Neetu said that the sum of −8+d is always positive when the value of d is greater than 0. • Is Neetu’s statement always true, sometimes true or never true?

neetu's statement is sometimes true

Step-by-step explanation:

We are told that Neetu said the sum of −8 + d is always positive when the value of d is greater than 0.

Values of d that are greater than zero is a set of positive real numbers like 1,2,3

Now, using d = 1, we will get;

-8 + 1 = -7

We can see that we got a negative number.

Using d = 2,we have; -8 + 2 = - 6

Again we got a negative number.

Now, by inspection, we will keep getting negative numbers until d = 7.

Thus, neetu's statement is not true for 0 < d < 8

Now,let's try d above 8. Let's use d = 9.

-8 + 9 = 1

This means from d > 8 and above we have the results as positive. Thus, neetu's statement is always true for d > 8

In summary, neetu's statement is sometimes true

the correct functions are

h(x) = 3x + 2  

g(x) = 2^x

step-by-step explanation:

i proceed to calculate the different values ​​of k (x)

case 1)  k(x) = g(x) ∘ h(x)

g(x) ∘ h(x)=g(h(x))

k(x) = [2]^[3x + > is the option  4. k(x) = 2^(3x + 2 )

case 2)  k(x) = g(x) + h(x)

k(x) = [2^x]+[3x + 2]=2^x+3x+> is the option  1.) k(x) = 2^x + 3x + 2

case 3)k(x) = h(x) ÷ g(x)

k(x) =  [3x + 2]/[2^x]=3x/(2^x)+2/(2^x)=3x*(2^-x)+2^(1-x)

is the option  3.)   k(x) = (3x)2^(-x) + 2^(-x + 1 )

case 4)k(x) = g(x) - h(x)

k(x) =  [2^x]-[3x + 2]=2^x-3x-> is the option  2.) k(x) = 2^x - 3x - 2 

case 5)  k(x) = g(x) × h(x)

k(x) = [2^x]*[3x + > is the option  5.) k(x) = 2^x(3x + 2)

case 6)k(x) = h(x) ∘ g(x)

h(x) ∘ g(x)=h(g(x))

k(x)=3*[2^x]+> is the option  6. k(x) = 3(2^x) +2

y-y1=4(x-x1)

step-by-step explanation:

this is point-slope form since you only have 4 as the slope it goes in place of m

hope this : )