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9 times 4 and 2/3 equals what

9 times 4 and 2/3 equals what

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What is 4 times 9 equals to - Answers.com


What is 4 times 9 equals to? SAVE CANCEL. already exists. Would ... 12.3 times 9 equals 110.7 1 person found this useful Edit. Share to: Ghost tigeress. 3,198 ...
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What is 9 times 3 over 4 - Answers.com


3/4 x 9/1 = 27/4 27 divided by 4 equals 6.75 or 6 3/4. Go. ... Answers.com ® WikiAnswers ® Categories Science Math and Arithmetic What is 9 times 3 over 4? What ...
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What is 9 times 3/4? - Quora


What is 9 times 3/4? Update Cancel. Answer Wiki. ... So 9 times 3/4 eventually means you have 6 and ... I mean why is 2 the only number with an equal sum and ...
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the difference of a number times 9 and 5 equals 2 . Use the ...


the difference of a number times 9 and 5 equals 2 . Use the variable c for the unknown number .
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What times what equals 9? - Research Maniacs


Below is a list of all the different ways that what times what equals 9. 1 times 9 equals 9 3 times 3 equals 9 9 times 1 equals 9 . What times what equals 10? Now ...
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Multiplying 2 fractions: 5/6 x 2/3 (video) | Khan Academy


Multiplying 2 fractions: 5/6 x 2/3. Practice: Multiplying fractions. ... So the numerator in our product is just 5 times 2. So it's equal to 5 times 2 over 6 times 3, ...
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Fractions - Grade 5 Math Questions With Solutions and ...


For example 2/3 is the fraction whose numerator is 2 and denominator is 3. ... 2/3 of 4 is equal to 2/3 × 4 = 8/3 = (6 + 2)/3 = 6/3 + 2/3 = 2 2/3
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Suggested Questions And Answer :


1. Set up the following word problem as an equation and solve. Then write your answer as a complete sentence. The ratio of two numbers is 3 to 5. Their sum is 136. Find the two numbers.

Let the 2 numbers be A and B. A/B=3/5; A+B=136. So B=136-A and we can substitute B in the other equation: A/(136-A)=3/5. Cross-multiply: 5A=3(136-A)=408-3A. 8A=408, A=408/8=51. B=136-51=85. Check the answer: A/B=51/85=3/5.  In words: This is a problem with two unknowns, A and B, and two equations. We use one equation to write one unknown in terms of the other (B is 136 less A), then we substitute for that unknown in the other equation to give A divided by (136 minus A). That gives us one equation and one unknown, A, where the quotient is equal to three fifths. By cross-multiplying to get rid of the fraction, we arrive at five times A is equal to 3 times B which is known to be 136 less A. Now we collect the A terms together to give 8 times A equal to the product of 3 and 136, that is 408. From this we simply divide 408 by 8 to get A equal to 51, then, by subtracting this value from 136 we get B equal to 85. 51 is 3 times 17 and 85 is 5 times 17 so the quotient is three fifths.
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2 times jim's age plus 3 times mary's age equals 34. Mary is 5 times jim's age. How old is Mary?

2 times jim's age plus 3 times mary's age equals 34. Mary is 5 times jim's age. How old is Mary?
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The perimeter of a rectangle swimming pool is 130 yards. Three times the length is equal to 10 times the width. Find the length and the width of the pool. Find the area of the pool.

The perimeter of a rectangle swimming pool is 130 yards. Three times the length is equal to 10 times the width. Find the length and the width of the pool. Find the area of the pool. Perimeter of a rectangle is 2*the length plus 2*the width P = 2L + 2W and P = 130, so L + W = 65 Three times the length is equal to 10 times the width, giving 3L = 10W Substituting into the above for L = 65 – W, 3(65 – W) = 10W 195 – 3W = 10W 195 = 13W W = 195/13 = 15 W = 15 yards, And L = (10/3)W = (10/3)*15 = 50 L = 50 yards
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5a+2=6b+3=7c+4

Assume that  those 3 clocks alarm at the same time when the 5-min-clock-hand reached the number 2 on its a-th turning, the 6-min-clock-hand reached the number 3 on its b-th turning, and the 7-min-clock- hand reached the number 4 on its c-th turning.  The time required to sound simultaneous alarming for each clock, T5 for the 5-min-clock, T6 for the 6-min-clock and T7 for the 7-min-clock, will be written as follows: T5=5(a-1)+2=5a-3 ··· Eq.1, T6=6(b-1)+3=6b-3 ··· Eq.2  and T7=7(c-1)+4=7c-3 ··· Eq.3  Here, a, b and c are unknown integers greater than or equal to 1.  They alarm at the same time. So, T5=T6=T7 ⇒ 5a-3=6b-3=7c-3 ⇒ 5a=6b=7c.  Therfore, the least common multiple(LCM) of 5a, 6b and 7c that satisfies the equation above is 210(=5x2x3x7).  Thus, a:b:c=(210/5):(210/6):(210/7)=42:35:30.  Therefore, the sets of {a, b, c} are as follows: {42, 35, 30}, {84, 70, 60}, {126, 105, 90}··· {42n, 35n, 30n}, n: integers greater than or equal to 1.  Plug a=42, b=35 and c=30 into Eq.1, Eq.2 and Eq.3 respectively.  T5=5x42-3=207, T6=6x35-3=207 and T7=7x30-3=207   Therefore, the clocks alarm simultaneously at 207 minute for the first time after they started at the same time from each one's origin (numbered 5, 6, 7 or 0?).  Then they repeat simultaneous alarming every 210(=LCM) minutes. CK: Plug a=84, b=70 and c=60 into Eq.1, Eq.2 and Eq.3 respectively.  T5= 5x84-3=417=207+210(LCM), T6=6x70-3=417=207+210(LCM) and T7=7x60-3=417=207+210(LCM)   CKD.
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what are common multiples of 2,4,5,and10?

2 times ten ls 20      5 times 4 equals 20          4 times 5 equals 20       10 times 2 equals 20
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can you get a negative answer when adding a positive number and several negative numbers?

Yes, you can. And no, you can't. It all depends on how many negatives there are in the equation. Like multiplying a positive with a negative will equal a negative. Or multiplying a positive times a negative times a negative times a negative will equal a negative, because a plus times a negative equals negative. Multiply the negative with a negative you get a positive. Multiply that positive with the last negative and you get a negative.
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Exponential Growth and Decay

Let's call 12:00 noon time zero.  Let t = time in minutes Amount of Quantity A at time t  = 100 + 10t Amount of Quantity B at time t = 100(1.05)^t These two will only be equal when t = 0 Hope that helps!
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At what time exactly after ten will the big and the small hands be equally distant from the 12?

analog klok with hands that go round at konstant speed minut hand go faster, so it get ahed av out hand, but then kum up behind & katch up both hands start at 12 & 12 ours later both bak up at 12 thus, minut hand go around 11 times in same time our hand go around 1 time Time tu katch up=12 ours/11=1.0909090909 ours=3927.2727272727272 sekonds time 10=10*1.09090909...10:54_32.7272727272 or 39,272.72727272 sekonds from zero
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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.
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If you spin a spinner 30 times, how many times would it land on red?

There isnt a specific number of times it will land on the red space, but the logical choice is to say that since the landing areas are equal, and you have 30 spins, that the spinner would land on red 10 times. But as I said, there is no way to precisely know probability.
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