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if i had 1,3,5,7,11,13,15 how can i get total 30 if i use 5 digits from the group

(1,3,5,7,11,13,15) to get total 30

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Mathematical Puzzles: What is () + () + () = 30 using 1,3,5,7 ...

This allows for the solution 11,3 + 15,7 + 3 = 30. Only use numbers ... Is it possible to make 10000 out of the digits 0, 1, 2, 3, 4, 5, 6, 7 ... (1,3,5,7,9,11,13,15 ...

Question: "If I had a group of odd numbers 1,3,5,7,9,11,13,15 ...

... 1,3,5,7,9,11,13,15 how can I get total sum ... Question: "if i had a group of odd numbers 1,3,5,7 ... 1,3,5,7,9,11,13,15 how can I get total sum = 30 if I ...

4 Ways to Add 5 Consecutive Numbers Quickly - wikiHow

How to Add 5 Consecutive Numbers Quickly. Bet someone you can add five consecutive integers faster than they can. ... Ex. 11, 12, 13, 14, 15; 11 x 5 = 55; 55 + 10 = 65;

Techniques for Adding the Numbers 1 to 100 – BetterExplained

296 Comments on "Techniques for Adding the Numbers 1 to 100" ... /2 = total (5(1+6))/2 = 15. 1 3 5 7 9 = 25 (5 ... 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15.

Count On

In any day before noon how many times is the sum of the digits of the time equal to 15? ... and 1 so they are 2, 3, 5, 7, 11, 13 ... 30, 31, 30... B) 7, 8, 5, 5, 3, 4 ...

Math Tricks for All Ages - University of Massachusetts ...

Math Tricks for All Ages ... 1: 15: 14: 4: 12: 6: 7: 9: 8: 10: 11: 5: 13: 3: 2: 16: ... Number everyone in the group from 1 to however many there are. Get a piece of ...

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Types of Numbers, Part I Compiled by ... Primes of two or more digits can only contain the digits 1, 3, 7 because, ... 1,2,3,4,5,7,8,9,10,11,13,14,15,16,17,19,21,22 ...

Homework Practice and Problem-Solving Practice Workbook

Homework Practice and Problem-Solving ... 2 3 4 5 6 7 8 9 10 079 17 16 15 14 13 12 11 10 09 ... 5. Which two numbers use the digits 3 and 1?

area of a region between two curves?

The picture below shows the required area between the two curves: y=e^(-x^2) (red) and y=1-cos(x) (blue). The vertical blue line shows the limit of the first quadrant (x=(pi)/2). Each grid square in the picture has an area of 0.04 and the area is roughly 14 complete squares: 14*0.04=0.56 is a rough estimate of the area between the curves. The points where the curves cross is the upper limit of the enclosed area which includes the y axis. I assume that the area beyond that up to x=(pi)/2 is not required because it is not enclosed by the y axis. The intersection is when x=0.941944 approx., that is, when e^(-x^2)=1-cos(x) (x in radians). At this point y=0.411783 approx. These define the upper limit for definite integration. Using S[low,high](...) to denote integration, we integrate to get the area between the two curves: S[0,0.941944]((e^(-x^2)-1+cos(x))dx). As far as I know, there's no solution to the indefinite integral of e^(-x^2), but the above definite integral can be evaluated by approximating dx. If dx=0.01, for example, then we can sum the areas of thin rectangles of width 0.01 over the range. This is tedious but it will yield a reasonable approximation. Let f(x)=e^(-x^2)-1+cos(x). Consider a starting point represented by a rectangle of width 0.01 and height f(0). Its area is 1*0.01=0.01. Now consider a different rectangle with the same width but height f(0.01) and area=0.0099985. The two rectangles "trap" the curve between them, so the true area under the curve between the limits 0 and 0.01 is somewhere between the two rectangular areas. If we take the average of these two areas we get 0.00999925. Then we move to f(0.01) and f(0.02). The two areas this time are 0.01f(0.01) and 0.01f(0.02) and the average area is 0.005(f(0.01)+f(0.02)). A third pair of rectangles will be averaged at 0.005(f(0.02)+f(0.03)). The last pair of rectangles will be narrower than 0.01: the first has a height of f(0.94) and a width of 0.001944, the second has a height of f(0.941944) which approximates to zero (to the nearest 10^-7). So the last average area is 0.000972f(0.94). The total area between the curves and the y axis can be expressed as a series: 0.005(f(0)+f(0.01) + f(0.01)+f(0.02) + ... + f(0.92)+f(0.93) + f(0.93)+f(0.94)) + 0.000972f(0.94). As you can see, f(0.01) to f(0.93) are all repeated, so we have 0.005(f(0)+f(0.94)) + 0.000972f(0.94) + 0.01(f(0.01)+...+f(0.93)). Since f(0)=1 and f(0.94)=).00308040 approx. So the first expression is 0.005*1.00308040=0.0050154020. 0.000972f(0.94)=0.000972*0.00308040=0.000003 approx.  To make matters more manageable, I'll divide the range of values into summed groups: 0.01-0.10, 0.11-0.20, 0.21-0.30, etc., up to 0.81-0.90, then 0.91-0.94. Here are the group results (these figures have a superfluous accuracy and will be rounded off later): 0.01-0.10:  9.946463769  0.11-0.20:  9.630991473 0.21-0.30:  9.036361176 0.31-0.40:  8.183088444 0.41-0.50:  7.104969671 0.51-0.60:  5.842140126 0.61-0.70:  4.437880465 0.71-0.80:  2.935475192 0.81-0.90:  1.375462980 0.91-0.93:  0.104328514 TOTAL: 58.59716181 AREA SUBTOTAL: 0.5859716181 TOTAL AREA: 0.585972+0.005018=0.591 approx. Control summation: 5.415713508. Control area subtotal: 0.5415713508 (based on increments of 0.1 across the range 0.1 to 0.9. We would expect this figure to be approximately the same as the more accurate summation.) It appears that the area between the curves and the y axis is of the order 0.591; this figure is close to the rough estimate given near the beginning of this answer.

using digits 0-9 only once, what combination of 3 digit numbers add to 1000, one digit will be spare

You have to have 3 sets of 3 digits each.  The first set totals 10 (for the ones place).  The second and third sets total 9 (for the tens and hundreds place).  The 1 carries over from the total of 10 in the ones place, moving the 9 up to 10 in the tens place and the 9 up to 10 in the hundreds place. Ways to make 10: 019, 028, 037, 046, 127, 136, 145, 235 Ways to make 9:  018, 027, 036, 045, 126, 135, 234 If 10 is 019, 9 leaves 234, not enough ways If 10 is 028, 9 leaves 135, not enough ways If 10 is 037, 9 leaves 126, not enough ways If 10 is 046, 9 leaves 135, not enough ways If 10 is 127, 9 leaves 036, 045, enough ways but 036 and 045 share 0 If 10 is 136, 9 leaves 027, 045, enough ways but 027 and 045 share 0 If 10 is 145, 9 leaves 027, 036, enough ways but 027 and 036 share 0 If 10 is 235, 9 leaves 018, not enough ways. We checked all of the possible combinations of 3 digits totaling 10 and it looks like none of them leave any combinations to make two sets of 3 digits totaling 9 each. Answer:  No solution.

Factoring by grouping?

Is there a typo here? Maybe it should be 0 = x^3 + 6x^2 -3x - 18 .  This can be solved by grouping like so:   0 = (x^3 + 6x^2) + (-3x - 18 )  This is the grouping step.  I groupred the first two terms, and the last two terms.   0 = x^2(x+6) - 3(x+6)   In this step, I factored out the greatest common factor from each set of parenthesis.   0 =  ( x+6 )(x^2 - 3)  In this step, I factored out the common (x+6) binomial factor from the right side of the equation. (a challenging notion to follow, but it goes something like this:  In  x^2(x+6) - 3(x+6) , think of x^2 as "a", 3 as "b", and (x+6) as "c",  Then x^2(x+6) - 3(x+6) is of the form  ac - bc.  I factored out the "c" to get c(a-b).  Substituting back, using  x^2 as "a", 3 as "b", and (x+6) as "c", get c(a-b) = (x+6)(x^2-3)   Continuing on to solve the equation, if 0 =  ( x+6 )(x^2 - 3) , then 0 = x+6   or    0 = x^2 -3 Solving these equations, get x = -6, x =sqrt(3), x = - sqrt(3)  (remember when taking the square root of each side you must include both the positive and negative results)

if i had 1,3,5,7,11,13,15 how can i get total 30 if i use 5 digits from the group

3 weeks + 7 days + 11hrs + 13hrs+ 1day =30... or (15-9)+ (13-7) +(11-5) +(9-3)+ (7-1)=30 okay.. :)

sqrt(3a+10) = sqrt(2a-1) + 2

sqrt(3a+10)=sqrt(2a-1)+2 To remove the radical on the left-hand side of the equation, square both sides of the equation. (~(3a+10))^(2)=(~(2a-1)+2)^(2) Simplify the left-hand side of the equation. 3a+10=(~(2a-1)+2)^(2) Squaring an expression is the same as multiplying the expression by itself 2 times. 3a+10=(~(2a-1)+2)(~(2a-1)+2) Multiply each term in the first group by each term in the second group using the FOIL method. FOIL stands for First Outer Inner Last, and is a method of multiplying two binomials.  First, multiply the first two terms in each binomial group.  Next, multiply the outer terms in each group, followed by the inner terms.  Finally, multiply the last two terms in each group. 3a+10=(~(2a-1)*~(2a-1)+~(2a-1)*2+2*~(2a-1)+2*2) Simplify the FOIL expression by multiplying and combining all like terms. 3a+10=(~(2a-1)^(2)+4~(2a-1)+4) Remove the parentheses around the expression ~(2a-1)^(2)+4~(2a-1)+4. 3a+10=~(2a-1)^(2)+4~(2a-1)+4 Raising a square root to the square power results in the expression inside the root. 3a+10=(2a-1)+4~(2a-1)+4 Add 4 to -1 to get 3. 3a+10=2a+3+4~(2a-1) Since a is on the right-hand side of the equation, switch the sides so it is on the left-hand side of the equation. 2a+3+4~(2a-1)=3a+10 Move all terms not containing ~(2a-1) to the right-hand side of the equation. 4~(2a-1)=-2a-3+3a+10 Simplify the right-hand side of the equation. 4~(2a-1)=a+7 Divide each term in the equation by 4. (4~(2a-1))/(4)=(a)/(4)+(7)/(4) Simplify the left-hand side of the equation by canceling the common terms. ~(2a-1)=(a)/(4)+(7)/(4) To remove the radical on the left-hand side of the equation, square both sides of the equation. (~(2a-1))^(2)=((a)/(4)+(7)/(4))^(2) Simplify the left-hand side of the equation. 2a-1=((a)/(4)+(7)/(4))^(2) Combine the numerators of all expressions that have common denominators. 2a-1=((a+7)/(4))^(2) Expand the exponent of 2 to the inside factor (a+7). 2a-1=((a+7)^(2))/((4)^(2)) Expand the exponent 2 to 4. 2a-1=((a+7)^(2))/(4^(2)) Simplify the exponents of 4^(2). 2a-1=((a+7)^(2))/(16) Multiply each term in the equation by 16. 2a*16-1*16=((a+7)^(2))/(16)*16 Simplify the left-hand side of the equation by multiplying out all the terms. 32a-16=((a+7)^(2))/(16)*16 Simplify the right-hand side of the equation by simplifying each term. 32a-16=(a+7)^(2) Since (a+7)^(2) contains the variable to solve for, move it to the left-hand side of the equation by subtracting (a+7)^(2) from both sides. 32a-16-(a+7)^(2)=0 Squaring an expression is the same as multiplying the expression by itself 2 times. 32a-16-((a+7)(a+7))=0 Multiply -1 by each term inside the parentheses. 32a-16-a^(2)-14a-49=0 Since 32a and -14a are like terms, add -14a to 32a to get 18a. 18a-16-a^(2)-49=0 Subtract 49 from -16 to get -65. 18a-65-a^(2)=0 Move all terms not containing a to the right-hand side of the equation. -a^(2)+18a-65=0 Multiply each term in the equation by -1. a^(2)-18a+65=0 For a polynomial of the form x^(2)+bx+c, find two factors of c (65) that add up to b (-18).  In this problem -5*-13=65 and -5-13=-18, so insert -5 as the right hand term of one factor and -13 as the right-hand term of the other factor. (a-5)(a-13)=0 Set each of the factors of the left-hand side of the equation equal to 0. a-5=0_a-13=0 Since -5 does not contain the variable to solve for, move it to the right-hand side of the equation by adding 5 to both sides. a=5_a-13=0 Set each of the factors of the left-hand side of the equation equal to 0. a=5_a-13=0 Since -13 does not contain the variable to solve for, move it to the right-hand side of the equation by adding 13 to both sides. a=5_a=13 The complete solution is the set of the individual solutions. a=5,13

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.

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