Answer:
Part 1) Helen's age is 32 years old and Jane's age is 24 years old
Part 2) 13 twenty-dollar bills
Step-by-step explanation:
Part 1) Helen is 8 years older than Jane. Twenty years ago Helen was three times as old as Jane. How old is each now and what is the equation?
Let
x----> Helen's age
y---> Jane's age
we know that
x=y+8 ----> equation A
(x-20)=3(y-20) -----> equation B
substitute equation A in equation B and solve for y
(y+8-20)=3(y-20)
y-12=3y-60
3y-y=60-12
2y=48
y=24 years
Find the value of x
x=y+8
x=24+8=32 years
Part 2)
Let
x-----> the number of five-dollar bills
y----> the number of twenty-dollar bills
we know that
5x+20y=305 -----> equation A
y=x+4 ------> x=y-4 ------> equation B
substitute equation B in equation A and solve for y
5(y-4)+20y=305
5y-20+20y=305
25y=325
y=13 twenty-dollar bills
Find the value of x
x=y-4
x=13-4=9 five-dollar bills
Answer:
864
Step-by-step explanation:
∵X-Y=5 and.
∵X+Y=59 add the two equations
∴2X=64
∴X=32 (substitute in any equation)
∴32 - Y=5
∴Y = 27
∴ So . X×Y=32 × 27=864
Answer:
<h2>80=5x+4y</h2>
Step-by-step explanation:
Step one:
given
let binders be x
and notebooks be y
let the total sale be T
the cost of binder is $5 each
the cost of notebook is $4 each.
the total sales expected is $80
step two:
the linear model for the total sales is given as
T=x+y
using the given data, the situation can be represented linearly as
80=5x+4y
Therefore the linear expression for the total sale is 80=5x+4y
f(h(x))= 2x -21
Step-by-step explanation:
f(x)= x^3 - 6
h(x)=\sqrt[3]{2x-15}
WE need to find f(h(x)), use composition of functions
Plug in h(x)
f(h(x))=f(\sqrt[3]{2x-15})
Now we plug in f(x) in f(x)
f(h(x))=f(\sqrt[3]{2x-15})=(\sqrt[3]{2x-15})^3 - 6
cube and cube root will get cancelled
f(h(x))= 2x-15 -6= 2 x-21
