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Tutor profile: Hosam K.

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Hosam K.
Transmission Engineer at Telecomax
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Questions

Subject: Pre-Calculus

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Question:

Find the zeroes of the following equation : y = x^2 + 3x + 2

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Hosam K.
Answer:

Looking to the given equation , we need to find zeroes of the equation , i.e: find values of x that makes y = 0. Assigning y = 0 = x^2 + 3x + 2, we will need to factorize this equation. As it's a 2nd degree equation , it can be factorized into 2 functions of 1st degree. Thus , x^2 + 3x + 2 = (x+1) (x+2) So (x+1) (x+2) = 0 , Either , x+1 = 0 >> x = -1 , or , x+2=0 >> x = -2 Finally , the zeroes of given equation are { -1 , -2 }

Subject: Trigonometry

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Question:

Verify the following identity : cos^2(x) - sin^2(x) = 1 - 2sin^2(x)

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Hosam K.
Answer:

Starting on the left hand-side , we know the pythagorean identity which is : sin^2(x) + cos^2(x) = 1 ----------------(1) So , moving [ sin^2(x) ] to the right-hand side , we get : cos^2(x) = 1 - sin^2(x) -------------------(2) Substituting this result in the main equation : [ 1 - sin^2(x)] - - sin^2(x) = 1 - sin^2(x) - sin^2(x) = 1 - 2sin^2(x) ------------- Solution

Subject: Algebra

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Question:

Find the slope of this line : y=5x+12

Inactive
Hosam K.
Answer:

Firstly, this is an equation of first degree ,since (x) has power of 1 , so it has a standard form : y = mx + b where , m: is the slope of straight line , and , b: is the y intercept of the line. Comparing both equations : y = 5x + 12 y = mx + b , we can see that slope = m = 5 (Solution!!!!!)

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