What is the sum of all of the odd numbers from 1 to 59?

sum = 900

Step-by-step explanation:

The sum of all odd number can be solved using arithmetic progression formula.

Odd numbers are numbers that cannot be exactly divided by 2 .That means odd number have remainder when divide by 2.

The first odd number is 1

The second is  3

The sequence goes like this

1, 3, 5, 7, 9, 11, 13, 15, 17, 19 , 21 59

a = 1

d = 2

where

a = first term

d = common difference

let us find the number of terms before finding the sum

nth term = a + (n - 1)d

59 = 1 + (n - 1)2

59 = 1 + 2n - 2

59 + 2 - 1 = 2n

60 = 2n

n = 60/2

n = 30

Sum = n/2(a+L)

where L = last term

sum = 30/2(1 + 59)

sum = 15(60)

sum = 900

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