f(x) increase by a factor of 3
Explanation:
Given that f(x)= 3* and the interval is x=4 to x=57
Now we put the value for x is 4 to 57 then value of f(x) increase with the multiply of 3.
Because the x is multiplied with 3 i.e., 3*
So f(x) increase by a factor of 3.
If we put x=4, then f(x)= 12 (∵ 3×4=12)
If we put x=5, the f(x)= 15 (∵ 3×5=15)
If we put x=6,the f(x)= 18 (∵ 3×6=18)
similarly., values of x= 7,8,9,...155.
Then,
If we put x=56, the f(x)=168
This process will continue until f(x)=171 for x=57.
1. Let
Shortest piece = x
Longest piece = y
Third piece = z
2.
Shortest piece = x
Longest piece = 40 + x
Third piece = (40 + x)/2 OR 20 + 0.5x
3. Add all the expressions since you know that the total of all the pieces is 120 inches
40 + x + 20 + 0.5x + x = 120.
4. Add the whole numbers first, then add the variables.
40+20=60
x+0.5=2.5x
60+2.5x=120
5. To get rid of the 60, do the opposite of addition which is subtraction and subtract 60 from both sides.
2.5x=120-60
2.5x=60
6. Do the same but rather than addition, do the opposite of multiplying which is dividing, on both sides.
x=60/2.5
x=24
7.
Shortest piece = 64-40=24 inches
Longest piece = 24+40=64
Third piece = 64/2=32
Answer:
The description is as follows:
Step-by-step explanation:
The description of the net of a rectangular prism is as follows:
The length i.e. 12 centimeters would be multiplied by width of 9 and then it added to the height i.e. 5 centimeters
The above represent the description of the net of a rectangular prism
Answer:
yes
Step-by-step explanation:
Answer:
Joe is correct
Step-by-step explanation:
Given equation:
![(x^2)^?=x^4 \cdot x^8](https://tex.z-dn.net/?f=%28x%5E2%29%5E%3F%3Dx%5E4%20%5Ccdot%20x%5E8)
The exponent outside the bracket is a question mark and the students are trying to determine the value of the question mark.
For ease of answering, let y be the unknown number (question mark):
![\implies (x^2)^y=x^4 \cdot x^8](https://tex.z-dn.net/?f=%5Cimplies%20%28x%5E2%29%5Ey%3Dx%5E4%20%5Ccdot%20x%5E8)
First, simplify the equation by applying exponent rules to either side of the equation:
![\textsf{Apply exponent rule} \quad (a^b)^c=a^{bc}\quad\textsf{to the left side}:](https://tex.z-dn.net/?f=%5Ctextsf%7BApply%20exponent%20rule%7D%20%5Cquad%20%28a%5Eb%29%5Ec%3Da%5E%7Bbc%7D%5Cquad%5Ctextsf%7Bto%20the%20left%20side%7D%3A)
![\implies (x^2)^y=x^4 \cdot x^8](https://tex.z-dn.net/?f=%5Cimplies%20%28x%5E2%29%5Ey%3Dx%5E4%20%5Ccdot%20x%5E8)
![\implies x^{2y}=x^4 \cdot x^8](https://tex.z-dn.net/?f=%5Cimplies%20x%5E%7B2y%7D%3Dx%5E4%20%5Ccdot%20x%5E8)
![\textsf{Apply exponent rule} \quad a^b \cdot a^c=a^{b+c} \quad \textsf{to the right side}:](https://tex.z-dn.net/?f=%5Ctextsf%7BApply%20exponent%20rule%7D%20%5Cquad%20a%5Eb%20%5Ccdot%20a%5Ec%3Da%5E%7Bb%2Bc%7D%20%5Cquad%20%5Ctextsf%7Bto%20the%20right%20side%7D%3A)
![\implies x^{2y}=x^4 \cdot x^8](https://tex.z-dn.net/?f=%5Cimplies%20x%5E%7B2y%7D%3Dx%5E4%20%5Ccdot%20x%5E8)
![\implies x^{2y}=x^{4+8}](https://tex.z-dn.net/?f=%5Cimplies%20x%5E%7B2y%7D%3Dx%5E%7B4%2B8%7D)
![\implies x^{2y}=x^{12}](https://tex.z-dn.net/?f=%5Cimplies%20x%5E%7B2y%7D%3Dx%5E%7B12%7D)
Now, apply the exponent rule:
![x^{f(x)}=x^{g(x)} \implies f(x)=g(x)](https://tex.z-dn.net/?f=x%5E%7Bf%28x%29%7D%3Dx%5E%7Bg%28x%29%7D%20%5Cimplies%20f%28x%29%3Dg%28x%29)
Therefore,
![x^{2y}=x^{12}\implies 2y=12](https://tex.z-dn.net/?f=x%5E%7B2y%7D%3Dx%5E%7B12%7D%5Cimplies%202y%3D12)
Finally, solve for y:
![\implies 2y=12](https://tex.z-dn.net/?f=%5Cimplies%202y%3D12)
![\implies 2y \div 2 = 12 \div 2](https://tex.z-dn.net/?f=%5Cimplies%202y%20%5Cdiv%202%20%3D%2012%20%5Cdiv%202)
![\implies y=6](https://tex.z-dn.net/?f=%5Cimplies%20y%3D6)
Therefore, <u>Joe is correct</u> as the unknown number is 6.