(will give brainliest)
the senior class is having a fundraiser to pay for the senior trip. selling a box of chocolates yields a profit of $2.45,
while selling a box of cookies yields a profit of $2.70. the demand for cookies is at least twice that of chocolates, but the
amount of cookies produced must be no more than 550 boxes plus 3 times the number of chocolates produced. assuming that
the senior class can sell every box that they order, how many boxes of each should they order to maximize profit if they
cannot order more than 1950 boxes combined?

350 choc     1599 of coockies

Step-by-step explanation:

h= box of chocolates

k=box of coockies

k< 550+3h

k-3h<550

k+h < 1950

I will multip  the second by 3

k -  3h <  550

3k +3h <  5850

ADD BOTH

4k  <  6400

k< 1600

1950-(<1600)=h

1950-1599=h

359-h

Chocolate   350

Cookies LESS  than 1600

1600 is AT LEAST twice than 350 (Twice or more)

its not b

step-by-step explanation:

let's add {f(x)=x+1}f(x)=x+1 and {g(x)=2x}g(x)=2x together to make a new function.

f(x)+g(x) =(x+1)+(2x)=x+1+2x=3x+1

let's call this new function hh. so we have:

{h(x)}={f(x)}+{g(x)}{=3x+1}h(x)=f(x)+g(x)=3x+1

we can also evaluate combined functions for particular inputs. let's evaluate function hh above for x=2x=2. below are two ways of doing this.

method 1: substitute x=2x=2 into the combined function hh.

h(x)

h(2)

​    

=3x+1

=3(2)+1

=7

​ since h(x)=f(x)+g(x)h(x)=f(x)+g(x), we can also find h(2)h(2) by finding f(2) +g(2)f(2)+g(2).

first, let's find f(2)f(2):

f(x)

f(2)

​    

=x+1

=2+1

=3

​  

now, let's find g(2)g(2):

g(x)

g(2)

​    

=2x

=2⋅2

=4

​  

so f(2)+g(2)=3+4=\greend7f(2)+g(2)=3+4=7.

notice that substituting x =2x=2 directly into function hh and finding f(2) + g(2)f(2)+g(2) gave us the same answer!