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Natalka [10]
3 years ago
14

How do I sketch the graph for y= 3/8x + 5 ​

Mathematics
2 answers:
tester [92]3 years ago
4 0

Answer:

Step-by-step explanation:

start at y intercept +5. Go up 3 points then go to the right 8 points and keep on doing that.  Formula for slope rise/run. Rise is 3. Run is 8. Then connect the line.

diamong [38]3 years ago
4 0

Answer:

go to desmos, copy and paste does numbers, and you'll get the answer for free, trust me is a good calculator

explanation:

sketch it exactly like it shows

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SERIOULSLY WOULD SOMEONE JUST HELP ME WITH THESE QUESTIONS FOR ONCE? BY THAT I MEAN NOT ONLY ANSWER 23 AND 22 ALSO 24, 25, 26,27
Fantom [35]

Answer:

Step-by-step explanation:

22. √125 is closer to √121 or 11

23. √23.5 is closer to √25 or 5

24. ∛59 is closer to ∛64 or 4 (cubed)

(4 cubed = 64, 3 cubed = 27)

25. ∛430 is closer to ∛512 or 8

(512 - 430 = 82

430- 343= 87)

26. y² = 55 is equal to y = √55,  Closer to √49 or 7

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4 0
3 years ago
Calculate pt3 such that a line from pt1 to pt3 is perpendicular to the line from pt1 to pt2, and the distance between pt1 and pt
Leni [432]
Let the point_1 = p₁ = (1,4)
and      point_2 = p₂ = (-2,1)
and      Point_3 = p₃ = (x,y)

The line from point_1 to point_2 is L₁ and has slope = m₁
The line from point_1 to point_3 is L₂ and has slope = m₂
m₁ = Δy/Δx = (1-4)/(-2-1) = 1
m₂ = Δy/Δx = (y-4)/(x-1)
L₁⊥L₂ ⇒⇒⇒⇒ m₁ * m₂ = -1
∴ (y-4)/(x-1) = -1 ⇒⇒⇒ (y-4)= -(x-1)
(y-4) = (1-x) ⇒⇒⇒⇒⇒⇒⇒⇒⇒⇒⇒⇒⇒⇒ equation (1)

The distance from point_1 to point_2 is d₁
The distance from point_1 to point_3 is d₂
d = \sqrt{Δx^2+Δy^2}
d₁ = \sqrt{(-2-1)^2+(1-4)^2}
d₂ = \sqrt{(x-1)^2+(y-4)^2}
d₁ = d₂
∴ \sqrt{(-2-1)^2+(1-4)^2} = \sqrt{(x-1)^2+(y-4)^2} ⇒⇒ eliminating the root
∴(-2-1)²+(1-4)² = (x-1)²+(y-4)²
 (x-1)²+(y-4)² = 18
from equatoin (1)  y-4 = 1-x
∴(x-1)²+(1-x)² = 18            ⇒⇒⇒⇒⇒ note: (1-x)² = (x-1)²
2 (x-1)² = 18
(x-1)² = 9
x-1 = \pm \sqrt{9} = \pm 3
∴ x = 4 or x = -2
∴ y = 1 or y = 7

Point_3 = (4,1)  or  (-2,7)












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