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ESTANCIA NATIONAL HIGH SCHOOL
Paper 3b
POWERPOINTPRESENTATION
By:
ADA MAE I. ROBLES
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ALGEBRA(Special Products and Factoring,
Rational Algebraic Expressions,
And Sets)
ESTANCIA NATIONAL HIGH SCHOOL
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Paper 3b - PowerPoint Presentation
DIRECTIONS: Answer the following then choose the letter of the correct answer.
1.) The simplest form of 18π₯3π¦5
27π₯4π¦2 is _________.
a.2π¦3
3π₯b.
6π¦3
9π₯
c.6π₯
9π¦3 d.2π₯
3π¦3
To reveal the solution CLICK HERE
ESTANCIA NATIONAL HIGH SCHOOL
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ESTANCIA NATIONAL HIGH SCHOOL
Paper 3b - PowerPoint Presentation (Answer Key)
ANSWER KEY:
1.) The simplest form of 18π₯3π¦5
27π₯4π¦2 is _________.
Solution:
18π₯3π¦5
27π₯4π¦2 =9β2βπ₯3βπ¦2βπ¦3
9β3βπ₯3βπ₯βπ¦2 =πππ
ππ
Answer: a
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ESTANCIA NATIONAL HIGH SCHOOL
Paper 3b - PowerPoint Presentation
2.) Factor 16π₯8 β 625π¦16 completely.
a. (4π₯2β25π¦4)(4π₯2 + 25π¦4)
b. (4π₯4β25π¦8)(4π₯4 + 25π¦8)
c. (2π₯2+5π¦4) 2π₯2 β 5π¦4 (4π₯4 + 25π¦8)
d. (2π₯2+5π¦4) 2π₯2 β 25π¦4
(2π₯2+5π¦4) 2π₯2 + 25π¦4
To reveal the solution CLICK HERE
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ESTANCIA NATIONAL HIGH SCHOOL
Paper 3b - PowerPoint Presentation (Answer Key)
2.) Factor 16π₯8 β 625π¦16 completely.
Solution:
To factor, use difference of two squares:
16π₯8 β 625π¦16 = 4π₯4 + 25π¦8 4π₯4 β 25π¦8
= πππ + ππππ πππ + πππ πππ β πππ
Answer: c
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ESTANCIA NATIONAL HIGH SCHOOL
Paper 3b - PowerPoint Presentation
3.) What is the surface area of the given cube? (x+3y)cm
a. 6π₯ + 18π¦ ππ2
b. (π₯2+6π₯π¦ +9π¦2)ππ2
c. (6π₯2+26π₯π¦ + 54π¦2)ππ2
d. (π₯3+9π₯2π¦ + 27π₯π¦2 + 27π¦3)ππ2
To reveal the solution CLICK HERE
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ESTANCIA NATIONAL HIGH SCHOOL
Paper 3b - PowerPoint Presentation (Answer Key)
3.) What is the surface area of the given cube?Solution: (x+3y)cm
Use the formula π΄ = π 2, we have, π΄ = (π₯ + 3π¦)2,
then, use special product βsquare of a binomialβ
π΄ = π₯2 + 2 π₯ 3π¦ + π¦2 thus, π¨ = ππ + πππ + ππ
Answer: b
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ESTANCIA NATIONAL HIGH SCHOOL
Paper 3b - PowerPoint Presentation
4.) Given the function f(x) = 4x2 + 1,
evaluate the following: f (-2), f (1), and f (1
2).
a. β17, 5, 2 respectively
b. 17, 5, 2 respectively
c. 17, 5, 5 respectively
d. β17, 5, 5 respectively
To reveal the solution CLICK HERE
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ESTANCIA NATIONAL HIGH SCHOOL
Paper 3b - PowerPoint Presentation (Answer Key)
4.) Given the function f(x) = 4x2 + 1, evaluate the following:
f (-2), f (1), and f (1
2).
Solution:
Substitute the value of x in the given function.
π π₯ = 4π₯2 + 1 π π₯ = 4π₯2 + 1 π π₯ = 4π₯2 + 1
π β2 = 4 4 + 1 π 1 = 4(1)2 + 1 π1
2= 4(
1
2)2 + 1
π β2 = 4(β2)2 + 1 π β2 = 4(1) + 1 π1
2= 4(
1
4) + 1
π β2 = 16 + 1 = ππ π β2 = 4 + 1 = π π β2 = 1 + 1 = π
Answer: b
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ESTANCIA NATIONAL HIGH SCHOOL
Paper 3b - PowerPoint Presentation
5.) Simplify: π¦β5
π¦+3=
π¦+6
π¦β2.
a. π¦ =7
4b. π¦ =
β7
4
c. π¦ =1
2d. π¦ =
β1
2
To reveal the solution CLICK HERE
G O O D L U C K !!!
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ESTANCIA NATIONAL HIGH SCHOOL
Paper 3b - PowerPoint Presentation (Answer Key)
5.) Simplify: π¦β5
π¦+3=
π¦+6
π¦β2
Solution: π¦ β 5 π¦ β 2 = π¦ + 3 π¦ + 6
π¦2 β 7π¦ + 10 = π¦2 + 9π¦ + 18
π¦2 β π¦2 β 7π¦ β 9π¦ = 18 β 10β16π¦ = 8
Answer: d π¦ =8
β16=
βπ
π
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THANK YOU!