Sq32/8 = sq4
It would be c. 2
Answer: ![{\boxed{h(x+4)=\frac{9+3x}{8+x}}}](https://tex.z-dn.net/?f=%7B%5Cboxed%7Bh%28x%2B4%29%3D%5Cfrac%7B9%2B3x%7D%7B8%2Bx%7D%7D%7D)
Concept:
In a function f(x), it represents a function in terms of x or for x. Therefore, to find the values in the function, substitute values within the parenthesis to solve.
Solve:
<u>Given information</u>
![h(x)=\frac{-3+3x}{4+x}](https://tex.z-dn.net/?f=h%28x%29%3D%5Cfrac%7B-3%2B3x%7D%7B4%2Bx%7D)
<u>Need to find</u>
![h(x+4)](https://tex.z-dn.net/?f=h%28x%2B4%29)
<u>Substitute (x + 4) to the position of x</u>
![h(x+4)=\frac{-3+3(x+4)}{4+(x+4)}](https://tex.z-dn.net/?f=h%28x%2B4%29%3D%5Cfrac%7B-3%2B3%28x%2B4%29%7D%7B4%2B%28x%2B4%29%7D)
<u>Distributive property on the numerator</u>
![h(x+4)=\frac{-3+3x+12}{4+(x+4)}](https://tex.z-dn.net/?f=h%28x%2B4%29%3D%5Cfrac%7B-3%2B3x%2B12%7D%7B4%2B%28x%2B4%29%7D)
<u>Combine like terms</u>
![h(x+4)=\frac{-3+12+3x}{4+4+x}](https://tex.z-dn.net/?f=h%28x%2B4%29%3D%5Cfrac%7B-3%2B12%2B3x%7D%7B4%2B4%2Bx%7D)
![\boxed{h(x+4)=\frac{9+3x}{8+x}}](https://tex.z-dn.net/?f=%5Cboxed%7Bh%28x%2B4%29%3D%5Cfrac%7B9%2B3x%7D%7B8%2Bx%7D%7D)
Hope this helps!! :)
Please let me know if you have any questions
Each persons ticket costs $34.00(D)
all you have to do is subtract $6.80 from $142.80
then you divide the answer by 4
Since it's a multiple of 24, it has to be a multiple of the factors of 24.
Factors of 24:
2,3,4,6,8,12
You can use some of this knowledge to help create the number.
Since the # needs to be a multiple off 2, the last digit needs to be an 8
All numbers that are multiples of 3 have the property that all of their digits added together have to be a number that is evenly divisible by 3.
so your number will look like:
_ _ _ _ _ 8
so start trying combinations for the other 5 digits that give you a number that is a multiple of 3: 3,6,9,12,15, ect. If you can't find one, then it's impossible
The annual return percentages will be evaluated using the formula:
A=P(1+r/100)^n
where:
A=amount
P=principle
r=rate
n=time
a] A=$500, P=$400, n=1 years
500=400(1+r)^1
solving for r we shall obtain:
1.25=1+r
hence
r=1.25-1
r==0.25
annual rate of investment is 25%
b] A=2500+100=$2600, P=$ 2000, n=1 year
hence
2600=2000(1+r)^1
2600/2000=1+r
1.3=1+r
r=1.3-1
r=0.3
annual rate of investment is 30%