{S}}

r=\frac{\sqrt{360AS}}{S\sqrt{\pi }}

r=\frac{6\sqrt{10AS}}{S\sqrt{\pi }}

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3 years ago
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