paper 3b - robles.pdf
TRANSCRIPT
ESTANCIA NATIONAL HIGH SCHOOL
Paper 3b
POWERPOINTPRESENTATION
By:
ADA MAE I. ROBLES
ALGEBRA(Special Products and Factoring,
Rational Algebraic Expressions,
And Sets)
ESTANCIA NATIONAL HIGH SCHOOL
Paper 3b - PowerPoint Presentation
DIRECTIONS: Answer the following then choose the letter of the correct answer.
1.) The simplest form of 18๐ฅ3๐ฆ5
27๐ฅ4๐ฆ2 is _________.
a.2๐ฆ3
3๐ฅb.
6๐ฆ3
9๐ฅ
c.6๐ฅ
9๐ฆ3 d.2๐ฅ
3๐ฆ3
To reveal the solution CLICK HERE
ESTANCIA NATIONAL HIGH SCHOOL
ESTANCIA NATIONAL HIGH SCHOOL
Paper 3b - PowerPoint Presentation (Answer Key)
ANSWER KEY:
1.) The simplest form of 18๐ฅ3๐ฆ5
27๐ฅ4๐ฆ2 is _________.
Solution:
18๐ฅ3๐ฆ5
27๐ฅ4๐ฆ2 =9โ2โ๐ฅ3โ๐ฆ2โ๐ฆ3
9โ3โ๐ฅ3โ๐ฅโ๐ฆ2 =๐๐๐
๐๐
Answer: a
ESTANCIA NATIONAL HIGH SCHOOL
Paper 3b - PowerPoint Presentation
2.) Factor 16๐ฅ8 โ 625๐ฆ16 completely.
a. (4๐ฅ2โ25๐ฆ4)(4๐ฅ2 + 25๐ฆ4)
b. (4๐ฅ4โ25๐ฆ8)(4๐ฅ4 + 25๐ฆ8)
c. (2๐ฅ2+5๐ฆ4) 2๐ฅ2 โ 5๐ฆ4 (4๐ฅ4 + 25๐ฆ8)
d. (2๐ฅ2+5๐ฆ4) 2๐ฅ2 โ 25๐ฆ4
(2๐ฅ2+5๐ฆ4) 2๐ฅ2 + 25๐ฆ4
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ESTANCIA NATIONAL HIGH SCHOOL
Paper 3b - PowerPoint Presentation (Answer Key)
2.) Factor 16๐ฅ8 โ 625๐ฆ16 completely.
Solution:
To factor, use difference of two squares:
16๐ฅ8 โ 625๐ฆ16 = 4๐ฅ4 + 25๐ฆ8 4๐ฅ4 โ 25๐ฆ8
= ๐๐๐ + ๐๐๐๐ ๐๐๐ + ๐๐๐ ๐๐๐ โ ๐๐๐
Answer: c
ESTANCIA NATIONAL HIGH SCHOOL
Paper 3b - PowerPoint Presentation
3.) What is the surface area of the given cube? (x+3y)cm
a. 6๐ฅ + 18๐ฆ ๐๐2
b. (๐ฅ2+6๐ฅ๐ฆ +9๐ฆ2)๐๐2
c. (6๐ฅ2+26๐ฅ๐ฆ + 54๐ฆ2)๐๐2
d. (๐ฅ3+9๐ฅ2๐ฆ + 27๐ฅ๐ฆ2 + 27๐ฆ3)๐๐2
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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 ๐ด = ๐ 2, we have, ๐ด = (๐ฅ + 3๐ฆ)2,
then, use special product โsquare of a binomialโ
๐ด = ๐ฅ2 + 2 ๐ฅ 3๐ฆ + ๐ฆ2 thus, ๐จ = ๐๐ + ๐๐๐ + ๐๐
Answer: b
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. โ17, 5, 2 respectively
b. 17, 5, 2 respectively
c. 17, 5, 5 respectively
d. โ17, 5, 5 respectively
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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.
๐ ๐ฅ = 4๐ฅ2 + 1 ๐ ๐ฅ = 4๐ฅ2 + 1 ๐ ๐ฅ = 4๐ฅ2 + 1
๐ โ2 = 4 4 + 1 ๐ 1 = 4(1)2 + 1 ๐1
2= 4(
1
2)2 + 1
๐ โ2 = 4(โ2)2 + 1 ๐ โ2 = 4(1) + 1 ๐1
2= 4(
1
4) + 1
๐ โ2 = 16 + 1 = ๐๐ ๐ โ2 = 4 + 1 = ๐ ๐ โ2 = 1 + 1 = ๐
Answer: b
ESTANCIA NATIONAL HIGH SCHOOL
Paper 3b - PowerPoint Presentation
5.) Simplify: ๐ฆโ5
๐ฆ+3=
๐ฆ+6
๐ฆโ2.
a. ๐ฆ =7
4b. ๐ฆ =
โ7
4
c. ๐ฆ =1
2d. ๐ฆ =
โ1
2
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G O O D L U C K !!!
ESTANCIA NATIONAL HIGH SCHOOL
Paper 3b - PowerPoint Presentation (Answer Key)
5.) Simplify: ๐ฆโ5
๐ฆ+3=
๐ฆ+6
๐ฆโ2
Solution: ๐ฆ โ 5 ๐ฆ โ 2 = ๐ฆ + 3 ๐ฆ + 6
๐ฆ2 โ 7๐ฆ + 10 = ๐ฆ2 + 9๐ฆ + 18
๐ฆ2 โ ๐ฆ2 โ 7๐ฆ โ 9๐ฆ = 18 โ 10โ16๐ฆ = 8
Answer: d ๐ฆ =8
โ16=
โ๐
๐
THANK YOU!