Guide :

342-173 solve this equation in base 8 (the answer in base 10 is 169)

solve in base 8 or octal

Research, Knowledge and Information :

The Number Base Calculator - Cleave Books

The Number Base Calculator. ... base 3 = ternary base 8 = octal base 10 = denary or decimal base 12 = duodecimal base 16 = hexadecimal The symbol & (called an ...

Converting to base 10 - Math Bits

Change a number from any base back to base 10: The computer is representing a number from the ... 1 + 2 + 0 + 8 = 11 Answer: 1011 2 = 11 10 . Example 3: ...

WEAK ACIDS AND BASES - University of Western Ontario

WEAK ACIDS AND BASES ... so the approximate answer 1.342 × 10 —3 M only differs from the ... base, defined by the equation, have been mixed.

Number Base Converter

Calculator that converts a base-10 number to other numeric bases between base-2 and base-36. ... Number Base Converter Convert a Base-10 Number to Other Bases: Base ...

Base Conversion Tool - Math is Fun

Base Conversion Tool. Click in either box and type. ... Hexadecimal (Base 16) Because there are more than 10 digits, hexadecimal is written using letters as well, ...

Online calculator: Simple math in any numeral system

Simple math in any numeral system from the online ... I used this to get the answers to a ... add a limit to the precision as 10 in base 16 with 66 precision ...

Base five numeration system - Basic mathematics

Base five numeration system The main difference between base five numeration system and our familiar base 10 numeration system is that grouping is done in groups of 5 ...

solving for x

I interpret the first number in brackets as the base.You need a calculator to do this because, although the bases 2 and 4 are related to one another because 4 is 2 squared, there's no easy relation between 2 and 3, 3 and 10, or 2 and 10. 10 in base 2 is approximately 3.322; log 2 in base 10 is approximately 0.3010 which is the reciprocal of 3.322. Log 3 in base 2 is approximately 1.585 and log 2 in base 3 is 0.6309 which is the reciprocal of 1.585. Log 4 to base 2 is 2 so log 2 to base 4 is 0.5. To convert log x in one base to another we use the conversion factor of the log of one base of the other base. Let's choose base 2 as the common base. This is what the equation becomes: 0.5logx+logx+logx/log3=logx/log10. Put y=logx (base 2) and we get 0.5y+y+0.6903y=0.3010y. If we were to divide through by y we would have an inequality because 2.1903 is not equal to .3010. Therefore y=0 and logx=0, so x=1. That was the longwinded way of proving the solution. We can see that x=1 makes all the terms in the original equation zero and 0=0 is always true! Why go to all the trouble of using a calculator when the only answer is x=1? It was just possible that the sum of the numbers on the left side of the equation came to the result on the right side of the equation, in which case we would have had an identity instead of an equation, which would have been true for all values of x (except x=0).

solve the equation 3 log(base 2)(x-1)+log(base 2)4=5

The equation is equivalent to log (base 2 ) (X-1)^3 (4) = 5 , therefore , log(base 2)[X^3-3X^2+3X-1](4) = 5 , so , 2^5 = 4X^3-12X^2+12X-4 , we simplify a little bit more , then 32 = 4(X^3 - 3X^2 + 3X - 1) , then : 8 = X^3 - 3X^2 + 3X - 1 , putting the whole thing in standard form yuo get X^3 - 3X^2 + 3X - 9 = 0 , which you factorize in this way : X^2(X-3) 3(X-3) = 0 , (X-3)(X^2 +3) =0 , implying that X = 3 , X = 3^(1/2) i and X = -3^ (1/2)i three roots because is a third degree equation . Because we were asked to express the irrational answer rounded up to the third place , then we add : 3^(1/2) i = 1.732 i and its pair ( every irrational root comes in pairs ) would be -1.732 i

x - z = -3, y + z = 9, -2x + 3y +5z = 33

Problem: x - z = -3, y + z = 9, -2x + 3y +5z = 33 1) x - z = -3 2) y + z = 9 3) -2x + 3y + 5z = 33 Add equation 2 to equation 1.      x     - z = -3 +(    y + z =  9) ------------------   x + y     = 6 4) x + y = 6 Multiply equation 2 by 5. 5(y + z) = 9 * 5 5) 5y + 5z = 45 Subtract equation 3 from equation 5.             5y + 5z = 45 -(-2x + 3y + 5z = 33) --------------------------     2x + 2y        = 12 6) 2x + 2y = 12 Multiply equation 4 by 2. 2(x + y) = 6 * 2 7) 2x + 2y = 12 Subtract equation 7 from equation 6.    2x + 2y = 12 -(2x + 2y = 12) --------------------   0x      = 0      We are solving for x, not y x = 0    <<<<<<<<<<<<<<<<<<< Use equation 4 to solve for y. x + y = 6 0 + y = 6 y = 6    <<<<<<<<<<<<<<<<<<< Use equation 3 to solve for z. -2x + 3y + 5z = 33 -2(0) + 3(6) + 5z = 33 0 + 18 + 5z = 33 5z = 15 z = 3    <<<<<<<<<<<<<<<<<<< Check the values. 1) x - z = -3    0 - 3 = -3    -3 = -3 2) y + z = 9    6 + 3 = 9    9 = 9 3) -2x + 3y + 5z = 33    -2(0) + 3(6) + 5(3) = 33    0 + 18 + 15 = 33    33 = 33 Those numbers work..... ------------- What if we had solved for y instead of x here:    2x + 2y = 12 -(2x + 2y = 12) -------------------        0y = 0      We are solving for y, not x y = 0    <<<<<<<<<<<<<<<<<<< Use equation 4 to solve for x. x + y = 6 x + 0 = 6 x = 6    <<<<<<<<<<<<<<<<<<< Use equation 3 to solve for z. -2x + 3y + 5z = 33 -2(6) + 3(0) + 5z = 33 -12 + 0 + 5z = 33 5z = 45 z = 9    <<<<<<<<<<<<<<<<<<< Check the values. 1) x - z = -3    6 - 9 = -3    -3 = -3 2) y + z = 9    0 + 9 = 9    9 = 9 3) -2x + 3y + 5z = 33    -2(6) + 3(0) + 5(9) = 33    -12 + 0 + 45 = 33    33 = 33 Those numbers work, too. That complicates the solution. Answer 1: x = 0, y = 6, z = 3 Answer 2: x = 6, y = 0, z = 9

find the truth set of

find the truth set of set of a. 3^(2x+3) -3^(x+2) -3^(x+1) +1=0 b.  log a base4=log b base 8+1 (a) we have here a function f(x) = 3^(2x+3) -3^(x+2) -3^(x+1) +1=0. What we are being asked to find is, in effect, the solution of f(x) = 0. i.e. find those values of x such that 3^(2x+3) -3^(x+2) -3^(x+1) +1=0. This will make the expression/equation true. Considering that the function is using x as an exponent, the simplest way of solving this is to plot the curve of y = 3^(2x+3) -3^(x+2) -3^(x+1) +1, and see where it crosses the x- axis, When plotted, it is noted that the x-axis is crossed at x = -1, and at x = -2. Thus the truth set is: Answer: x = {-2.-1} (b) log_4(a) = log_8(b) + 1 Let log_4(a) = x, then a = 4^x Let log_8(b) = y, then b = 8^y We can rewrite the eqn log_4(a) = log_8(b) + 1 as x = y + 1, or x – y = 1 We are creating a truth set in which we are only interested in integers. This makes the eqn x - y = 1 a Diophantine equation for which the solution is, x = 2 + t y = 1 + t t e Z We can thus write down, a = 4^(2 + t) b = 8^(1 + t) which gives us our truth set. Answer: {(a, b): a = 4^(2 + t), b = 8^(1 + t), t e Z}

solve a system of linear equations with three variables:

solve a system of linear equations with three variables: x+y-3z=1 2x-y+z=9 3x+y-4z=8 1) x + y - 3z = 1 2) 2x - y + z = 9 3) 3x + y - 4z = 8 Multiply equation 2 by 3. 3(2x - y + z) = 9 * 3 4) 6x - 3y + 3z = 27 Add equation 4 to equation 1.      x +   y - 3z =   1 +(6x - 3y + 3z = 27) --------------------------    7x - 2y        = 28 5) 7x - 2y = 28 Multiply equation 2 by 4. 4(2x - y + z) = 9 * 4 6) 8x - 4y + 4z = 36 Add equation 3 to equation 6.    8x - 4y + 4z = 36 +(3x +  y - 4z =   8) ------------------------   11x - 3y       = 44 7) 11x - 3y = 44 Multiply equation 5 by 3. 3(7x - 2y) = 28 * 3 8) 21x - 6y = 84 Multiply equation 7 by 2. 2(11x - 3y) = 44 * 2 9) 22x - 6y = 88 Subtract equation 9 from equation 8.    21x - 6y = 84 -(22x - 6y = 88) ---------------------      -x        =  -4 x = 4  <<<<<<<<<<<<<<<<< Use equation 5 to solve for y. 7x - 2y = 28 7(4) - 2y = 28 28 - 2y = 28 -2y = 0 y = 0  <<<<<<<<<<<<<<<<< Use equation 3 to solve for z. 3x + y - 4z = 8 3(4) + 0 - 4z = 8 12 - 4z = 8 -4z = -4 z = 1  <<<<<<<<<<<<<<<<< Check your answers. 1) x + y - 3z = 1    4 + 0 - 3(1) = 1    4 - 3 = 1    1 = 1 2) 2x - y + z = 9    2(4) - 0 + 1 = 9    8 + 1 = 9    9 = 9 3) 3x + y - 4z = 8    3(4) + 0 - 4(1) = 8    12 - 4 = 8    8 = 8 Everything checks. Answer: x = 4, y = 0, z = 1

3x + 6y – 6z = 9 2x – 5y + 4z = 6 -x +16y + 14z = -3 what is the answer

3x + 6y – 6z = 9 2x – 5y + 4z = 6 -x +16y + 14z = -3 what is the answer how do you solve it and what are the answers ? 1)  3x + 6y – 6z = 9 2)  2x – 5y + 4z = 6 3)  -x +16y + 14z = -3 The first objective is to eliminate z so we can solve for x and y. It's a multi-step process, so follow along. Multiply equation one by 4. 4 * (3x + 6y – 6z) = 9 * 4 4)  12x + 24y - 24z = 36 Multiply equation 2 by 6. 6 * (2x – 5y + 4z) = 6 * 6 5)  12x - 30y + 24z = 36 Now, we have two equations with a "24z" term. Add the equations and the z drops out. Add equation five to equation four.    12x + 24y - 24z = 36 +(12x - 30y + 24z = 36) ----------------------------------    24x -  6y           = 72 6)  24x - 6y = 72 The same process applies to equations two and three. Multiply equation two by 7 this time. 7 * (2x – 5y + 4z) = 6 * 7 7)  14x - 35y + 28z = 42 Multiply equation three by 2. 2 * (-x + 16y + 14z) = -3 * 2 8)  -2x + 32y + 28z = -6 Subtract equation eight from equation seven.   14x - 35y + 28z = 42 -(-2x + 32y + 28z = -6) ---------------------------------   16x - 67y          = 48 9)  16x - 67y = 48 Looking at equations six and nine, it would be simpler to eliminate the x. The multipliers are smaller. Multiply equation six by 2. 2 * (24x - 6y) = 72 * 2 10)  48x - 12y = 144 Multiply equation nine by 3. 3 * (16x - 67y) = 48 * 3 11)  48x - 201y = 144 Subtract equation eleven from equation 10.   48x -   12y = 144 -(48x - 201y = 144) ---------------------------          189y  =   0 189y = 0 y = 0  <<<<<<<<<<<<<<<<<<<<<<<<<<<<<< Substitute that value into equations six and nine to solve for x and verify. Six: 24x - 6y = 72 24x - 6(0) = 72 24x - 0 = 72 24x = 72 x = 3  <<<<<<<<<<<<<<<<<<<<<<<<<<<<<< Nine: 16x - 67y = 48 16x - 67(0) = 48 16x - 0 = 48 16x = 48 x = 3    same answer for x To solve for z,substitute the x and y values into the three original equations. One: 3x + 6y – 6z = 9 3(3) + 6(0) – 6z = 9 9 + 0 - 6z = 9 9 - 6z = 9 -6z = 9 - 9 -6z = 0 z = 0  <<<<<<<<<<<<<<<<<<<<<<<<<<<<<< Two: 2x – 5y + 4z = 6 2(3) – 5(0) + 4z = 6 6 - 0 + 4z = 6 6 + 4z = 6 4z = 6 - 6 4z = 0 z = 0     same answer Three: -x +16y + 14z = -3 -(3) +16(0) + 14z = -3 -3 + 0 + 14z = -3 -3 + 14z = -3 14z = -3 + 3 14z = 0 z = 0     once again, same answer x = 3, y = 0, z = 0

what is the answer to ... -2x+y=0 -3x-2y=-35

1) -2x + y = 0 2) -3x - 2y = -35 We can eliminate y; multiply equation 1 by 2. 2 * (-2x + y) = 0 * 2 3) -4x + 2y = 0 Add equation 2 to equation 3.    -4x + 2y =    0 +(-3x - 2y = -35) ----------------------    -7x        = -35 -7x = -35 -7x/-7 = -35/-7 x = 5 With that, we can use equation 1 to solve for y. -2x + y = 0 -2(5) + y = 0 -10 + y = 0 -10 + y + 10 = 0 + 10 y = 10 Use equation 2 to check the answers. -3x - 2y = -35 -3(5) - 2(10) = -35 -15 - 20 = -35 -35 = -35 Answer: x = 5, y = 10 Problem statement: what is the answer to ... -2x+y=0 -3x-2y=-35 it has to do with "solving systems of equations:elimination method

how do you solve this using a gausian reduction using the three equations? 4x+z=3,2x-y=2,and 3y+2z=0

how do you solve this using a gausian reduction using the three equations? 4x+z=3,2x-y=2,and 3y+2z=0 I think gaussian reduction is similar to back substitution, but with all the equation shaving different variables... just not sure how to do it. You create a matrix using the constants in the equations. Below, you will see approximately what the matrix would look like. Unfortunately, it is impossible to get it to display properly on this page. ┌                       ┐ │ 4   0   1  |   3  │ │ 2  -1  0   |  2   │ │ 0   3   2  |   0   │ └                        ┘ The idea is to perform the same math procedures on these matrix rows that you would perform on the full equations. We want the first row to be  1 0 0 | a, meaning that whatever value appears as the a entry is the value of x. The second row has to be  0 1 0 | b,  and the third row has to be  0 0 1 | c. Multiply row 2 by 2. ┌                      ┐ │ 4   0   1  |   3  │ │ 4  -2   0  |  4   │ │ 0   3   2  |   0   │ └                       ┘ Now subtract row 2 from row 1, and replace row 2 with the result. ┌                       ┐ │ 4   0   1  |   3   │ │ 0  2   1   |  -1   │ │ 0   3   2  |   0   │ └                       ┘ Multiply row 2 by 2. ┌                       ┐ │ 4   0   1  |   3   │ │ 0  4    2  |  -2   │ │ 0   3   2  |   0   │ └                       ┘ Subtract row 3 from row 2, replacing row 2. ┌                        ┐ │ 4   0   1  |   3   │ │ 0  1    0  |  -2   │ │ 0   3   2  |   0   │ └                        ┘ Multiply row 2 by 3 and subtract row 3, replacing row 3. ┌                       ┐ │ 4   0   1  |   3   │ │ 0  1    0  |  -2   │ │ 0   0  -2  |   -6  │ └                       ┘ Divide row 3 by -2. ┌                       ┐ │ 4   0   1  |   3   │ │ 0  1    0  |  -2   │ │ 0   0   1  |   3   │ └                       ┘ Subtract row 3 from row 1, replacing row 1. ┌                      ┐ │ 4   0   0  |   0  │ │ 0  1    0  |  -2  │ │ 0   0  1  |   3   │ └                      ┘ Divide row 1 by 4. ┌                      ┐ │ 1   0   0  |   0  │ │ 0  1    0  |  -2  │ │ 0   0   1  |   3  │ └                      ┘ Row 1 shows that x = 0 Row 2 shows that y = -2 Row 3 shows that z = 3 Plug those values into the original equations to check the answer. 4x + z = 3 4(0) + 3 = 3 0 + 3 = 3 3 = 3 2x – y = 2 2(0) – (-2) = 2 0 + 2 = 2 2 = 2 3y + 2z = 0 3(-2) + 2(3) = 0 -6 + 6 = 0 0 = 0

find the ordered triple of these equations.

find the ordered triple of these equations. x-2y+2z=-2 2x+3x-z=0 3x+2x+3z=-15 1) x - 2y + 2z = -2 2) 2x + 3x - z = 0 3) 3x + 2y + 3z = -15 ---- surely that's 2y, not 2x Multiply equation 2 by 2. 2(2x + 3x - z) = 0 * 2 4) 4x + 6y - 2z = 0 Add equation 4 to equation 1.      x -  2y + 2z = -2 +(4x + 6y - 2z =   0) -------------------------    5x + 4y        =  -2 5) 5x + 4y = -2 Multiply equation 2 by 3. 3(2x + 3y - z) = 0 * 3 6) 6x + 9y - 3z = 0 Add equation 3 to equation 6.    6x +  9y -  3z =    0 +(3x +  2x + 3z = -15) ---------------------------     9x + 11y      = -15 7) 9x + 11y = -15 Multiply equation 5 by 11. 11(5x + 4y) = -2 * 11 8) 55x + 44y = -22 Multiply equation 7 by 4. 4(9x + 11y) = -15 * 4 9) 36x + 44y = -60 Subtract equation 9 from equation 8.   55x + 44y = -22 -(36x + 44y = -60) ------------------------   19x           =  38 19x = 38 x = 2     <<<<<<<<<<<<<<<< Use equation 5 to solve for y. 5x + 4y = -2 5(2) + 4y = -2 10 + 4y = -2 4y = -12 y = -3     <<<<<<<<<<<<<<<< Use equation 3 to solve for z. 3x + 2y + 3z = -15 3(2) + 2(-3) + 3z = -15 6 - 6 + 3z = -15 3z = -15 z = -5     <<<<<<<<<<<<<<<< Check the answer with equation 1. 2 - 2(-3) + 2(-5) = -2 2 + 6 - 10 = -2 8 - 10 = -2 -2 = -2 Answer: x = 2, y = -3, z = -5