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\(3x^2+24x+48=11x+4\)

\(\Rightarrow 3x^2+13x+44=0\)

Solve with quadratic equation:

\(\Rightarrow \frac{-13\pm \sqrt{13^2-4\cdot \:3\cdot \:44}}{2\cdot \:3}\)

\(\Rightarrow \frac{-13\pm \sqrt{359}i}{6}\)

\(x =\boxed{ -\frac{13}{6} + \frac{\sqrt{359}}{6}i , -\frac{13}{6} - \frac{\sqrt{359}}{6}i }\)

|w| = 1

After some extremely long computations, |w| works out to 2.

Since z² = 156 + 65i, we must have |z²| = |156 + 65i| = |13(12+5i)| = 13|12+5i| = 13(13) = 169. 

We also have |z|² = |z| * |z| = |(z)(z)| = |z²|, so |z|² = 169, which gives us |z| = \(\sqrt{169}\) = \(\boxed{13}\).

A

        14

                         20

               D         

                  6

B                   C

Since BD is an altitude....angles BDA  and BDC are right angles......so triangles BDA and BDC are right triangles

BD^2 = AB^2  - 14^2   ( 1)

BD^2  = BC^2 - 6^2   (2)          subtract (1)  from (2)

BC^2  - AB^2  = 14^2  - 6^2

BC^2  - AB^2  =  160

AB^2  = BC^2  - 160

So....by the Pythagorean Theorem

AB^2  +  BC^2 =  AC^2

(BC^2  - 160)  + BC^2 =  20^2

2BC^2  - 160  =  400

2BC^2  =  560

BC^2 =  280

BC =  sqrt (280) =   2sqrt (70) ≈  16.733

z = sqrt(165 + 65i) = 6.194... + 7.851...*i, so

|z| = sqrt(6.194...^2 + 7.851...^2) = 10.

How did you get that? It is incorrect.

By brother currently has 22 dollars.

The sum of all possible values of g(25) is 15.

If  you count them, you should see this pattern:

1, 4, 9, 16, 25.........etc. They are squares of the counting numbers, 1, 2, 3, 4, 5.........etc.

If you sum up all the squares up 17, you should get:

1785 + {18 x 13} ==2019 + 2 (14s) ==2021 terms

So, the 2021st term should be: 14

We have the form  Ax + By  = C

The slope of the given line =    -A / B  =  -2/3

So...a parallel line  will  have the same  slope

(-17 - - 9)  / ( j  - 2)  =   -2/3   

-8 / ( j - 2)   = -2/3     cross-multiply

-8 (3) =  -2 ( j - 2)

-24  =  -2j + 4

-24 - 4  =  -2j

-28  = -2j

j = -28/-2   =   14

x =2  y=2  z  = 6

x+ y + z =10

Slope  =    ( 7 - -5) / ( 6 - -3)  =  12/9  = 4/3

Using either point  we can write the equation  as

y = (4/3) (x - 6) + 7          multiply through by 3

3y  = 4(x - 6) + 21

3y =  4x - 24 + 21

3y  = 4x  - 3             rearrange as

4x - 3y  =  3

A  +  B + C  =   4   -3  + 3   =    4

There are \({13 \choose 3} = 286\) ways to choose the committee

Of these, there are \({5 \choose 1} = 5\) ways to choose a Republican and \({6 \choose 2} = 15\) ways to choose a Democrat, making for \(5 \times 15 = 75\) choices.

So, the probability is \(\color{brown}\boxed{75 \over 286}\)

BuilderBoi do you like to \(\text{su} \text{ck} \ \text{co} \text{ck}\)?

Subtract \(411_6\) from \(1215_6\) and convert the result to base 6.

There are 3 red lamps to put as the first lamp.

There are 6 lamps to put as the first lamp.

So, tell me, what is the probability?

We already know 2 of the digits, so there are 4 choices for the final digit. 

Now, we have to order them. Divide this into 2 cases.

Case 1: All 3 digits are distinct. 

There are \(3! = 6\) ways to do this, and there are 2 options (C or D), which makes for \(6 \times 2 = 12\) options.

Case 2: The 3rd digit is A or B

There are \(3! \div 2! = 3\) ways to order them, and there are 2 options (A or B), which makes for \(3 \times 2 = 6\) options.

So, there are \(12 + 6 = \color{brown}\boxed{18}\) words. 

BuilderBoi do you like to \(\text{su} \text{ck} \ \text{co} \text{ck}\)?

The sum of all possible vaues of f(1) is 3.

Here's another way: 

Each ball has 2 choices to go into a box. There are 5 boxes, so there are \(2^5 = 32\) combinations.

But, because the boxes are indistinguishable, we divide by 2. (Putting balls A, B, and C in 1 and D and E in the other is the same putting D and E in 1, and A, B, and C in the other)

So, there are \(32 \div 2 = \color{brown}\boxed{16}\)

The 2021st number is 17.

The circumradius is 15.

3 choices for the first number

3 choices for the second number

3 choices for the third number

How many numbers are there in total (hint: multiply them)