Idon't get it! > .< which number can each term of the equation be multiplied by to eliminate the decimals before solving? 5.6j – 0.12 = 4 + 1.1j a. 0.01 b. 0.1 c. 10 d. 100

100

Step-by-step explanation:

To Find: Which number can each term of the equation be multiplied by to eliminate the decimals before solving? 5.6j – 0.12 = 4 + 1.1j

Solution :

5.6j – 0.12 = 4 + 1.1j

Since we can see that in the given equation the constants and the coefficients  are in decimals

All the number are in decimal up to tenth place except 0.12

0.12 is in decimal up to hundredth place

So if we want to eliminate the decimal in 0.12. so we need to multiply it with 100

So, the given equation must be multiplied by 100 to make it decimal free

⇒100*( 5.6j – 0.12 = 4 + 1.1j)

⇒56j-12=4+11j

Thus Option D is correct : 100

b. x- 48 =120

step-by-step explanation:

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