If triangle CDE is dilated by a scale factor of 1/5 with a center of dilation at vertex E, what is the area of triangle C'D'E'?

Mathematics

2 answers:

timurjin [86]3 years ago

8 0

The correct answer would be Choice B: 4 square units.

When the scale factor is 1/5, that means the lengths of the sides are 1/5 of the original size. So instead of having a base of 20 and height of 10, the new triangle has a base of 4 and a height of 2.

The area of the triangle is: (4 x 2) / 2 = 4

Rainbow [258]3 years ago

3 0

Answer:

B) 4 units2

Step-by-step explanation:

The resultant triangle frmo the dilatation would have 1/5 of the size on each size, in the original triangle the base is 10 units and the height is 20 units, so 1/5 of 10 would be 2, and 1/5 of 20 would be 4, the resultant triangle from the dilatation at 1/5 would have an area of 4 squared units.

The area of the triangle is calculated with the next formula:

$area=\frac{basexheight}{2} \\area=\frac{4x2}{2} \\Area=4u^{2}$

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Pls explain how you figured out both questions

Mariana [72]

Answer:

A=0.125

B=0.125

Step-by-step explanation

for both you would divide 0.5 by 4. you would cut 0.125 off of each side, but two sides together would 0.25

6 0

3 years ago

Solve the equation 6w2 – 7w – 20 = 0.

azamat

For this case we have the following quadratic equation:

$6w ^ 2-7w-20 = 0$

Where:

$a = 6\\b = -7\\c = -20$

Its roots will be given by:

$w = \frac {-b \pm \sqrt {b ^ 2-4 (a) (c)}} {2 (a)}\\w = \frac {- (- 7) \pm \sqrt {(- 7) ^ 2-4 (6) (- 20)}} {2 (6)}\\w = \frac {7 \pm \sqrt {49 + 480}} {12}\\w = \frac {7 \pm \sqrt {529}} {12}\\w = \frac {7 \pm23} {12}$

The roots are:

$w_ {1} = \frac {7 + 23} {12} = \frac {30} {12} = \frac {15} {6} = \frac {5} {2}\\w_ {2} = \frac {7-23} {12} = \frac {-16} {12} = - \frac {8} {6} = - \frac {4} {3}$