How do you solve |-26+10|=?
│-26+10│ = absolute value of -26+10 The absolute value ││ is always a positive (+). So what you need to do is solve the equation inside the ││and add a positive sign to it │-26+10│ = │-16│ = 16
x+4=3then, x=3–4x=-1:so x=-1verification:substitute -1 as xso,-1+4=3=3=3hence verified
We have:[math]P = \begin{bmatrix}1 & 1 & 1\\ 0 & 1 & 1 \\ 0 & 0 & 1\end{bmatrix}[/math]We have: [math]P^2 = \begin{bmatrix}1 & 1 & 1\\ 0 & 1 & 1 \\ 0 & 0 & 1\end{bmatrix}.\begin{bmatrix}1 & 1 & 1\\ 0 & 1 & 1 \\ 0 & 0 & 1\end{bmatrix} = \begin{bmatrix}1 & 2 & 3\\ 0 & 1 & 2 \\ 0 & 0 & 1\end{bmatrix}[/math]We have [math]P^3 = \begin{bmatrix}1 & 2 & 3\\ 0 & 1 & 2 \\ 0 & 0 & 1\end{bmatrix}.\begin{bmatrix}1 & 1 & 1\\ 0 & 1 & 1 \\ 0 & 0 & 1\end{bmatrix} = \begin{bmatrix}1 & 3 & 6\\ 0 & 1 & 3 \\ 0 & 0 & 1\end{bmatrix}[/math]Do you start seeing a pattern here? Let’s do one more.[math]P^4 = \begin{bmatrix}1 & 3 & 6\\ 0 & 1 & 3 \\ 0 & 0 & 1\end{bmatrix}.\begin{bmatrix}1 & 1 & 1\\ 0 & 1 & 1 \\ 0 & 0 & 1\end{bmatrix} = \begin{bmatrix}1 & 4 & 10\\ 0 & 1 & 4 \\ 0 & 0 & 1\end{bmatrix}[/math]By a rough guess, we can say from this pattern that:[math]P^n = \begin{bmatrix}1 & n & \frac{n(n+1)}{2}\\ 0 & 1 & n \\ 0 & 0 & 1\end{bmatrix}[/math]As it turns out, this is actually not a bad guess. The above result can be proved quite easily using induction, and I won’t be presenting the proof here.But to conclude, we have:[math]P^{50} = \begin{bmatrix}1 & 50 & 1275\\ 0 & 1 & 50 \\ 0 & 0 & 1\end{bmatrix}[/math]You’re welcome.
How do you solve: 5!?
Those are called factorials (not related to factors), and just means that you take the product of every integer, starting at the given number and going down until you end at 1. So 5! = 5 * 4 * 3 * 2 * 1=120 9! = 9 * 8 * 7 * 6 * 5 * 4 * 3 * 2 * 1=362,880 9! = 9 * 8 * 7 * 6 * 5! --see what I did here? This line and the one above it are basically the same. Similarly (9!)/(5!)=9*8*7*6 You could prove this by doing it the long way (9!)/(5!)=( 9 * 8 * 7 * 6 * 5 * 4 * 3 * 2 * 1)/(5 * 4 * 3 * 2 * 1) You see that all of the numbers below 6 cancel out. These don't have anything to do with integrals (directly at least).
How do you solve this? 1/3 + x = 3/4?
1/3 + x = 3/4 Just subtract 1/3 from both sides (to isolate x): x = 3 / 4 - 1 / 3 That is the answer for the value of x, but it's an ugly expression and we need to simplify it into one rational fraction. First, multiply each numerator by the OTHER denominator, to get the new numerator: (3*3) - (1*4) so the new numerator is 9-4 or 5 Now, multiply the denominator, to get a new, common denominator: 4 * 3 is 12 So the fraction is: = 5 / 12 (If you use this method, don't forget to reduce the fraction, but here 5 and 12 have no common denominator, so that's the answer.)
How do you Solve for Q?
The reason why you seem confused is because you do not have enough equations to explicitly solve for Q and P, and get a unique solution. In this case, you are going to get an infinite solution. This is right now just like the old linear line y = mx+b. So you get a linear line, and depending on what P is, you get a certain Q. Note that if you have n variables, then you need n distinct equations to get a single unique answer.
How do you solve x+3+2x=x+5?
Sharon, You want to get all the numbers on one side of the equal sign and the x on the other side of the equal sign. Let's move the 3 to the right side of the equal sign. Remember that whatever we do to one side of the equal sign we must also do to the other side of the equal sign. x (+3 -3) +2x=x (+5 -3) Now let's move the x. (x -x) +2x = (x -x) +2 2x = 2 What has been done to the x? It has been multiplied by 2. What is the opposite of multiplication? It is division 2x = 2 ---__--- 2___2 x = 1 Let's test this by substituting 1 for x. 1+3+(2 * 1)=1+5 4 + (2*1) = 6 6 = 6
How do you solve sec 4x = 2?
All other answers don't seem to understand what sec means! Sec4x = 2 is the same as 1/cos(4x) = 2 Therefore cos(4x) = 1/2 I know from memory cos(root3/2) = 1/2 So root3/2 = 4x Root3/8 = x You didn't specify a range of x values however, so there could possibly be an infinitite number of values for x.
How do you solve (6/x+3) + (1/2x+5) =3?[math]\left (\frac{6}{x}+3 \right )+\left (\frac{1}{2}x+5 \right )=3[/math]1) Get ride of the parenthesis, they are really superfluous here (or used wrong, I will come to this later).[math]\frac{6}{x}+3+\frac{1}{2}x+5=3[/math]2) Shift the known values to one side.[math]\frac{6}{x}+\frac{1}{2}x=-5[/math]3) Multiply both sides with [math]x[/math][math]6+\frac{1}{2}x^2=-5x[/math]4) Multiply with [math]2[/math][math]12+x^2=-10x[/math]5) Make one side zero (add [math]10x[/math] to both sides):[math]x^2+10x+12=0[/math]6) Complete the square:[math](x+5)^2-25+12=0[/math]7) Isolate [math](x+5)^2[/math] by adding [math]13[/math] to both sides ([math]25-12=13[/math]):[math](x+5)^2=13[/math]8) Take the square root:[math]x+5=\pm \sqrt{13}[/math]9) Subtract 5[math]\boxed{x=-5 \pm \sqrt{13}}[/math]Now If you meant[math]\frac{6}{x+3}+\frac{1}{2x+5}=3[/math]This is how the parenthesis would be used right here:(6/(x+3)) + (1/(2x+5)) =31) Unify the fractions:[math]\frac{6 \cdot (2x+5)+(x+3)}{(x+3) \cdot (2x+5)}=3[/math]2) Undo the fraction by multiplying with its denominator[math]6 \cdot (2x+5)+(x+3)=3 \cdot (x+3) \cdot (2x+5)[/math]3) Expand[math]13x+33=6x^2+33x+45[/math]4) Make one side zero (minus [math]13x+33[/math])[math]6x^2+20x+12=0[/math]5) Divide by [math]6[/math][math]x^2+\frac{10}{3}x+2=0[/math]6) Complete the square:[math]\left (x+\frac{5}{3} \right )^2-\frac{100}{36}+2=0[/math]7) Isolate [math]\left (x+\frac{10}{6} \right )^2[/math] by adding [math]\frac{7}{9}[/math] ([math]\frac{100}{36}-2=\frac{7}{9}[/math])[math]\left (x+\frac{5}{3} \right )^2=\frac{7}{9}[/math]8) Take the square root:[math]x+\frac{5}{3}=\pm \frac{\sqrt{7}}{3}[/math]9) Subtract [math]\frac{5}{3}[/math][math]\boxed{x=\frac{-5 \pm \sqrt{7}}{3}}[/math]Third option, really quick:((6/x)+3) + ((1/(2x)+5) =3?[math]\left (\frac{6}{x}+3 \right )+\left (\frac{1}{2x}+5 \right )=3[/math]1) The parenthesis are superfluous again.[math]\frac{6}{x}+3+\frac{1}{2x}+5=3[/math]2) Bringing known values to one side:[math]\frac{6}{x}+\frac{1}{2x}=-5[/math]3) Multiply with [math]2x[/math][math]13=-10x[/math]4) Divide by [math]-10[/math][math]\boxed{x=\frac{-10}{13}}[/math]
Thanks for asking. Remember, to solve an equation for X we must collect the terms with X on one side and the constant terms on the other side of the equation. We must remember that the term is changing sign when transferred from one side to the other.Here we have an expression in brackets, we need to do that first. Observe that we are required to multiply the expression in the bracket by 3 and then divide by 3. So that just cancels out. 3/3=1, so it is equivalent to just multiplying by 1.Then we may rewrite the equation:3X+3X-3=3. Doing the transfer of the constant term6x=6 or X=1Test it. Put 1 in for X in the original equation:3+3(3–3)/3=3 Indeed 3–3 in the bracket=0 Multiplied and then divided by 3 is still 0 So we get 3=3.