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Radda [10]
2 years ago
14

Need help solving pleaseee!!!

Mathematics
1 answer:
natta225 [31]2 years ago
5 0

Answer:

Step-by-step explanation:

50.

line AB that has the points (2,5) and (9,2) hhas the slope

m= (y2-y1) / (x2-x1)= 5-2 / 2-9 = 3/-7 = -3/7

Parallel line have the same slope so m= -3/7 for the line that has the point (-3,4).

Point-slope equation is y-y1 = m (x-x1), y -4 = -3/4 (x+3) or

you can writte it as y= (-3/4)x - 9/4 +16/4 so y = (-3/4)x + 7/4

51.

line PQ has slope m= -6+1 / 4-6 = -5/-2 = 5/2

Perpendicular lines have their slope negative reciprocal of eachother so the line that goes trough the point (1,3) has the slope m= -2/5

y-3 = -2/5 (x-1)

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Answer:

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Step-by-step explanation:

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8 0
3 years ago
-10x-15y=-30 find y<br><br>show work please
Effectus [21]

-10x - 15y = -30

-10x + 30 = 15y

(30 - 10x) / 15 = y

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Ms. Masse's biology class is conducting an experiment
olga_2 [115]
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2 years ago
Will give Brainliest to whoever helps me.
il63 [147K]

Answer:

  1. B
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Step-by-step explanation:

8 0
2 years ago
Read 2 more answers
‼️‼️Help anyone ‼️‼️‼️
Debora [2.8K]
Method 1
You can use <span>The equilateral triangle </span>(look at the picture)
\dfrac{a\sqrt3}{2}=1\ \ \ \cdot2\\\\a\sqrt3=2\ \ \ |\cdot\sqrt3\\\\3a=2\sqrt3\ \ \ |:3\\\\a=\dfrac{2\sqrt3}{3}\\\\\boxed{|AC|=\dfrac{2\sqrt3}{3}}


Method 2
You can use the <span>trigonometric function</span>
\sin60^o=\dfrac{1}{|AC|}\\\\\sin60^o=\dfrac{\sqrt3}{2}\\\\\dfrac{1}{|AC|}=\dfrac{\sqrt3}{2}\ \ \ |\cross\ multiply\\\\|AC|\sqrt3=2\ \ \ |\cdot\sqrt3\\\\3|AC|=2\sqrt3\ \ \ \ |:3\\\\\boxed{|AC|=\dfrac{2\sqrt3}{3}}
<span />

4 0
3 years ago
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