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2 years ago

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15. The vertices of A DEF are D(0,0), E(0,7), and F(6,3).

1 answer:

Marat540 [252]2 years ago

6 0

Orthocenter of a triangle with one vertex at the origin D(0,0), E(0,7), F(6,3)

The orthocenter is the meet of the altitudes.

DE is the line x=0. That's the y axis. The altitude is the perpendicular through F(6,3), so is the horizontal line y=3.

DF is the line

6y=3x

y=(1/2)x

Perpendicular through E(0,7) is

y - 7 = (-2)(x - 0)

y = -2x + 7

That's the other altitude. Intersecting with y=3

3 = -2x + 7

-4 = -2x

x = 2

y = 3

Answer: orthocenter is (2,3)

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Help it’s for a quiz

Vilka [71]

Answer:

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2 years ago

Use cross products to demonstrate whether the two ratios are equivalent in this problem: 4 over 6 times 14 over 21

ExtremeBDS [4]

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3 years ago

Steve likes to entertain friends at parties with "wire tricks." Suppose he takes a piece of wire 60 inches long and cuts it into

Alex_Xolod [135]

Answer:

a) the length of the wire for the circle = $(\frac{60\pi }{\pi+4}) in$

b)the length of the wire for the square = $(\frac{240}{\pi+4}) in$

c) the smallest possible area = 126.02 in² into two decimal places

Step-by-step explanation:

If one piece of wire for the square is y; and another piece of wire for circle is (60-y).

Then; we can say; let the side of the square be b

so 4(b)=y

b= $\frac{y}{4}$

Area of the square which is L² can now be said to be;

$A_S=(\frac{y}{4})^2 = \frac{y^2}{16}$

On the otherhand; let the radius (r) of the circle be;

2πr = 60-y

$r = \frac{60-y}{2\pi }$

Area of the circle which is πr² can now be;

$A_C= \pi (\frac{60-y}{2\pi } )^2$

$=( \frac{60-y}{4\pi } )^2$

Total Area (A);

A = $A_S+A_C$

= $\frac{y^2}{16} +(\frac{60-y}{4\pi } )^2$

For the smallest possible area; $\frac{dA}{dy}=0$

∴ $\frac{2y}{16}+\frac{2(60-y)(-1)}{4\pi}=0$

If we divide through with (2) and each entity move to the opposite side; we have:

$\frac{y}{18}=\frac{(60-y)}{2\pi}$

By cross multiplying; we have:

2πy = 480 - 8y

collect like terms

(2π + 8) y = 480

which can be reduced to (π + 4)y = 240 by dividing through with 2

$y= \frac{240}{\pi+4}$

∴ since $y= \frac{240}{\pi+4}$ , we can determine for the length of the circle ;

60-y can now be;

= $60-\frac{240}{\pi+4}$

= $\frac{(\pi+4)*60-240}{\pi+40}$

= $\frac{60\pi+240-240}{\pi+4}$

= $(\frac{60\pi}{\pi+4})in$

also, the length of wire for the square (y) ; $y= (\frac{240}{\pi+4})in$

The smallest possible area (A) = $\frac{1}{16} (\frac{240}{\pi+4})^2+(\frac{60\pi}{\pi+y})^2(\frac{1}{4\pi})$

= 126.0223095 in²

≅ 126.02 in² ( to two decimal places)

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3 years ago

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vitfil [10]

Z= -7
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2 years ago

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maksim [4K]

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2 years ago