Maria is ordering trophies for school. Company Christopher charges $3.50 for each trophy and a one-time engraving fee of $25. Company Asya charges $7.50 for each trophy and a one-time engraving fee of $17 Develop an equation where the two companies would be the same.

An equation where the two companies would be the same is given as:

25 + 3.50x = 17 + 7.50x

Step-by-step explanation:

Let the number of trophies ordered be represented by x

Maria is ordering trophies for school. Company Christopher charges $3.50 for each trophy and a one-time engraving fee of $25.

$25 + $3.50 × x

25 + 3.50x

Company Asya charges $7.50 for each trophy and a one-time engraving fee of $17

$17 + $7.50× x

17 + 7.50x

An equation where the two companies would be the same is given as:

Company Christopher = Company Asya

25 + 3.50x = 17 + 7.50x

Simplifying further:

Collect like terms

25 - 17 = 7.50x - 3.50x

8 = 4.00x

x = 8/4

x = 4

Hence, the two companies would be the same after 4 trophies.

the right system of equations to describe the situation would

be on the form:

x1 = 8000 + y1*t

and

x2 = 8000 + y2*t

where x1 and x2 represents the total money of imogene and her

friend respectively at the end of t years.

now for the value of amount earned, y1 and y2:

y1=8000*0.08

y2=2000*√(t-2)

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