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# what is 13/56 plus 5/7 equals

13/56 + 5/7 =

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### 13 plus 7 equals 13 plus 7 - Answers.com

13 plus 7 equals 13 plus 7? SAVE CANCEL. already exists. Would you like to merge this question into it? MERGE CANCEL. already exists ... What is 13 over 56 plus 5 over 7?

### What is 56 plus 7 equal - Answers.com

What is 56 plus 7 equal? SAVE CANCEL. already exists. ... What is 13 over 56 plus 5 over 7? 13/56 + 40/56 = 53/56 3 people found this useful Edit.

### 13 56 5 7 equals_Yaelp Search

13 56 5 7 equals. Web; News; Img; Video; Shop; Music; Website Navigation . 13 56 5 7 equals. Results for 13 56 5 7 equals: Also try: ... what is 13/56 plus 5/7 equals ...

### Math Forum: Ask Dr. Math FAQ: Dividing Fractions - Don't ...

Dividing Fractions - Don't change ... 3/5 1 3/5 3/5 _____ * _____ = _____ = _____ = 3/5 7/8 8/7 56/56 1 ... Is 3/5 divided by 7/8 equal to 3/5 ...

### What plus what equals 56 - Brainly.com

What plus what equals 56 1. Ask for details ; Follow; ... The answer is 55 +1 or 43 + 13 Comments; Report; 1 5 1. ... for different inputs, x: Input (x) 3 4 5 6 ...

### Plus Equals | Plus Size Fashion Sizes 14-42 & Custom Made

Plus Equals offers handmade, plus size fashion for women sizes 14-42. ... [email protected] © 2016 PlusEquals. All rights registered ...

### Quick MathSpeak Tutorial

Quick MathSpeak Tutorial. Without Semantic Interpretation, ... 3 plus 4 times 2 minus negative 2 equals 3 plus 8 plus 2 equals 13 ...

## Suggested Questions And Answer :

### 5 over x+2 PLUS 6 over x-2 EQUALS 14 over x-1

5 over x+2 PLUS 6 over x-2 EQUALS 14 over x-1 How do I solve this equation!? 5/(x + 2)  + 6/(x - 2) = 14/(x - 1) tranfer "5/(x + 2)  + 6/(x - 2)" to the other side & change the sign 0 = 14/(x - 1) - 5/(x + 2)  - 6/(x - 2) LCD = (x - 1) ;(x + 2) ; (x - 2) 0 = 14(x + 2)(x - 2) - 5(x - 1)(x - 2) - 6(x - 1) (x + 2) 0 = 14(x² - 4) - 5(x² - 3x + 2) - 6(x² + x - 2) 0 = 14x² - 56 - 5x² + 15x - 10 - 6x² - 6x + 12 combine terms 0 = 14x² - 5x² - 6x² + 15x - 6x - 56 - 10 + 12 0 = 3x² + 9x  - 54 divide by 3 0/3 = 3x²/3 + 9x/3 - 54/3 0 = x² + 3x - 18 0 = (x + 6)(x - 3) ◄ Ans

### integral from 0 to infinity of (cos x * cos(x^2)) dx

The behaviour of this function f(x)=cos(x)cos(x^2) is interesting. The integral is the area between the curve and the x axis. If the functions cos(x) and -cos(x) are plotted on the same graph, the latter form an envelope for f(x). Between x=(pi)/2 and 3(pi)/2, the curve has 4 maxima and 3 minima; between x=3(pi)/2 and 5(pi)/2 there are 7 maxima and 6 minima; between x=(2n-1)(pi)/2 and (2n+1)(pi)/2 there are 3n+1 maxima and 3n minima (integer n>0), a total of 6n+1. (These are based purely on observation, and need to be supported by sound mathematical deduction.) As n becomes large the envelope appears to fill as the extrema become closer together. As x tends to infinity n also tends to infinity. The envelope apparently has as much area above (positive) as below (negative) the x axis so the total area will be zero as the positive and negative areas cancel out. The question is: do the areas cancel out exactly? As x gets larger, the curve starts to develop irregularities and patterns, but it stays within the envelope, and positive irregularities appear to be balanced by negative irregularities, so the overall symmetry appears to be preserved. f(x)=0 when cos(x)=0 or cos(x^2)=0, which means that x=(2n-1)(pi)/2 or sqrt((2n-1)(pi)/2), where n>0. Between 3(pi)/2 and 5(pi)/2, for example, we have sqrt(3(pi)/2), sqrt(5(pi)/2), ..., sqrt(13(pi)/2), because sqrt(13(pi)/2)<3(pi)/20, and this lies between (2n-1) and (2n+1); so n is defined by 2n-1<(2m+1)^2(pi)/2<2n+1.  For m-1 we have 2(n-z)-1<(2m-1)^2(pi)/2<2(n-z)+1, where z is related to the number of zeroes in the current "batch". For example, take m=3: 2n-1<49(pi)/2<2n+1; 49(pi)/2=76.97 approx., so 2n-1=75, and n=38. Also 2(n-z)-1<25(pi)/2<2(n-z)+1 so, because 25(pi)/2=39.27 approx., 2(n-z)+1=41, n-z=20, and z=18. When m=2, 2n-1=39, n=20; 2(20-z)-1<9(pi)/2<2(20-z)+1; 2(20-z)-1=13, 20-z=7, z=13. The actual number of zeroes, Z, including the end points is 2 more than this: Z=z+2. Now we have an exact way to calculate the number of zeroes in each batch. So Z and n are both related to m. The number of extrema, E=Z-1=z+1. In fact, E=int(2(pi)(m-1)+1), where int(a) means the integer part of a, so as m increases, there are proportionately more extrema over the range (2m-1)(pi)/2 to (2m+1)(pi)/2. The figure of 6n+1 deduced earlier by observation approximates to the mathematical findings, because 2(pi) is approximately equal to 6. But we still need to show, or disprove, that the areas above and below the x axis are equal and therefore cancel out. Unfortunately, if we consider the area under the first maximum (between x=(pi)/2 and sqrt(3(pi)/2)), and the area above the first minimum (between x=sqrt(3(pi)/2) and sqrt(5(pi)/2)), they are not the same, so do not cancel out. More...

### Can you simplify it step by step as well

Answered earlier: 26r-2=3r^2. We need to put this into standard form. The standard quadratic form is ax^2+bx+c=0, where x is the unknown and a, b and c are numbers. The unknown is r in this problem, rather than x, so let's put the equation into quadratic form: 3r^2-26r+2=0. How did I get this? Subtract 26r from each side: -2=3r^2-26r. Now, add 2 to each side: 0=3r^2-26r+2, which is the same as 3r^2-26r+2=0. (There's no mystery here. The equals sign shows us that we have equality of quantity. If, for example, an apple cost 30 cents or pennies, then 30 cents will buy an apple, so we can say apple=30 or 30=apple.) This equation doesn't factorise, so we need the formula to solve it: r=(26+sqrt(26^2-4*3*2))/6 (this is the quadratic formula x=(-b+sqrt(b^2-4ac))/2a. All I've done is put b=-26, a=3 and c=2 into the formula. The plus-or-minus sign (+) means there are two answers, one using plus and the other using minus.) So r=(26+sqrt(676-24))/6=(26+sqrt(652))/6=8.589 or 0.0776. Because b=-26, -b=26: that's why the minus has disappeared. I hope the extra info helps you to understand.

### how can I simplify the last problem

26r-2=3r^2. We need to put this into standard form. The standard quadratic form is ax^2+bx+c=0, where x is the unknown and a, b and c are numbers. The unknown is r in this problem, rather than x, so let's put the equation into quadratic form: 3r^2-26r+2=0. How did I get this? Subtract 26r from each side: -2=3r^2-26r. Now, add 2 to each side: 0=3r^2-26r+2, which is the same as 3r^2-26r+2=0. (There's no mystery here. The equals sign shows us that we have equality of quantity. If, for example, an apple cost 30 cents or pennies, then 30 cents will buy an apple, so we can say apple=30 or 30=apple.) This equation doesn't factorise, so we need the formula to solve it: r=(26+sqrt(26^2-4*3*2))/6 (this is the quadratic formula x=(-b+sqrt(b^2-4ac))/2a. All I've done is put b=-26, a=3 and c=2 into the formula. The plus-or-minus sign (+) means there are two answers, one using plus and the other using minus.) So r=(26+sqrt(676-24))/6=(26+sqrt(652))/6=8.589 or 0.0776. Because b=-26, -b=26: that's why the minus has disappeared. I hope the extra info helps you to understand.

### If x(9) divided by 6 minus y plus 44 divided by 297 divided by 6 equals 0, what is the value of x if y equals 35?

?????????? (x/6) -[(y+44) /(297/6)]=0  & y=35 ?????????? 297/6=49.5 y+44 bekum 35+44=79 (x/6) -(79 / 49.5)=0....79/49.5=1.5959595959595959... (x/6) -1.59595959595...=0 (x/6)=1.59595959595... x=6*1.59595959595... x=9.57575757575757...

### 6 minus 1 times 0 plus 2 divided by 2 equals

The answer is 1. Anything multiplied by 0 is 0. So the only relevant part of that equation is plus 2 divided by 2 and that equals 1.

### What is x - 2y equals 2 and 3x plus y equals 6?

x - 2y = 2 --- (1) 3x + y = 6 --- (2) From (1), we have: x = 2y + 2 --- (3) Substituitng (3) into (2), we will get: 3(2y + 2) + y = 6 6y + 6 + y = 6 7y = 6 - 6 7y = 0 y = 0 / 7 y = 0 Substituting y = 0 into (3), we have: x = 2(0) + 2 x = 0 + 2 x = 2 Hence, x = 2 and y = 0.

### 12x cubed minus 10x squared plus 18x

Problem: 12x cubed minus 10x squared plus 18x how to solve 12x^3 - 10x^2 + 18x Factor out 2x. 2x(6x^2 - 5x + 9) If we set 2x equal to zero, we get x = 0. When we set the remaining polynomial equal to zero, solving with the quadratic formula, we find that the roots are imaginary. Let's look at the discriminant. (b^2 - 4*a*c) ((-5)^2 - 4*6*9) (25 - 216) (-191) Taking the square root of that is what produces the imaginary numbers. x = 0 x = 13.82i x = -13.82i

### what is the number of square units in the region satisfying |x|+|y| less than equal to 9 and -5 less than equal to x less than equal to 4?

When x=-5, |y|<4, so y<4 or -y<4, y>-4, making -4 Read More: ...

### log6 (x plus 5) plus log6 x equals 2

log6 (x+5) + log6 x = 2 log6((x( x + 5)) = 2 x( x + 5) = 6^2 x^2 + 5x = 36 x^2 + 4x - 36 = 0 ( x - 4)(x + 9 ) = 0 x - 4 = 0 or x + 9 = 0 x =  4 or x = - 9