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ivann1987 [24]
3 years ago
14

NEED HELP ASAP AGAIN

Mathematics
2 answers:
balu736 [363]3 years ago
8 0
4)
A=1/2 bh
A = 1/2(6)(3)
A = 9 in^2 

5)
A=1/2 bh
A=1/2(7)(10)
A = 35 m^2

6)
A=1/2 bh
A=1/2(5)(8)
A=20 ft^2

7)
A=1/2 (b1+b2)h
A=1/2 (5+10)8
A=60 in^2 

8)
A=1/2 (b1+b2)h
A=1/2 (13+19) 10
A=160 mm^2

9)
A=1/2 (b1+b2)h
A=1/2 (3+6) 2
A = 9 ft^2


Reptile [31]3 years ago
7 0
4)A=1/2 bh
A = 1/2(6)(3)
A = 9 in^2 

5)A=1/2 bh
A=1/2(7)(10)
A = 35 m^2

6)A=1/2 bh
A=1/2(5)(8)
A=20 ft^2

7)A=1/2 (b1+b2)h
A=1/2 (5+10)8
A=60 in^2 

8)A=1/2 (b1+b2)h
A=1/2 (13+19) 10
A=160 mm^2

9)A=1/2 (b1+b2)h
A=1/2 (3+6) 2
A = 9 ft^2

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Verify that 2,-1 and 1⁄2 are the zeroes of the cubic polynomial
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Answer:

i) Since P(2), P(-1) and P(½) gives 0, then it's true that 2,-1 and 1⁄2 are the zeroes of the cubic polynomial.

ii) - the sum of the zeros and the corresponding coefficients are the same

-the Sum of the products of roots where 2 are taken at the same time is same as the corresponding coefficient.

-the product of the zeros of the polynomial is same as the corresponding coefficient

Step-by-step explanation:

We are given the cubic polynomial;

p(x) = 2x³ - 3x² - 3x + 2

For us to verify that 2,-1 and 1⁄2 are the zeroes of the cubic polynomial, we will plug them into the equation and they must give a value of zero.

Thus;

P(2) = 2(2)³ - 3(2)² - 3(2) + 2 = 16 - 12 - 6 + 2 = 0

P(-1) = 2(-1)³ - 3(-1)² - 3(-1) + 2 = -2 - 3 + 3 + 2 = 0

P(½) = 2(½)³ - 3(½)² - 3(½) + 2 = ¼ - ¾ - 3/2 + 2 = -½ + ½ = 0

Since, P(2), P(-1) and P(½) gives 0,then it's true that 2,-1 and 1⁄2 are the zeroes of the cubic polynomial.

Now, let's verify the relationship between the zeros and the coefficients.

Let the zeros be as follows;

α = 2

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The coefficients are;

a = 2

b = -3

c = -3

d = 2

So, the relationships are;

α + β + γ = -b/a

αβ + βγ + γα = c/a

αβγ = -d/a

Thus,

First relationship α + β + γ = -b/a gives;

2 - 1 + ½ = -(-3/2)

1½ = 3/2

3/2 = 3/2

LHS = RHS; So, the sum of the zeros and the coefficients are the same

For the second relationship, αβ + βγ + γα = c/a it gives;

2(-1) + (-1)(½) + (½)(2) = -3/2

-2 - 1½ + 1 = -3/2

-1½ - 1½ = -3/2

-3/2 = - 3/2

LHS = RHS, so the Sum of the products of roots where 2 are taken at the same time is same as the coefficient

For the third relationship, αβγ = -d/a gives;

2 * -1 * ½ = -2/2

-1 = - 1

LHS = RHS, so the product of the zeros(roots) is same as the corresponding coefficient

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