Answer:
4,2 is tge answer due to the interaction point plus it says the solution is at the interaction.
Answer:
-1+5z, -25+20y, 4d+14, 2n-2, 40k+8, 8b+5, -9+22p
Step-by-step explanation:
1) -4+7z+3-2z
-1+5z
2) 15+4(5y-10)
15+20y-40
-25+20y
3) 2d+17-3-2d+4d
4d+14
4) 12n-8-2n+10-4
2n-2
5) 8(2k+1+3k)
16k+8+24k
40k+8
5) 4(2b+2)-3
8b+8-3
8b+5
6) -4+8p-6p-5+20p
-9+22p
Answer:
Step-by-step explanation:b
Answer:

Step-by-step explanation:
meants when
, that the value for
is 5.
So this gives us this equation:

meants when
, that the value for
is 10.
So this gives us this equation:

So I take equation 2 and divide it be equation 1 I get:

Simplifying:

Since the base for an exponential function can't be negative then
.
So plugging into one of my equations I began with gives me an equation to solve for the initial value,
:


Divide both sides by 2:

The function is:
