3.14 is the answer
hope it helped!
Answer:
1. 2(3x−2)(2x−3)
2. 3(3x+1)(2x+5)
3. 2(4x+1)(2x−5)
4. 4(2x+1)(2x+7)
5. 5(3x+1)(2x−5)
6. 3(3x−7)(2x+1)
7. 3(5x+3)(3x−2)
8. 2(7x−2)(2x−5)
9. 2(7x−3)(2x−3)
10. 4(5x−11)(2x+1)
11. 5(x−2)(x+12)
12. (x−1)(x−8)
15. x=1 or x=4
Step-by-step explanation:
1. Factor 12x2−26x+12
12x2−26x+12
=2(3x−2)(2x−3)
2. 3(3x+1)(2x+5)
3. Factor 16x2−36x−10
16x2−36x−10
=2(4x+1)(2x−5)
4. Factor 16x2+64x+28
16x2+64x+28
=4(2x+1)(2x+7)
5. Factor 30x2−65x−25
30x2−65x−25
=5(3x+1)(2x−5)
6. Factor 18x2−33x−21
18x2−33x−21
=3(3x−7)(2x+1)
7. Factor 45x2−3x−18
45x2−3x−18
=3(5x+3)(3x−2)
8. Factor 28x2−78x+20
28x2−78x+20
=2(7x−2)(2x−5)
9. Factor 28x2−54x+18
28x2−54x+18
=2(7x−3)(2x−3)
10. Factor 40x2−68x−44
40x2−68x−44
=4(5x−11)(2x+1)
11. Factor 5x2+50x−120
5x2+50x−120
=5(x−2)(x+12)
12. Let's factor x2−9x+8
x2−9x+8
The middle number is -9 and the last number is 8.
Factoring means we want something like
(x+_)(x+_)
Which numbers go in the blanks?
We need two numbers that...
Add together to get -9
Multiply together to get 8
Can you think of the two numbers?
Try -1 and -8:
-1+-8 = -9
-1*-8 = 8
Fill in the blanks in
(x+_)(x+_)
with -1 and -8 to get...
(x-1)(x-8)
15. Let's solve your equation step-by-step.
(x−2)(x−3)=2
Step 1: Simplify both sides of the equation.
x2−5x+6=2
Step 2: Subtract 2 from both sides.
x2−5x+6−2=2−2
x2−5x+4=0
Step 3: Factor left side of equation.
(x−1)(x−4)=0
Step 4: Set factors equal to 0.
x−1=0 or x−4=0
x=1 or x=4
Sorry I wasn't able to do 13 and 14 but hope this helps! :)
Answer: The mean of the list given is 10.
To find the mean of a list of numbers, first add up all the numbers.
10 + 10 + 11 + 9 = 40
Now, divide the list by the total number of people.
40 / 4 = 10
The mean of this list is 10.
Answer:
1 and −1 are the only solutions to the equation x^2 = 1.
Step-by-step explanation:
We shall proceed as he suggests
![x=a+bi](https://tex.z-dn.net/?f=x%3Da%2Bbi)
Given ![x^{2}=1](https://tex.z-dn.net/?f=x%5E%7B2%7D%3D1)
substitute a+bi in x, we get
![(a+bi)^{2}=1](https://tex.z-dn.net/?f=%28a%2Bbi%29%5E%7B2%7D%3D1)
Rewriting the both sides in standard form for a complex number
![(a^{2}-b^{2})+2abi=1+0i](https://tex.z-dn.net/?f=%28a%5E%7B2%7D-b%5E%7B2%7D%29%2B2abi%3D1%2B0i)
Equating the real parts on each side of the equation, and equating the imaginary parts on each side of the equation.
and ![2ab=0](https://tex.z-dn.net/?f=2ab%3D0)
So either a=0 or b=0. If a=0 then
![0^{2}-b^{2}=1](https://tex.z-dn.net/?f=0%5E%7B2%7D-b%5E%7B2%7D%3D1)
. has no real solution.
If b=0 then
![a^{2}-0^{2}=1](https://tex.z-dn.net/?f=a%5E%7B2%7D-0%5E%7B2%7D%3D1)
![a^{2}=1](https://tex.z-dn.net/?f=a%5E%7B2%7D%3D1)
![a^{2}-1^{2}=0](https://tex.z-dn.net/?f=a%5E%7B2%7D-1%5E%7B2%7D%3D0)
![(a-1)(a+1)=0](https://tex.z-dn.net/?f=%28a-1%29%28a%2B1%29%3D0)
. or ![a=-1](https://tex.z-dn.net/?f=a%3D-1)
Hence proved.