keeping in mind that perpendicular lines have negative reciprocal slopes, let's check for the slope of the equation above

well then therefore

so we're really looking for the equation of a line with slope of -1/3 and that passes through (1, -3 )

Answer:
6
Step-by-step explanation:
Solve x then y after..
Answer:
45/28
Step-by-step explanation:
Convert them to mixed fractions
(2 * 4 + 1) / 4 ÷ (1 * 5 + 2) / 5
8 + 1 / 4 ÷ 5 + 2 / 5
9/4 ÷ 7/5
Flip the second fraction and flip the sign - to multiply.
9/4 × 5/7
9 × 5 = 45
4 × 7 = 28
45/28
This could be left as a fraction or converted to a number:
1.60714285714
I believe your answer would be <span>d.
BAEAB</span>
Answer:
1) 30
2) 60
Step-by-step explanation:
1. 90 + 2x + x = 180
solve for x
x = 30
plug in 30 for x
<em>hope this helps. I am in algebra two so you can trust my answer. Happy holidays stay safe</em>
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