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blagie [28]
3 years ago
6

Andre constructs the image of line n under a dilation centered at point O that is not on line n. Becca construct a different ima

ge of line n under a dilation centered at point O. what can you conclude about andre's and becca's lines?
Mathematics
1 answer:
GenaCL600 [577]3 years ago
6 0
<em /><u><em /></u>I think that both lines will show only 1 line because 1 line will be on top of the other. I believe this because both lines are dilated with its center at point O which does not lie on line n. Whatever dilation Andre uses may also be the dilation used by Becca since their lines centered at point O. Both lines will have the same slope and share the same coordinates.
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Answer:

The y-intercept is 15 and the slope is 5.5

y=5.5x+15

Step-by-step explanation:

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Find the critical numbers of the function. (Enter your answers as a comma-separated list. If an answer does not exist, enter DNE
Viktor [21]

Answer:

<h2> (2, -1)</h2>

Step-by-step explanation:

Given the function f(x) = 8x³ − 12x² − 48x, <em>the critical point of the function occurs at its turning point i,e at f'(x) = 0</em>

First we have to differentiate the function as shown;

f'(x)= 3(8)x^{3-1}- 2(12)x^{2-1} - 48x^{1-1}\\  \\f'(x) = 24x^2 - 24x-48x^0\\\\f'(x) = 24x^2 - 24x-48\\\\At \ the\turning\ point\ f'(x)= 0\\24x^2 - 24x-48 = 0\\\\\\

Dividing \ through \ by \ 24\\\\x^2-x-2 = 0\\\\On \ factorizing\\\\x^2-2x+x-2 = 0\\\\x(x-2)+1(x-2) = 0\\\\(x-2)(x+1) = 0\\\\x-2 = 0 \ and \ x+1 = 0\\\\x = 2 \ and \ -1

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8 0
3 years ago
What equation can be used to find the length of AC
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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)

4 0
3 years ago
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