Find the equation of the line passing through the points (3,-2) and (4,6) in slope intercept form.
1 answer:
Hey!
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They already give us the slope, so we can solve the y-intercept.
We can use (3, -2).
y = 8x + b
-2 = 8(3) + b
-2 = 24 + b
-2 - 24 = 24 + b - 24
-26 = b
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Hence, the answer is
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Hope This Helped! Good Luck!
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I am too lazy to do question 6 if u want the answer tell me
Answer:
11 ;
Step-by-step explanation:
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30 27 22 25 24 25 24 15 35 35 33 52 49 10 27 18 20 23 24 25 30 24 24 24 18 20 25 27 24 32 13 13 21 2 37 35 32 33 29 3 28 28 25 29 31
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Class width = (52 - 2) / 5 = 50/5 = 10 + 1 = 11
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Answer: ∠A=48°,∠B=48°,∠C=84°.
Step by-step explanation:
Given: AD and BE are the angle bisectors of ∠A and ∠B
i.e ∠6=∠7 ( ∵ Angles formed after AD bisected ∠A)
∠4=∠5 ( ∵ Angles formed after BE bisected ∠B)
Also, DE║AB
⇒ ∠2=∠7 (∵ Alternate interior angles)
∠3=∠6 (∵ Alternate interior angles)
And ∠ADE : ∠ADB =∠2:∠3= 2:9 =2x : 9x ..(1)
To Find: ∠A,∠B,∠C.
Solution: ∠2=∠7 (∵ Given) ...(2)
∠2=∠4 (∵ angles on the same segment) ...(3)
∠4=∠5 =∠B/2 (∵ Given) ...(4)
∴ In Δ ABD
∠3+∠4+∠5+∠7 = 180 (∵ Sum of interior angles of a triangle)
From equation 2,3,4,5, Put values
9x+2x+2x+2x =180°
⇒15x = 180°
⇒x=12°
Putting values in equation (4) ⇒ ∠ B =2*(2*12) = 48°
Also, <u>∠B=∠A=48°</u>
Now,in Δ ABC
∠C+∠B+∠A= 180°
⇒48°+48°+∠C= 180°
<u><em>⇒∠C=84°</em></u>
