Moore's law says that the number of transistors in a dense integrated circuit increases by 41% every year.
In 1974. a dense integrated circuit was produced with 5000 transistors.
Which expression gives the number of transistors in a dense integrated circuit in 1979?
Choose 1

5000 - 0.415

5000(1 +0.41)

5000+ 0.41

5000 + (1 +0.41)5

Solution:

The increasing function is given as:

Where,

y is the future value

a is the initial value

r is the growth rate

t is the number of years

From given,

a = 5000

t = 1974 to 1979 = 5 years

Substituting the values in formula,

Thus the expression gives the number of transistors in a dense integrated circuit in 1979 is found

answer:

(-8,7)

explanation:

we know that  

f

(

−

3

)

=

7

therefore, if we wish to get the output as 7, in the function, we need to have our input to be -3.

looking at the function  

f

(

x

+

5

)

, what should the value of x be, such that the input in the function becomes 7?

we simply equate the terms,

x

+

5

=

−

3

x

=

−

8

thus, the point on the new function would be  

(

−

8

,

7

)