Jony is 2 now and is 2x older than Chika. So Chika is 1 year old. This means Chika is 1 year younger than Jony. When Jony is 30, Chika is 29
So here is the answer. Initially, Ed's toy cars compared to Pete's toy cars was 5:2. So for every 5 cars that Ed has, Pete has 2. Now that Ed gave 30 cars to Pete. So here it goes. The total number of ratio units is 5+2=7, so each will have an equal number if they both have 3.5 ratio units. That is, if Ed transfers to Pete 1.5 ratio units, their car counts will be equal. Thus 1.5 ratio units = 30 cars, or 1 ratio unit = 20 cars. Therefore, this makes <span> 7*20 cars = 140 cars.
</span>Hope this helps.
Answer:
Step-by-step explanation:
Starting from the y-intercept of ![\displaystyle [0, -8],](https://tex.z-dn.net/?f=%5Cdisplaystyle%20%5B0%2C%20-8%5D%2C)
you do 
by either moving eight blocks <em>south</em><em> </em>over three blocks <em>west</em><em> </em>or eight blocks <em>north</em> over three blocks <em>east</em><em> </em>[<em>west</em> and <em>south</em> are negatives].
I am joyous to assist you anytime.
* 2⅔ = 8⁄3
2
3
−
1
1
=
3
+
3
2
3
x
−
11
=
x
3
+
3
32x−11=3x+3
2
3
−
1
1
=
3
+
3
2
x
3
−
11
=
x
3
+
3
32x−11=3x+3
2
Find common denominator
2
3
−
1
1
=
3
+
3
2
x
3
−
11
=
x
3
+
3
32x−11=3x+3
2
3
+
3
(
−
1
1
)
3
=
3
+
3
2
x
3
+
3
(
−
11
)
3
=
x
3
+
3
32x+33(−11)=3x+3
3
Combine fractions with common denominator
2
3
+
3
(
−
1
1
)
3
=
3
+
3
2
x
3
+
3
(
−
11
)
3
=
x
3
+
3
32x+33(−11)=3x+3
2
+
3
(
−
1
1
)
3
=
3
+
3
2
x
+
3
(
−
11
)
3
=
x
3
+
3
32x+3(−11)=3x+3
4
Multiply the numbers
2
+
3
(
−
1
1
)
3
=
3
+
3
2
x
+
3
(
−
11
)
3
=
x
3
+
3
32x+3(−11)=3x+3
2
−
3
3
3
=
3
+
3
2
x
−
33
3
=
x
3
+
3
32x−33=3x+3
5
Find common denominator
2
−
3
3
3
=
3
+
3
2
x
−
33
3
=
x
3
+
3
32x−33=3x+3
2
−
3
3
3
=
3
+
3
⋅
3
3
2
x
−
33
3
=
x
3
+
3
⋅
3
3
32x−33=3x+33⋅3
6
Combine fractions with common denominator
2
−
3
3
3
=
3
+
3
⋅
3
3
2
x
−
33
3
=
x
3
+
3
⋅
3
3
32x−33=3x+33⋅3
2
−
3
3
3
=
+
3
⋅
3
3
2
x
−
33
3
=
x
+
3
⋅
3
3
32x−33=3x+3⋅3
7
Multiply the numbers
2
−
3
3
3
=
+
3
⋅
3
3
2
x
−
33
3
=
x
+
3
⋅
3
3
32x−33=3x+3⋅3
2
−
3
3
3
=
+
9
3
2
x
−
33
3
=
x
+
9
3
32x−33=3x+9
8
Multiply all terms by the same value to eliminate fraction denominators
2
−
3
3
3
=
+
9
3
2
x
−
33
3
=
x
+
9
3
32x−33=3x+9
3
⋅
2
−
3
3
3
=
3
(
+
9
3
)
3
⋅
2
x
−
33
3
=
3
(
x
+
9
3
)
3⋅32x−33=3(3x+9)
9
Cancel multiplied terms that are in the denominator
3
⋅
2
−
3
3
3
=
3
(
+
9
3
)
3
⋅
2
x
−
33
3
=
3
(
x
+
9
3
)
3⋅32x−33=3(3x+9)
2
−
3
3
=
+
9
2
x
−
33
=
x
+
9
2x−33=x+9
10
Add
3
3
33
33
to both sides of the equation
2
−
3
3
=
+
9
2
x
−
33
=
x
+
9
2x−33=x+9
2
−
3
3
+
3
3
=
+
9
+
3
3
2
x
−
33
+
33
=
x
+
9
+
33
2x−33+33=x+9+33
11
Simplify
Add the numbers
Add the numbers
2
=
+
4
2
2
x
=
x
+
42
2x=x+42
12
Subtract
x
x
from both sides of the equation
2
=
+
4
2
2
x
=
x
+
42
2x=x+42
2
−
=
+
4
2
−
2
x
−
x
=
x
+
42
−
x
2x−x=x+42−x
13
Simplify
Combine like terms
Multiply by 1
Combine like terms
=
4
2
x
Answer: P(t)=25000(1.12)^t
Step-by-step explanation:
