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Try to expand and you will get a linear equation. After that you can solve it by moving terms around and simplifying it.

(x + A)(x + B) = (x + C)(x + D)

x² + (A + B)x + AB = x² + (C + D)x + CD

(A + B)x + AB = (C + D)x + CD

(A + B)x - (C + D)x = CD - AB

(A + B - C - D)x = CD - AB

x = ____________________

One step left for you to complete.

Thank you very much Max. You are as brilliant as ever. Congrats.

A story:

Imagine a wall being placed on the right of a binary number.

Whenever a number gets left shifted, a zero comes out of the wall to fill the gap between the wall and the number.

Whenever a number gets right shifted, the last digit just goes through the wall and never come back again.

Now, let's try left shifting and right shifting some numbers.

right shift of two places on 00110100.

The bar denotes the "wall", and blue digits means the digit is gone.

\(00110100\Big|\)

right shift once

\(0011010\Big|\color{blue}0\)

right shift once more

\(001101\Big|\color{blue}00\)

The result should be \(001101\), but because you needs 8 bits for a byte, it becomes \(00001101\). We add zeroes in front when we don't have enough digits.

Hints for the others:

Q2) Think of what a left shift and a right shift really does on the value of the binary number by trying out few examples.

Q3) Similar to Q1.

EDIT: Italicized text is what I changed.

Since it is a BINARY number ....every position to the left is TWICE the one to the right    (edited)

example    010    is twice  001       100 is twice 010 

so shift the number to the left ONE space to double it

00100100    to double it   shift it one space left   

01001000

Binary shift right two places

00110100

00001101

See my answer here:
https://web2.0calc.com/questions/please-help-question-not-answered#r2

Group the terms in pairs. Using the formula a² - b² = (a - b)(a + b), we have

\(\quad(a +(2n+1)d)^2- (a + (2n)d)^2 +(a + (2n-1)d)^2 - (a+(2n-2)d)^2 + \cdots + (a+d)^2 - a^2\\ =\left((a +(2n+1)d)^2- (a + (2n)d)^2\right) +\left((a + (2n-1)d)^2 - (a+(2n-2)d)^2\right) + \cdots + \left((a+d)^2 - a^2\right)\\ =(\color{red}(a + (2n + 1)d)\color{black} - \color{blue}(a + (2n)d)\color{black})(\color{red}(a + (2n + 1)d)\color{black} + \color{blue}(a + (2n)d)\color{black}) \\\quad\quad+ (\color{red}(a + (2n-1)d)\color{black} - \color{blue}(a + (2n-2)d)\color{black})(\color{red}(a + (2n-1)d)\color{black} + \color{blue}(a + (2n-2)d)\color{black}) + \cdots \\\quad\quad\qquad+(\color{red}(a + d)\color{black} - \color{blue}a\color{black})(\color{red}(a + d)\color{black} + \color{blue}a\color{black})\)

The last line may be a bit long, but if we simplify it:

\(= d(2a + \color{red}(4n + 1)\color{black}d) + d(2a + \color{red}(4n - 3)\color{black}d) + \cdots + d(2a + \color{red}1\color{black}d)\)

Now we see that the original expression is nothing other than the sum of A.S. in disguise.

Factorising out a common factor for a more easy-to-spot A.S.:

\(= d((2a + (4n + 1)d) + (2a + (4n - 3)d) + \cdots + (2a + d))\)

We can now see the expression inside the bracket is an A.S. with first term = (2a + (4n + 1)d), common difference = -4d, number of terms = (n + 1).

We can plug those into the formula:

\(\quad\text{sum of A.S.} \\= \dfrac{\text{number of terms}}{2} \left(2\times \text{first term} + (\text{number of terms} - 1)\times \text{common difference}\right) \\= \dfrac{n + 1}{2}\left(2(2a + (4n + 1)d) + n(-4d)\right) \\= \dfrac{n + 1}2\left(4a + (4n + 2)d\right) \\= (n + 1)(2a+(2n + 1)d)\)

multiply the equation by 5  to get

5x - 5y - 5z = -15         now since   5y = 3x

5x - 3x - 5z = -15          multiply THIS equation by 1.6

8x -4.8x - 8z = -24        now since 8z = 3x

8x - 4.8x -3x = -24

x = -120

Since the height of the cone is the same as the hieght of the scoop of ice cream, the height of the

cone =  2r,  where r is the radius of both the cone and the sphere of ice cream.

The volume of the sphere:  V  =  (4/3)·pi·radius³  =  V  =  (4/3)·pi·r³​ 

The volume of the cone:  V  =  (1/3)·pi·radius²·(height)   =  (1/3)·pi·r²·(2r)  =  (2/3)·pi·r³

The volume of the sphere is twice the volume of the cone, so it can fill two cones.

I believe that the number 2 is found only once; in the line 1  2   1.

Unless what is meant the number of times a digit is found (which can be part of a longer number) ...

To put it into standard form (and to make it look better), multiply by -1. This is the resulting equation: \(-1(4x^2+3x-2)\). For the whole next part we will totally disregard the -1.Then we complete the square. But to do that we need to divide by four. \(4(x^2+\frac{3}{4}x-\frac{1}{2})\). Now we can easily complete the square. \(4[(x^2+\frac{3}{4}x+(\frac{3}{8})^2)-\frac{1}{2}]-\frac{9}{16}\). Then write it out into a simpler form: \(4[(x+\frac{3}{8})^2-\frac{1}{2}]-\frac{9}{16}\). Now expand what is within the parentheses. \(4(x+\frac{3}{8})^2-2-\frac{9}{16}\). Combine the constants \(\Rightarrow 4(x+\frac{3}{8})^2-\frac{41}{16}\).

But wait!!! If you remember what I said: we have to multiply by the negative one since we disregarded it back in the first paragraph. So we multiply the whole thing by negative one. This is what it will look like: \(-4(x+\frac{3}{8})^2+\frac{41}{16}\). By the Trivial Inequality, we know that that square right in the middle must be positive or 0. But if you look back at the equation: you have to multiply by a negative number (-4). So the smaller the square is the better. If the square can only be 0 or positive. We definitely want it to be 0. If you plug that into the equation, you will realize that it is only just \(\frac{41}{16}\). 

Edit: I realized I made 2 stupid mistakes! (Thanks EP!!!!)
\(\)

From 5th to 32nd is 27 terms      27 of the common diference, d

From  9   to  -84   is  - 93   and is 27 of d     -93/27 = d

from 5th to 23rd = 18 terms    or   18 d's added to  9   

   9 + 18 (-93/27) = -53  

-4 x^2 -3x+2    is a dome shaped parabola        max will occur at x = -b/2a        b = -3   a = -4

   use this value of 'x' to calculate the max value

Use the equation for a term of an arithemetic sequence: a+d(n-1), where a is the first term, r is the common difference, and n is the term you are trying to find the value of.

From the first sentence we form the equation:

a+d(4)=9--->a+4d=9

a+d(31)=-84--->a+31d=-84

Now we have a system of equations and solve to get a=205/9 and d=-31/9.

So the 23rd term is

205/9-31/9(23-1)

=205/9-31/9*22

=-53

The details are left to you.

Notice that the complex numbers are \(\omega = \dfrac{-1 + i\sqrt 3}{2}\) and \(\omega^2 = \dfrac{-1 - i\sqrt 3}{2}\) respectively, where \(\omega\) is the cube root of unity.

The original expression equates to \(\omega^{2021} + \left(\omega^2\right)^{2021} = \omega^{2021} + \omega^{4042}\).

Noticing that \(\omega^3 = 1\), \(\omega^{2021} + \omega^{4042} = \omega^2 + \omega = \dfrac{-1-i\sqrt 3}2+ \dfrac{-1+i\sqrt 3}2 = \boxed{-1}\).

If you look at the "first terms", the 101, 93, ... 5, these have a difference of 8.

We can use this difference to express not only the "first terms" (101, 93, ... 5) but also the "second terms"

(97, 89, ... 1):

If n = 0:     8n + 5  =  5        8n + 1  =  1

...

If n = 11,   8n + 5  =  93      8n + 1  =  89

If n = 12,  8n + 5  =  101     8n + 1  =  97

Also note that if you multiply the expressions together, you get a difference of squares:

  (101 + 97)(101 - 97)  =  101² - 97²  =  792

  (93 + 89)(93 - 89)      =  93² - 89²     =  729

  ...

  (5 + 1)(5 - 1)  =  5² - 1²  =  24

If we rewrite the numbers using the  8n + 5  and  8n + 1  forms;

    [ (8n + 5) + (8n + 1) ] ·[ (8n + 5) - (8n + 1) ]  

=    (8n + 5)² - (8n + 1)² 

=    64n² + 80n + 25) - (64n² + 16n + 1)

=    64n + 24    [this clearly indicates that it will be an arithmetic progression with a common difference of 64]

First term:  n = 0     --->   value  = 24

Last term:  n = 12   --->   value = 792         [there are 13 term]

Sum  =  (13)(24 + 792) / 2  =  5304

Explain this problem a little better please!

See how the exponents are the same and so are both of the parts. Try to maybe simplify it. This is an answer you have to get using cheeky simplifying. See what you can cancel out, etc...

-11/4!

Thank you so much!

Put the biggest numbers in the  one's position:

9   +  * 8  + * * 7  +  *** 6         then put the next biggest numbers in the 10's place

9 + 58 + * 47 + **36                continue with the 100's place

9 + 58 + 247 + 0136               and then add them .....

20 pages/4 hr   *   x  hrs    +  20 pages/5hr  * x hr   = 900 pages

20/4  x      +  20/5  x   = 900      solve for 'x'  the number of hours

                                       then slower (typer) must type        20/5  * x   pages   .....

For it to have two real roots, the disciminant portion of the Quadratic Formula must = 0        (b²-4ac)=0

a = 8

b = 15  

c = c                   You should be able to take it from here......

tan(30^o)  =  5 / x

           x  =  5 / tan(30^o)

If you complete the square on  x² - 3x

                                                 x² - 3x + 9/4                 (divide -3 by 2 and square the result)

                                                 ( x - 3/2 )²

So:  x² - 3x + 1  =  ( x² - 3x ) + 1

                          =  ( x² - 3x + 9/4 ) + 1 - 9/4                 (if you add 9/4 to a side, you must also subtract it)

                          =  ( x - 3/2 )² - 5/4

Left for you to finish ...

csc(x)  =  1 / sin(x)

--->   csc(x) / sin(x)  =  [ 1 / sin(x) ] / sin(x)  =  1 / sin²(x) 

cot(x)  =  cos(x) / sin(x)

tan(x)  =  sin(x) / cos(x)

--->   cot(x) / tan(x)  =  [ cos(x) / sin(x) ] / [ sin(x) / cos(x) ]  =  cos²(x) / sin²(x)

So, we now have:  [ 1 / sin²(x) ] - [ cos²(x) / sin²(x) ]

--->                         [ 1 - cos²(x) ] / sin²(x)

--->                                  sin²(x) / sin²(x)  =  1