![\bf \begin{cases} x=3\implies &x-3=0\\ x=1+3i\implies &x-1-3i=0\\ x=1-3i\implies &x-1+3i=0 \end{cases} \\\\[-0.35em] ~\dotfill\\\\ (x-3)(x-1-3i)(x-1+3i)=0 \\\\\\ (x-3)\underset{\textit{difference of squares}}{([x-1]-3i)([x-1]+3i)}=0\implies (x-3)([x-1]^2-[3i]^2)=0 \\\\\\ (x-3)([x^2-2x+1]-[3^2i^2])=0\implies (x-3)([x^2-2x+1]-[9(-1)])=0](https://tex.z-dn.net/?f=%5Cbf%20%5Cbegin%7Bcases%7D%20x%3D3%5Cimplies%20%26x-3%3D0%5C%5C%20x%3D1%2B3i%5Cimplies%20%26x-1-3i%3D0%5C%5C%20x%3D1-3i%5Cimplies%20%26x-1%2B3i%3D0%20%5Cend%7Bcases%7D%20%5C%5C%5C%5C%5B-0.35em%5D%20~%5Cdotfill%5C%5C%5C%5C%20%28x-3%29%28x-1-3i%29%28x-1%2B3i%29%3D0%20%5C%5C%5C%5C%5C%5C%20%28x-3%29%5Cunderset%7B%5Ctextit%7Bdifference%20of%20squares%7D%7D%7B%28%5Bx-1%5D-3i%29%28%5Bx-1%5D%2B3i%29%7D%3D0%5Cimplies%20%28x-3%29%28%5Bx-1%5D%5E2-%5B3i%5D%5E2%29%3D0%20%5C%5C%5C%5C%5C%5C%20%28x-3%29%28%5Bx%5E2-2x%2B1%5D-%5B3%5E2i%5E2%5D%29%3D0%5Cimplies%20%28x-3%29%28%5Bx%5E2-2x%2B1%5D-%5B9%28-1%29%5D%29%3D0)
[ correction added, Thanks to @stef68 ]
![\bf (x-3)([x^2-2x+1]+9)=0\implies (x-3)(x^2-2x+10)=0 \\\\\\ x^3-2x^2+10x-3x^2+6x-30=0\implies x^3-5x^2+16x-30=f(x) \\\\\\ \stackrel{\textit{applying a translation with a -2f(x)}}{-2(x^3-5x^2+16x-30)=f(x)}\implies -2x^3+10x^2-32x+60=f(x)](https://tex.z-dn.net/?f=%5Cbf%20%28x-3%29%28%5Bx%5E2-2x%2B1%5D%2B9%29%3D0%5Cimplies%20%28x-3%29%28x%5E2-2x%2B10%29%3D0%20%5C%5C%5C%5C%5C%5C%20x%5E3-2x%5E2%2B10x-3x%5E2%2B6x-30%3D0%5Cimplies%20x%5E3-5x%5E2%2B16x-30%3Df%28x%29%20%5C%5C%5C%5C%5C%5C%20%5Cstackrel%7B%5Ctextit%7Bapplying%20a%20translation%20with%20a%20-2f%28x%29%7D%7D%7B-2%28x%5E3-5x%5E2%2B16x-30%29%3Df%28x%29%7D%5Cimplies%20-2x%5E3%2B10x%5E2-32x%2B60%3Df%28x%29)
D = sq.root of((x2 - x1)^2 + (y2 - y1)^2)
sq.root of 13 = sq.root of((x - (-2))^2 + (1 - 3)^2)
= sq.root of((x + 2)^2 + (-2)^2)
= sq.root of (x^2 + 4x + 4 + 4)
= sq.root of (x^2 + 4x + 8)
Now if we square both sides:
x^2 + 4x + 8 = 13
x^2 + 4x - 5 = 0
(x + 5)(x - 1) = 0
x = -5 or x = 1
Answer:
For example, [latex]\text{60%}=\frac{60}{100}[/latex] and we can simplify [latex]\frac{60}{100}=\frac{3}{5}[/latex]. Since the equation [latex]\frac{60}{100}=\frac{3}{5}[/latex] shows a percent equal to an equivalent ratio, we call it a percent proportion.
Step-by-step explanation:
Answer:
False
Step-by-step explanation:
In order to perform matrix multiplication, the number of columns in the first matrix must be the same as the number of rows in the second matrix.
In this case, the number of columns in the first matrix is 5 and the number of rows in the second column is 3 therefore they aren't the same value, hence, cannot be multiplied
Answer:
<em>The total change in his account is represented by the integer -5</em>
Step-by-step explanation:
<u>Operations With Integers</u>
We need to perform addition and subtraction to complete the answer to this problem.
Derrick first withdrew $17 from his account. Since the money is coming out from the account, it's considered as negative. Thus this operation is recorded as -$17.
The next day he deposited $12, which is considered as positive. This operation is considered as $12 or +$12.
The net operation or total change in the account is: -$17 + $12.
Subtracting and simplifying: -$17 + $12 = -$5
The total change in his account is represented by the integer -5
