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Gemiola [76]
3 years ago
14

The two triangles shown below are simiar. Which statement is true of the transformation from ABC to A' B' C'

Mathematics
1 answer:
AfilCa [17]3 years ago
8 0

Answer: Thus when transforming from ABC to A'B'C', the lengths are scaled by a factor of 0.5 .

Step-by-step explanation:

Since the triangles are similar, the ratio of their sides are equal.

And we can count the number of blocks over which AC and A'C' is drawn and take them to be their length,

Therefore,

AC = 16

A'C'= 8

Thus when transforming from ABC to A'B'C', the lengths are scaled by a factor of 0.5 .

Measuring the tans of the angles by taking the ratio of opposite by adjacent, we get,

tanA = \frac{10}{4}

tanA'=\frac{5}{2}

which means tanA= tanA'

The angles do not change.

Thus when transforming from ABC to A'B'C', the lengths are scaled by a factor of 0.5 .

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To discourage guessing on the multiple-choice exam a professor assigned 4 points for a correct answer, -4 points for an incorrec
Zanzabum

Answer:

35

Step-by-step explanation:

4 x 15 = 60

-4 x 5 = - 20

-1 x 5 = -5

60 - 20 - 5 = 35

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Suppose f(x) = x2 and g(x) = (3x)2. Which statement best compares the graph
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Answer:

Step-by-step explanation:

Please use " ^ " to indicate exponentiation:  f(x) = x^2 and g(x) = (3x)^2.

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3 years ago
Name 2 same side interior angles
nasty-shy [4]

Answer:

Lines a and b are parallel to each other! ... The same-side interior angle theorem states that when two lines that are parallel are intersected by a transversal line, the same-side interior angles that are formed are supplementary, or add up to 180 degrees.

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3 0
2 years ago
1.) What is the equation of the path of firework #1? Write your equation in general form.
valina [46]

1. I'm assumig that the paths are perfect parabolas

this means that their general forms can be written in y=ax^2+bx+c

it's easier to find vertex form first then expand to get general form

vertex form is y=a(x-h)^2+k where the vertex is (h,k) and a is a constant


firework #1

vertex is (10,50), so (h,k)=(10,50) and h=10, k=50

h_1=a(t-10)^2+50

to find the value of a, subsitute another point

(0,0)

0=a(0-10)^2+50)

0=100a+50

a=\frac{-1}{2}

so the equation in vertex form is h_1=\frac{-1}{2}(t-10)^2+50

expand to get general form

h_1=\frac{-1}{2}(t^2-20t+100)+50

h_1=\frac{-1}{2}t^2+10t-50+50

h_1=\frac{-1}{2}t^2+10t





2.

same as last time

vertex is (10,72) so (h,k)=(10,72) so h=10 and k=72

equation is h_2=a(t-10)^2+72

find a

use another point

(0,22)

22=a(0-10)^2+72

22=100a+72

-50=100a

a=\frac{-1}{2}

so the equation in vertex form is h_2=\frac{-1}{2}(t-10)^2+72




3.

range is the numbers that h is allowed to be

think about what h represents. it represents the height of the rocket

from the graph, we can see that the lowest possible height is 0yd and the highest height is 50yd

so range is 0 to 50 or 0≤h≤50


domain is the numbers that t is allowed to be

think about what t represents. it represents how long the rocket has been flying

it will stop flying when it hits the ground or at t=20

it starts flying at t=0

so domain is from 0 to 20 or 0≤t≤20

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6x^2 is the correct answer to the problem.
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