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AlekseyPX
3 years ago
14

Solve A=1/2h(b1+b2) Solve for b1

Mathematics
1 answer:
IgorLugansk [536]3 years ago
6 0
A =  \frac{1}{2} h (b_1+b_2) \\  \\  \frac{2A}{h} = b_1+b_2 \\  \\ b_1 =  \frac{2A}{h} -b_2
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Evaluate the function f(x)=x^2+2x+3 at the given values of the independent variable and simplivy
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Answer: i) f(-9)= 66

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Step-by-step explanation:

The given function : f(x)=x^2+2x+3.

i) For f(-9) , the independent variable is x= -9.

Put x= -9 in given function , we get

f(-9)=(-9)^2+2(-9)+3

=81-18+3=66

Thus ,  f(-9)= 66

ii) For f(x+1) , the independent variable is x= x+1.

Replace x  by x+1 in given function , we get

f(x+1)=(x+1)^2+2(x+1)+3.

=(x^2+1+2x)+2x+2+3  [\because\ (a+b)^2=(a^2+b^2+2ab)]

=x^2+1+2x+2x+5

=x^2+4x+6

Thus ,  f(x+1)=x^2+4x+6.

iii)For f(-x) ,  the independent variable is x=-x.

Replace x  by -x in given function , we get

f(-x)=(-x)^2+2(-x)+3.

=x^2-2x+3  [∵ (+)(-)=(-)]

Thus , f(-x)=x^2-2x+3

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