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nirvana33 [79]
2 years ago
5

Triangle \triangle A'B'C'△A

Mathematics
1 answer:
Harrizon [31]2 years ago
6 0

From the information given, the rotated triangle is same as the original triangle. This is because, vertices of the rotation remain at the origin - (0,0).

<h3>What is the proof the above?</h3>

Notice that the question states:  △A'B'C' is the image of △ABC. Given that the △ without primes is the original and the one with primes depict the rotated triangle,

under a rotation about the origin, (0,0) (0,0) (0, 0), indicating no movement,

△A'B'C' is in the same position with △ABC.

Learn more about rotation at:
brainly.com/question/27915888
#SPJ1

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Hey can someone help me :)
Arada [10]
1.
Convert 22 miles to feet.


1 mile = 5280 feet

22 × 5280 = 116,160 feet

2.
To convert this you divide the speed by 60 since there's 60 minutes in an hour
125/60 = 2.0833....

Since you round the answer would be
2.1 miles per minute
6 0
3 years ago
Bob decided to give up a full-time salary of $45000 a year to go to school for 4 years. The total cost of going to school will n
den301095 [7]

Answer:

Step-by-step explanation:

Semester Costs = 8*2858 =                     22864

Books / semester= 8 * 391 =                        <u>3128</u>

Total                                                            25992

If he wants to repay all this in six years the answer would be

45000 + 25992/6 = 45000 + 4332 = 49332

3 0
3 years ago
F(x) = –16x2 24x 16
evablogger [386]
The x-intercepts are: -0.5 and 2.
3 0
3 years ago
An equation parallel and perpendicular to 4x+5y=19
UNO [17]

Answer:

Parallel line:

y=-\frac{4}{5}x+\frac{9}{5}

Perpendicular line:

y=\frac{5}{4}x-\frac{1}{2}

Step-by-step explanation:

we are given equation 4x+5y=19

Firstly, we will solve for y

4x+5y=19

we can change it into y=mx+b form

5y=-4x+19

y=-\frac{4}{5}x+\frac{19}{5}

so,

m=-\frac{4}{5}

Parallel line:

we know that slope of two parallel lines are always same

so,

m'=-\frac{4}{5}

Let's assume parallel line passes through (1,1)

now, we can find equation of line

y-y_1=m'(x-x_1)

we can plug values

y-1=-\frac{4}{5}(x-1)

now, we can solve for y

y=-\frac{4}{5}x+\frac{9}{5}

Perpendicular line:

we know that slope of perpendicular line is -1/m

so, we get slope as

m'=\frac{5}{4}

Let's assume perpendicular line passes through (2,2)

now, we can find equation of line

y-y_1=m'(x-x_1)

we can plug values

y-2=\frac{5}{4}(x-2)

now, we can solve for y

y=\frac{5}{4}x-\frac{1}{2}


4 0
3 years ago
I need help. Please answer if u do know
monitta

Answer:

i think is (-3,2) because it look like in negative and posted

3 0
2 years ago
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