Step-by-step explanation:
Hey there!
The coordinates of a triangle are;
- A (1,2)
- B (1,-2)
- C (-2,-2)
- Scale factor = -4 and centre at (0,0)
Now, We have;
P(x,y)----------- P'(kx, ky)
<u>Keep</u><u> </u><u>formula</u><u>;</u>
<u>A</u><u>(</u><u>1</u><u>,</u><u>2</u><u>)</u><u>----------</u><u> </u><u>A'</u><u>(</u><u>-4</u><u>×</u><u>1</u><u> </u><u>,</u><u>-4</u><u>×</u><u>2</u><u>)</u>
= A'(-4, -8)
B(1,-2)------------- B'(-4,8)
C(-2,-2)-------------C'(8,8)
[ <u>The</u><u> </u><u>A'</u><u> </u><u>should</u><u> </u><u>be</u><u> </u><u>kept</u><u> </u><u>in</u><u> </u><u>(</u><u>-4</u><u>,</u><u>-</u><u>8</u><u>)</u><u>.</u><u> </u><u>There</u><u> </u><u>was</u><u> </u><u>no</u><u> </u><u>place</u><u> </u><u>in</u><u> </u><u>this</u><u> </u><u>graph</u><u> </u><u>so</u><u>,</u><u> </u><u>i</u><u> </u><u>kept</u><u> </u><u>it</u><u> </u><u>just</u><u> </u><u>below</u><u> </u><u>-7</u><u>.</u><u> </u><u>Please</u><u> </u><u>while</u><u> </u><u>plotting</u><u> </u><u>remember</u><u> </u><u>to</u><u> </u><u>keep</u><u> </u><u>it</u><u> </u><u>on</u><u> </u><u>"</u><u>-8"</u><u>.</u><u>]</u>
<em><u>Hope it helps</u></em><em><u>.</u></em><em><u>.</u></em>
Answer:
6b+3m
Step-by-step explanation:
I hope this helped.
Answer:
see explanation
Step-by-step explanation:
All of these questions use the external angle theorem, that is
The external angle of a triangle is equal to the sum of the 2 opposite interior angles.
18
∠3 = 43° + 22° = 65°
19
∠2 + 71 = 92 ( subtract 71 from both sides )
∠2 = 21°
20
90 + ∠4 = 123 ( subtract 90 from both sides )
∠4 = 33°
21
2x - 15 + x - 5 = 148
3x - 20 = 148 ( add 20 to both sides )
3x = 168 ( divide both sides by 3 )
x = 56
Hence ∠ABC = x - 5 = 56 - 5 = 51°
22
2x + 27 + 2x - 11 = 100
4x + 16 = 100 ( subtract 16 from both sides )
4x = 84 ( divide both sides by 4 )
x = 21
Hence ∠JKL = 2x - 11 = (2 × 21) - 11 = 42 - 11 = 31°
C = 3.5t....when t = 15
c = 3.5(15)
c = 52.50 <==
Add 4 to both sides
x < -5
Closed circle at x = 5, line points left