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# How do I find the answer to (10/11)² - 1/11 • 1 1/12 ? No matter how I work it I'm told my answer is wrong.

How do I find the answer to (10/11)² - 1/11 • 1 1/12= ? No matter how I work it I'm told my answer is wrong. Note: 99 119/132 isn't the answer. I got that by (10/11)² = 100/121, 1/11 • 1 1/12= 13/132, so 100/121 - 13/132 = 99 119/132

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### How do I find the answer to (10/11)² - 1/11 • 1 1/12 ? No matter how I work it I'm told my answer is wrong.

How do I find the answer to (10/11)² - 1/11 • 1 1/12= ? No matter how I work it I'm told my answer is wrong. Note: 99 119/132 isn't the answer. I got that by (10/11)² = 100/121, 1/11 • 1 1/12= 13/132, so 100/121 - 13/132 = 99 119/132 Please note that your last computation, 100/121 - 13/132 = 99 119/132 is where you went wrong. You final result, 99 119/132, you would get from, 100 - 13/132 = 99 119/132. It looks as though you simply forgot to divide by 121! The answer is: 100/121 - 13/132 = (100*132 – 13*121) / (121 * 132) 100/121 - 13/132 = (13200 – 1573) / (15,972) 100/121 - 13/132 = (11,627) / (15,972) = 1057/1452  (take out a factor of 11, top and bottom) Answer: 1057/1452

### How to Find Square Root

98=49*2, so sqrt(98)=sqrt(49)*sqrt(2)=7sqrt(2)=7*1.4142=9.8994 approx. There's another way using the binomial theorem. 98=100-2=100(1-0.02). sqrt(100)=10 so sqrt(98)=10(1-0.02)^(1/2) because square root is the same as power 1/2. (1+x)^n expands to 1+nx+(n(n-1)/1*2)x^2+(n(n-1)(n-2)/1*2*3)x^3+... Putting n=1/2 and  x=-0.02, we get sqrt(98)=10(1-0.02)^(1/2)=10[1-(1/2)0.02+((1/2)(-1/2)/2)0.0004+...]. This gives us: 10(1-0.01-0.00005+...)=10*0.98995=9.8995. A third method is to use an iterative process, which means you keep repeating the same action over and over again. Look at this: x=10-(2/(10+x)). If we solve for x we get x=sqrt(98); but we're going to find x in an iterative way. Start with x=0 and work out the right hand side: 10-2/10=9.8. This gives us a new value for x, 9.8, which we feed back into the right hand side: 10-(2/(10+9.8))=10-2/19.8=9.8989..., giving us another value for x, 9.8989... which we feed back into the right hand side: 10-(2/(10+9.8989...))=9.89949..., giving us yet another value for x and so on. Very quickly we build up accuracy with each x. You can do this on a calculator, a basic one that doesn't even have square roots, using the memory to hold values for you. Here's a very simple program, where STO means store in memory (if your calculator doesn't have STO use MC (memory clear) followed by M+ (add to memory)); MR means read memory (the steps show what calculator keys to press in order; / may be ÷ on your calculator): 0= +10=STO 10-2/MR= GO TO STEP 2 OR STOP (display should show the answer for sqrt(98)) Note: In STEP 3 the division must be carried out before subtracting from 10, otherwise you get the wrong answer. If your calculator doesn't do this you need to replace STEP 3 with: 0-2=/MR=+10= You should only have to go round the loop a few times before you get a really accurate result. To find the square root of 2 directly the iteration equation is x=1+1/(1+x) and the program is: 0= +1=STO 1/MR+1= GO TO STEP 2 OR STOP STEP 3 should work on all calculators.

### What is Mike's speed given the information below

In the first part of the problem, Andrew, traveling at a speed of v1, travels 4 miles, while Mike, traveling at speed v2, travels (d - 4) miles. They leave from their respective starting points at the same time, so the time it takes for them to meet and pass is the same for both. t = d / s 1. t1 = 4/v1 = (d - 4) / v2 Multiply both sides by v1 to eliminate the denominator on the left side, and multiply both sides by v2 to eliminate the denominator on the right side. 2. (4 / v1) * v1 * v2 = ((d - 4) / v2) * v1 * v2 3. 4v2 = (d - 4) v1 Divide both sides by four to get the value of v2 4. v2 = ((d - 4)v1) / 4 In the second part of the problem, Andrew has reached Simburgh (d) and turned around, travelling another 2 miles, or (d + 2), while Mike has reached Kirkton and turned around, travelling another (d - 2) miles, for a total of d + (d - 2) = (2d - 2) miles. Again, their times are equal when they meet and pass. 5. t2 = (d + 2) / v1 = (2d - 2) / v2 As in the first part, multiply both sides by v1 to eliminate the denominator on the left side, and multiply both sides by v2 to eliminate the denominator on the right side. 6. ((d + 2) / v1) * v1 * v2 = ((2d - 2) / v2) * v1 * v2 7. (d + 2)v2 = (2d - 2)v1 Divide both sides by (d + 2) go get the value of v2 8. v2 = ((2d - 2)v1) / (d + 2) We have two equations for v2, equation 4 and equation 8. The problem states that v2 remains the same throughout the journey. Therefore: ((d - 4)v1) / 4 = ((2d - 2)v1) / (d + 2) Once again, multiply both sides by both denominators. (((d - 4)v1) / 4) * 4 * (d + 2) = (((2d - 2)v1) / (d + 2)) * 4 * (d + 2) v1 * (d - 4) * (d + 2) = v1 * (2d - 2) * 4 Divide both sides by v1, eliminating speed from this equation. (d - 4) * (d + 2) = (2d - 2) * 4 d^2 - 4d + 2d - 8 = 8d - 8 d^2 - 2d - 8 = 8d - 8 Subtract 8d from both sides and add 8 to both sides. (d^2 - 2d - 8) - 8d + 8 = (8d - 8) + 8d + 8 = 0 d^2 - 10d = 0 Factor out a d on the left side. d * (d - 10) = 0 One of those factors is equal to 0 (to give a zero answer). d = 0 doesn't work; we already know the distance is more than 4 miles. d - 10 = 0 d = 10    <<<<<   That's the answer to the first question, how far is it? We'll substitute that into equation 4 to find v2 in relation to v1. v2 = ((d - 4)v1) / 4 = ((10 - 4)v1) / 4 v2 = 6v1 / 4 = (6/4)v1 v2  = 1.5 * v1   <<<<<< That's the answer to the second question No matter what speed you choose for Andrew (v1), Mike's speed is one-and-a-half times faster. Let's set Andrew's speed to 6mph and solve equation 1. t1 = 4/v1 = (d - 4) / v2 t1 = 4mi / 6mph = (10 - 4) / (1.5 * 6mph) 4/6 hr = 6/9 hr 2/3 hr = 2/3 hr With Andrew travelling at 4 mph, and Mike travelling at 6 mph, it took both of them 2/3 of an hour to reach a point 4 miles from Kirkton.

### How to find equation of the circle passing through (9,1), (-3,-1) and (4,5)

The equation of a circle is more specific: (x-h)^2+(x-k)^2=a^2, where (h,k) is the centre and a the radius. Plug in the points: (9-h)^2+(1-k)^2=(-3-h)^2+(-1-k)^2=(4-h)^2+(5-k)^2=a^2 This can be written (9-h)^2+(1-k)^2=(3+h)^2+(1+k)^2=(4-h)^2+(5-k)^2=a^2 Leaving a out of it for the moment we can use pairs of equations, using difference of squares: 12(6-2h)+2(-2k)=0, 36-12h-2k=0, 18-6h-k=0, so k=18-6h. 7(-1-2h)+6(-4-2k)=0, -7-14h-24-12k=0, -31-14h-12k=0 or 31+14h+12k=0, 31+14h+12(18-6h)=0, 31+14h+216-72h=0. 247-58h=0 so h=247/58 This looks suspiciously ungainly. So rather than continuing, I'm going to call the three points: (Q,R), (S,T), (U,V) to provide a general answer to all questions of this sort. (Q-h)^2+(R-k)^2=(S-h)^2+(T-k)^2=(U-h)^2+(V-k)^2=a^2 [(equation 1)=(equation 2)=(equation 3)=a^2] Take the equations in pairs and temporarily ignore a^2. Equations 1 and 2: (Q-S)(Q+S-2h)+(R-T)(R+T-2k)=0 Q^2-S^2-2h(Q-S)+R^2-T^2-2k(R-T)=0 2k(R-T)=Q^2+R^2-(S^2+T^2)-2h(Q-S), so k=(Q^2+R^2-(S^2+T^2)-2h(Q-S))/(2(R-T)) Equations 2 and 3: S^2-U^2-2h(S-U)+T^2-V^2-2k(T-V)=0, so k=(S^2+T^2-(U^2+V^2)-2h(S-U))/(2(T-V)) At this point, we can substitute for k and end up with an equation involving the unknown h only: (Q^2+R^2-(S^2+T^2)-2h(Q-S))/(2(R-T))=(S^2+T^2-(U^2+V^2)-2h(S-U))/(2(T-V)) (T-V)(Q^2+R^2-(S^2+T^2)-2h(Q-S))=(R-T)(S^2+T^2-(U^2+V^2)-2h(S-U)) (T-V)(Q^2+R^2-(S^2+T^2))-2h(T-V)(Q-S)=(R-T)(S^2+T^2-(U^2+V^2))-2h(R-T)(S-U) 2h((R-T)(S-U)-(T-V)(Q-S))=(R-T)(S^2+T^2-(U^2+V^2))-(T-V)(Q^2+R^2-(S^2+T^2)) h=((R-T)(S^2+T^2-(U^2+V^2))-(T-V)(Q^2+R^2-(S^2+T^2)))/((R-T)(S-U)-(T-V)(Q-S)). Once h is found we can calculate k, and then we can substitute the values for h and k in any equation to find a^2 which is equal to each of the three equations. When we use Q=9, R=1, S=-3, T=-1, U=4, V=5 or -5, we appear to get very ungainly solutions. One way to find which values need to be changed may be to plot the values and work out where the circle should fit and where its centre has less complex values. If V=6 or -6, there is a simple solution: (x-3)^2+y^2=37 (centre at (3,0), a^2=37=6^2+1^2. (-3-3)^2=(9-3)^2=6^2; (-6)^2=6^2; (2-3)^2=(4-3)^2=1; (-1)^2=1^2 shows the combination of points that would lie on the circle: (9,1), (9,-1), (-3,1), (-3,-1), (4,6), (4,-6), (2,6), (2,-6) and this includes points A and B, but not C. Note that the equation k=18-6h is valid for h=3, k=0.

### Rolles theorem on f(x) = Xsqrt(64-X^2) on the interval [-8,8]?

your derivative should have been sqrt(64-x^2) - x^2/sqrt(64-x^2) The minus sign coming from the derivative of (-x^2) Setting the derivative to zero, sqrt(64-x^2) - x^2/sqrt(64-x^2) = 0   multiply both terms by sqrt(64 - x^2) (64 - x^2) - x^2 = 0 64 = 2x^2 32 = x^2 x = +/- 4.sqrt(2) Answer: option b

### i cant remember how to factor a quadratic equation

There are two possibilities.  One is that this factors nicely (we know it does in this case).  The other is that it doesn't factor nicely (givings us irrational answers like x = 3.584538257389025 etc.).  The steps below should work if the equation factors nicely, but won't work if the answers are irrational / not friendly. Note:  Once you get used to it, you won't have to follow all of these steps and do all of this stuff every time.  You'll get used to it, like how you don't have to do a multiplication problem on paper to know that 10 * 10 = 100.  Here's the step by step procedure though. 6x^2 + 7x - 3 = 0 We're looking for something like (?x + ??)(???x + ????) The number on the x^2 is 6, so the numbers in front the x's (? and ???) have to multiply together and make 6.  That means ? and ??? should be 1 and 6, 6 and 1, 2 and 3, or 3 and 2.  That means we should have something like: (x + ??)(6x + ????) or (2x + ??)(3x + ????) Now look at the number without an x in the equation:  -3.  The only way to factor 3 is 1 and 3.  But, since we're starting with -3, the factors can be 1 and -3 or -1 and 3.  That means we now have something like this: (x + 1)(6x - 3) or (2x + 1)(3x - 3) (x - 1)(6x + 3) or (2x - 1)(3x + 3) (x + 3)(6x - 1) or (2x + 3)(3x - 1) (x - 3)(6x + 1) or (2x - 3)(3x + 1) So that's all eight possible combinations.  Which one is right? In the equation we have 7x in the middle.  We have to get a 7 to sit on the x. Look at (x + 1)(6x - 3).  The number sitting on the x in the equation is made by x * -3 + 1 * 6x = -3x + 6x = 3x.  That didn't make 7 to sit on the x, so we know (x + 1)(6x - 3) is wrong. (x + 1)(6x - 3) or (2x + 1)(3x - 3) (x - 1)(6x + 3) or (2x - 1)(3x + 3) (x + 3)(6x - 1) or (2x + 3)(3x - 1) (x - 3)(6x + 1) or (2x - 3)(3x + 1) Let's go through the other possibilities and see what we get (x + 1)(6x - 3): 3x or (2x + 1)(3x - 3): -3x (x - 1)(6x + 3): -3x or (2x - 1)(3x + 3): 3x (x + 3)(6x - 1): 17x or (2x + 3)(3x - 1): 7x (x - 3)(6x + 1): -17x or (2x - 3)(3x + 1): 3x The only possibility that makes a 7 to sit on the x is (2x + 3)(3x - 1), so that's our answer. 6x^2 + 7x - 3 = (2x + 3)(3x - 1)

### seven times a two digit number

Short answer:  The two digit number is 36. Long answer: 7 * x1x2 = 4 * x2x1 "If the difference between the number is 3. . ." Last digits: x2: 0, 1, 2, 3, 4, 5, 6, 7, 8, 9 7*x2:  0, 7, 4, 1, 8, 5, 2, 9, 6, 1 x1:  0, 1, 2, 3, 4, 5, 6, 7, 8, 9 4*x1:  0, 4, 8, 2, 6, 0, 4, 8, 2, 6 The only way this works is when these last digits for 7*x2 and 4*x1 are the same.  That means the 7*x2 line can only be: 7*x2:  0, 4, 8, 2, 6 Which means the possible values for x2 are: x2:  0, 2, 4, 6, 8 Now let's look at x1.  Right now the possible values for x1 are: x1:  0, 1, 2, 3, 4, 5, 6, 7, 8, 9 But we have to end up with a choice from x1 and x2 having a difference of 3 (odd number).  There is no way to get an odd number by subtracting an even number from an even number.  That means x1 has to be odd.  Our possible values for x1 are now: x1:  1, 3, 5, 7, 9 And our possible values for x2 are: x2:  0, 2, 4, 6, 8 The possible combinations for x1x2 and x2x1 are: 10, 01 12, 21 14, 41 16, 61 18, 81 30, 03 32, 23 34, 43 36, 63 38, 83 50, 05 52, 25 54, 45 56, 65 58, 85 70, 07 72, 27 74, 47 76, 67 78, 87 90, 09 92, 29 94, 49 96, 69 98, 89 We want 7*x1x2 = 4*x2x1, so we can do 7 * the first column and 4 * the second column: 70, 4 84, 84 98, 164 112, 244 126, 324 210, 12 224, 92 238, 172 252, 252 266, 332 350, 20 364, 100 378, 180 392, 260 406, 340 490, 28 504, 108 518, 188 532, 268 546, 348 630, 36 644, 116 658, 196 672, 276 686, 356 But since we want 7*x1x2 to equal 4*x2x1, that list reduces to: 84, 84 252, 252 The corresponding x1 and x2 values are: 12, 21 36, 63 But the difference between x1 and x2 is 3, so we can't use x1 = 1, x2=2.  We have to use x1 = 3, x2 = 6. Answer:  The two digit number is 36. Check:  7 * 36 = 4 * 63 252 = 252 good. 6 - 3 = 3 good.

### whats the answer to -2x + 3 > 3(2x - 1)

We need to expand the brackets on the right: -2x+3>6x-3. Add 2x to each side of the inequality: 3>8x-3 and add 3 to each side: 6>8x. Another way of writing this is 8x<6. Divide through by 2: 4x<3 and divide through by 4: x<3/4. We added 2x to each side because that helps us to bring the x's together. We added 3 to each side to bring the numbers together. Now we've ended up with x's on one side and a number on the other. Just what we want. 2 is common to 6 and 8 so we simplify the inequality by dividing through by it. Then to get x instead of 4x we divided through by 4. Why is 6>8x the same as 8x<6? Well, think about it. 2<3, isn't it? So 3>2. That's true. If Jack is taller than Jill, then Jill is shorter than Jack. See how it works? You just reverse the inequality when the quantities swap sides. It's a good idea to check your answer by substituting a value for x in the original inequality, just to make sure we haven't made a mistake. The answer was that x must be less than 3/4 so let's try x=1/2. -2x+3 is -1+3=2. Now the right side: 2x-1 is 1-1=0 multiplied by 3 is still zero. Is the inequality correct? Yes it is, because 2 is greater than zero. If we put x=1 we should find that the inequality doesn't work out because x is bigger than 3/4. Let's see. The left is 3-2=1 and the right is 3. 1 isn't bigger than 3, so the inequality is false, as expected. If we put x=3/4 we will find that the left equals the right when it should be bigger than it. Looks like x<3/4 is right!