choose the correct answer with Reason
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) Option (A) and option (B) are not polynomials.
Beacuse thier degree in non positive.
2) Option (C) - 4.
We know that highest power of a polynomial is its degree.
3) Option (A) and Option (D)
a²x = 0
x = 0/a²
x = 0.
4) Option (A)
The remaining has the power in negative.
5) Option (B) - 9
=5x²-3x+7
=5(1)²-3(1)+7
=9
6) x³+1 = (x+1) (x²-x+1)
Option (A)
7) we know that x³+y³+z³=3xyz
x³+y³+2³=3(x)(y)(2) = 6xy
So, option (C)
8) Option (D)
f(2) = 2(2²)+3(2)-p = 8+6-p = 14-p
So, p = 14.
9) Option (D) Not a polynomial.
Beacuse its degree is in non positive.
10) Option (C) -3 and 7.
11) X-2 is a factor. x-2 = 0
x=2.
So, p(2). Option (B)
12) 2(a²+b²) - a²-2ab-b² =0
2a²+2b²-a²-2ab-b²=0
a²+b²-2ab=0
(a-b)²=0
a=b
So, option (B).
13) (x-1) is a factor. so x = 1
-1²+3(-1)²+3(-1)-1
= 1+3-3+1
= 0
So, option (C)
14) It's in the form a²-b²
= a+b(a-b)
=525+475(525-475)
= 1000(50)
= 50000.
There is no option. This is the right answer.
15) a³+b³=(a+b)³-3 ab(a + b).
a³+b³-3ab=(a+b)³-3 ab(a + b)-3ab
=-1²-3ab(-1)-3ab
=-1+3ab-3ab
= -1
So, Option (D)
16) Option (C)
a+b+c=6
2+2+2=6
Substituting in that we get 0.
17) a/b+b/a =1
a²+b²/ab = 1
a²+b²=ab
a²+b²-ab=0
a³-b³=(a-b) ( a²+b²-ab)
= 0
So, Option (D)
18) x= 1/ 2-√3
= 1(2+√3)/2²-√3²
= 2+√3
x²-4x+1 = (2+√3)²-4(2+√3)+1
= 4+3+2(2)(√3)-8-4√3+1
=7+4√3-4√3-7
= 0
Option (C)
19) Option (D)- 3 Zeros.
It's a cubic polynomial,so it has 3 Zeros.
20)
It means 2 and -2 are the zeros.
p(2) ax⁴+2x-3x²+bx-4=0
a2⁴+2(2)-3(2)²+b(2)-4 =0
16a+4-12+2b-4=0
16a+2b=12
2(8a+b)=12
8a+b=6 ----------(1)
p(-2) a(-2)⁴+2(-2)-3(-2)²+b(-2)-4=0
16a-4-12-2b-4=0
16a-2b = 20
8a-b = 10 -------(2)
Solving them we get,
a=1 and b=6
a+b = 1+6=7
:)
) Option (A) and option (B) are not polynomials.
Beacuse thier degree in non positive.
2) Option (C) - 4.
We know that highest power of a polynomial is its degree.
3) Option (A) and Option (D)
a²x = 0
x = 0/a²
x = 0.
4) Option (A)
The remaining has the power in negative.
5) Option (B) - 9
=5x²-3x+7
=5(1)²-3(1)+7
=9
6) x³+1 = (x+1) (x²-x+1)
Option (A)
7) we know that x³+y³+z³=3xyz
x³+y³+2³=3(x)(y)(2) = 6xy
So, option (C)
8) Option (D)
f(2) = 2(2²)+3(2)-p = 8+6-p = 14-p
So, p = 14.
9) Option (D) Not a polynomial.
Beacuse its degree is in non positive.
10) Option (C) -3 and 7.
11) X-2 is a factor. x-2 = 0
x=2.
So, p(2). Option (B)
12) 2(a²+b²) - a²-2ab-b² =0
2a²+2b²-a²-2ab-b²=0
a²+b²-2ab=0
(a-b)²=0
a=b
So, option (B).
13) (x-1) is a factor. so x = 1
-1²+3(-1)²+3(-1)-1
= 1+3-3+1
= 0
So, option (C)
14) It's in the form a²-b²
= a+b(a-b)
=525+475(525-475)
= 1000(50)
= 50000.
There is no option. This is the right answer.
15) a³+b³=(a+b)³-3 ab(a + b).
a³+b³-3ab=(a+b)³-3 ab(a + b)-3ab
=-1²-3ab(-1)-3ab
=-1+3ab-3ab
= -1
So, Option (D)
16) Option (C)
a+b+c=6
2+2+2=6
Substituting in that we get 0.
17) a/b+b/a =1
a²+b²/ab = 1
a²+b²=ab
a²+b²-ab=0
a³-b³=(a-b) ( a²+b²-ab)
= 0
So, Option (D)
18) x= 1/ 2-√3
= 1(2+√3)/2²-√3²
= 2+√3
x²-4x+1 = (2+√3)²-4(2+√3)+1
= 4+3+2(2)(√3)-8-4√3+1
=7+4√3-4√3-7
= 0
Option (C)
19) Option (D)- 3 Zeros.
It's a cubic polynomial,so it has 3 Zeros.
20)
It means 2 and -2 are the zeros.
p(2) ax⁴+2x-3x²+bx-4=0
a2⁴+2(2)-3(2)²+b(2)-4 =0
16a+4-12+2b-4=0
16a+2b=12
2(8a+b)=12
8a+b=6 ----------(1)
p(-2) a(-2)⁴+2(-2)-3(-2)²+b(-2)-4=0
16a-4-12-2b-4=0
16a-2b = 20
8a-b = 10 -------(2)
Solving them we get,
a=1 and b=6
a+b = 1+6=7
:)
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