Consider the following functions. F(x)= 2x^2+4x+10
G(x)= 3x-11
H(x)= 3x^2+x^2-21

The value of H(F(-2)) is

The value of H(F(-2)) is 379

Step-by-step explanation:

Soln,

Given,

F(x)= 2x^2+4x+10

G(x)= 3x-11

H(x)= 3x^2+x^2-21

Now,

F(-2) = 2(-2)^2 + 4×(-2) + 10

= 2×4 + (-8) + 10

= 8-8+10

= 18 - 8

= 10

Again,

H(F(-2)) = 3x^2 + x^2 - 21

Putting the value of F(-2) , we get

H(10) = 3×10^2 + 10^2 - 21

= 3×100 + 100 - 21

= 300 + 100 - 21

= 400 - 21

= 379

Therefore, H(F(-2)) = 379.

since proposition states that a number is greater than 25, if is true then the number must actually be greater than 25.

this implies that all possible options, except 24, can be accepted

see explanation

step-by-step explanation:

since the sequence is geometric there is a common ratio r between consecutive terms, that is

r = 4 ÷ - 2 = - 2

r = - 8 ÷ 4 = - 2

r = 16 ÷ 8 = - 2

the next term - 13 must be incorrect as it should be - 32 for the sequence to be geometric.

the first 4 terms are therefore geometric with common ratio r = - 2