How do you solve this

Mathematics

1 answer:

morpeh [17]2 years ago

5 0

D50% because its half off right?

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Help ( correct answer ) 16÷(4)(2)-3(2)÷2+1

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16 ÷ (4)(2) - 3(2) <span>÷ 2+1
16 </span>÷ 8 - 6 <span>÷ 3
2 - 2
0

Answer: 0</span>

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2 years ago

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Hey can you help me answer this question by giving me the answer?

Leto [7]

The value of f(a)=4-2a+6 $a^{2}$ , f(a+h) is $6a^{2} +6h^{2} -2a-2h+12ah$ , [f(a+h)-f(a)]/h is 6h+12a-2 in the function f(x)=4-2x+6 $x^{2}$ .

Given a function f(x)=4-2x+6 $x^{2}$ .

We are told to find out the value of f(a), f(a+h) and [f(a+h)-f(a)]/hwhere h≠0.

Function is like a relationship between two or more variables expressed in equal to form.The value which we entered in the function is known as domain and the value which we get after entering the values is known as codomain or range.

f(a)=4-2a+6 $a^{2}$ (By just putting x=a).

f(a+h)== $4-2(a+h)+6(a+h)^{2}$

=4-2a-2h+6( $a^{2} +h^{2} +2ah$ )

=4-2a-2h+6 $a^{2} +6h^{2} +12ah$

= $6a^{2} +6h^{2}-2a-2h+12ah$

[f(a+h)-f(a)]/h=[ $6a^{2} +6h^{2}-2a-2h+12ah$ -(4-2a+6 $a^{2}$ )]/h

= $(6a^{2} +6h^{2} -2a-2h+12ah)/h$

= $(6h^{2} -2h+12ah)/h$

=6h+12a-2.

Hence the value of function f(a)=4-2a+6 $a^{2}$ , f(a+h) is $6a^{2} +6h^{2} -2a-2h+12ah$ , [f(a+h)-f(a)]/h is 6h+12a-2 in the function f(x)=4-2x+6 $x^{2}$ .

Learn more about function at brainly.com/question/10439235

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