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2 years ago

7

Triangle RST was dilated by a scale factor of One-half. The image, triangle R'S'T', is an isosceles triangle, with each leg meas

uring 8 units.
Point Q is the center of dilation. Isosceles triangle R S T was dilated to smaller triangle R prime S prime T prime. Sides R S and R T are 8 units long.

What is the length of a leg of the pre-image, triangle RST?

4 units
8 units
16 units
20 units

2 answers:

andrew11 [14]2 years ago

8 0

Answer:

The length side of the pre-image is 16 units

Step-by-step explanation:

we know that

The length side of the image is equal to the length side of the pre-image multiplied by the scale factor

or

The length side of the pre-image is equal to the length side of the image divided by the scale factor

in this problem we have that

The scale factor is 1/2

The length side of the image is 8 units

therefore

8/(1/2)=16 units

The length side of the pre-image is 16 units

DiKsa [7]2 years ago

4 0

Answer:

16

Step-by-step explanation:

just took the test (:

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A cardboard box without a lid is to have a volume of 16,384 cm3. Find the dimensions that minimize the amount of cardboard used.

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Answer:

The dimensions that minimize the amount of cardboard used is

x = 31 cm , y = 34 cm & Z = 15.54 cm

Step-by-step explanation:

Volume of the cardboard = 16,384 $cm^{3}$

The function that represents the area of the cardboard without a lid is given by

$f (x,y,z) = xy + 2xz + 2yz$ ------ (1)

Volume of the cardboard with sides x, y & z is

$xyz = 16384$

$z = \frac{16384}{xy}$

Put this value of z in equation (1) we get

$f (x,y,z) = xy + 2x(\frac{16384}{xy} ) + 2y(\frac{16384}{xy} )$

$f (x,y,z) = xy + \frac{32768}{y} + 2y(\frac{32768}{x} )$

Differentiate above equation with respect to x & y we get

$f_{x} = y - \frac{32768}{x^{2} }$

$f_{y} = x - \frac{32768}{y^{2} }$

Take $f_{x} = 0 \ and \ f_{y} = 0$

$y - \frac{32768}{x^{2} } = 0$

$y = 32768 \ x^{-2}$ ------ (2)

$x - \frac{32768}{y^{2} } = 0$

$x = 32768 \ y^{-2}$ ------- (3)

By solving equation (2) & (3) we get

$x^{3} = 32768$

x = 31 cm

From equation 2

$y = 32768 \ x^{-2}$

y = 32768 ( $31^{-2}$ )

y = 34 cm

$z = \frac{16384}{xy}$

$z = \frac{16384}{(34)(31)}$

Z = 15.54 cm

Thus the dimensions that minimize the amount of cardboard used is

x = 31 cm , y = 34 cm & Z = 15.54 cm

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