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find two numbers 3 x 567

Exact answer in between

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Find two numbers the exact number is between 3 x 567 ...

Find two numbers the exact number is between 3 x 567 - 1722542. 1. Log in ... Find two numbers the exact number is between 3 x 567 1. Ask for details ; Follow;
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Find two numbers the exact answer is between. 3 X 567

math - find two numbers the exact answer is between 3*567 ; math - Find two numbers the ... Math - Find two numbers the exact answer is between 4 x 94; math - find ...
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Lesson 2.4 Go Math Flashcards | Quizlet

Find two numbers the exact answer is between 3 x 567. ... ANSWER: Yes, 2,920 is reasonable because when we find two numbers the exact answer is between, ...
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find two numbers the exact answer is between 3 x 567 ...

find two numbers the exact answer is between 3 x 567 - 630875
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Find two numbers the exact answer is between 3x567

Find two numbers the exact answer is between. 3 X 567 ... Find two numbers the exact answer is between 3*567 Need ... Find two numbers exact is between 3 x 567 please ...
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Q Find two numbers the exact answer is between - Boodom

Find two numbers the exact answer is between.3 X 567 ... Find two numbers the exact answer is between. 3 X 567. 0 0. Edit. ... Give two numbers between which the ...
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2 - Wikipedia

Two and three are the only two consecutive prime numbers. 2 is the first Sophie ... Two is the only number x such that the sum of the reciprocals of the powers of ...
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Systems of Equations Answer Key - hoffman

The sum of the two numbers is 30. Find the numbers. x = y - 4, x + y = 30 11. ... Find the numbers. x = 3y + 3, 2y + 4 = x + 8 Angle Problems 1.
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Grade 4: Distributive Property of Multiplication: Overview

Distributive Property of Multiplication: Overview. ... in a product of 3 or 4 digits. For example, in 4 x 567, ... (3 x 2). Mathematicians say that the number 3 has ...
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Suggested Questions And Answer :

how many ways are there to add and get the sum of 180

There are an infinite number of ways to get 180 from two numbers, if we count decimals and fractions as well as other real numbers; but if we are limited to positive integers greater than zero and just the sum of two of them, we are limited to x and 180-x. If we also exclude 90+90 because the numbers are the same, then we have 1 to 89 combined with 179 to 91, which is 89 pairs. Moving on to the sum of three different numbers, let's make 1 plus another two different numbers adding up to 179. So we have 2+177, 3+176, ..., 87+92, 88+91, 89+90, which is 88 groups combined with 1. Move on to 2 plus another two different numbers adding up to 178: 3+175, ..., 87+91, 88+90, which is 86 groups. Then we move on to 3 plus 177: 4+173, ..., 86+91, 87+90, 88+89, 85 groups. And so on, with reducing numbers, until we get to 59, 60 and 61. Let's divide the numbers into two groups A and B. In A we start with 1 and in B we put 2 and (180-A-B)=177 as a pair (2,177). Then we put the next pair in group B: (3,176), then (4,175) and keep going till we have used up all the numbers, ending up with (88,90). Then we count how many pairs there are in group B and pair it up with the number in group A, so we start with (1,88) which covers all the combinations of numbers in group B. Now we move to 2 in group A, put all the pairs adding up to 178 in group B, and finally put the count of these pairs with 2 in group A: (2,86). We then move on to 3, and so on, putting in the counts to make up the number pair in group A. When we've finished by putting the last count in group A, which is (59,1), we can forget about group B and look at the pattern in group A. What we see is this: (1,88), (2,86), (3,85), (4,83), (5,82), (6,80), (7,79), ... See how the counts come in pairs with a gap? All the multiples of 3 are missing in the counts sequence (e.g., 87, 84, 81). We find there are 29 pairs and one odd count, 88, which is unpaired. Number the pairs 0 to 28 and refer to the pair number as N. Add the counts in the pairs together so we start with pair 0 as 86+85=171, pair 1 as 165, pair 2 as 159, and so on. The sequence 171, 165, 159, ..., 3 is an arithmetic sequence with a start of 171 and a difference of 6 between each term in the sequence. [Note also that the terms in the series are all multiples of 3: 3*57, 3*55, 3*55, ...] The rule for the Nth term is 171-6N. When N=0 we have the first term 171 and when N=28 the last term is 3. There is one more term at the end which is unpaired made up of the numbers 59, 60 and 61. We can combine this with the unpaired (1,88). We can find the sum of the terms in the series, which will tell us how many ways there are of adding three different integers so that their sum is 180 (like the sum of the angles of a triangle).  To find the sum of the terms of the series we note that there are 29 terms (0 to 28) and they all contain 171, so that's 171*29=4959. We also have to subtract 6(0+1+2+3+...+28)=6*28*29/2=2436. So 4959-2436=2523. [The sum of the series is also 3(57+55+53+...+5+3+1)=2523.] To this we add the "odd couple" 88+1=89 and 2523+89=2612. Add also the 89 which is the number of pairs of integers adding up to 180 we calculated at the beginning. The total so far is 2612+89=2701 ways of adding 2 or 3 positive integers so that their sum is 180. If you want to go further, please feel free to do so!
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show that the function f(x)= sqrt (x^2 +1) satisfies the 2 hypotheses of the Mean Value Theorem

f(x) = sqrt(x^2 + 1) ; [(0, sqrt(8)] Okay, so for the Mean Value Theorem, two things have to be true: f(x) has to be continuous on the interval [0, sqrt(8)] and f(x) has to be differentiable on the interval (0, sqrt(8)). First find where sqrt(x^2 + 1) is continous on. We know that for square roots, the number has to be greater than or equal to zero (definitely no negative numbers). So set the inside greater than or equal to zero and solve for x. You'll get an imaginary number because when you move 1 to the other side, it'll be negative. So, this means that the number inside the square root will always be positive, which makes sense because the x is squared and you're adding 1 to it, not subtracting. There would be no way to get a negative number under the square root in this situation. Therefore, since f(x) is continuos everywhere, (-infinity, infinity), then f(x) is continuous on [0, sqrt(8)]. Now you have to check if it is differentiable on that interval. To check this, you basically do the same but with the derivative of the function. f'(x) = (1/2)(x^2+1)^(-1/2)x2x which equals to f'(x) = x/sqrt(x^2+1). So for the derivative of f, you have a square root on the bottom, but notice that the denominator is exactly the same as the original function. Since we can't have the denominator equal to zero, we set the denominator equal to zero and solve to find the value of x that will make it equal to zero. However, just like in the first one, it will never reach zero because of the x^2 and +1. Now you know that f'(x) is continous everywhere so f(x) is differentiable everywhere. Therefore, since f(x) is differentiable everywhere (-infinity, infinity), then it is differentiable on (0, sqrt(8)). So the function satisfies the two hypotheses of the Mean Value Theorem. You definitely wouldn't have to write this long for a test or homework; its probably one or two lines of explanation at most. But I hope this is understandable enough to apply to other similar questions!
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complex, rational and real roots

The quintic function should have 5 roots.  The changes of sign (through Descartes) tell us the maximum number of positive roots. Since there are two changes of sign there is a maximum of 2 positive roots. To find the number of negative roots we negate the terms with odd powers and check for sign changes: that means that there is at most one negative root, because there's only one sign change between the first and second terms. Complex roots always come in matching or complementary pairs, so that means in this case 2 or 4.  Put x=-1: function is positive; for x=-2 function is negative, so there is a real root between -1 and -2, because the x axis must be intercepted between -1 and -2. This fulfils the maximum for negative roots. That leaves 4 more roots. They could be all complex; there could be two complex and two positive roots. Therefore there are at least two complex roots. To go deeper we can look at calculus and a graph of the function: The gradient of the function is 10x^4-5. When this is zero there is a turning point: x^4=1/2. If we differentiate again we get 40x^3. When x is negative this value is negative so the turning point is a maximum when we take the negative fourth root of 1/2; at the positive fourth root of 1/2 the turning point is a minimum, and these are irrational numbers. The value of the function at these turning points is positive. The fourth root of 1/2 is about 0.84. and once the graph has crossed the x axis between -1 and -2, it stays positive, so all other roots must be complex. The function is 12 when x=0, so we can now see the behaviour: from negative values of x, the function intersects the x axis between -1 and -2 (real root); at x=4th root of 1/2 (-0.84) it reaches a local maximum, intercepts the vertical axis at 12 until it reaches 4th root of 1/2 (0.84) and a local minimum, after which it ascends rapidly at a steep gradient.  [Incidentally, one way to find the real root is to rearrange the equation: x^5=2.5x-6=-6(1-5x/12); x=-(6(1-5x/12))^(1/5)=-6^(1/5)(1-5x/12)^(1/5) We can now use an iterative process to find x. We start with x=0, so x0, the first approximation of the solution, is the fifth root of 6 negated=-1.430969 approx. To find the next solution x1, we put x=x0 and repeat the process to get -1.571266. We keep repeating the process until we get the accuracy we need, or the calculator reaches a fixed value. After just a few iterations, my calculator gave me -1.583590704.]
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the product of...

The product of two consecutive integers is 306. Find the integers. The hardest part is understanding what the question is asking and setting up the problem.  Once you have this, it is a piece of cake.  Here's what we do: First what its asking: (Break the question down into segments) The product: Ok this means Multiply. Two consecutive integers: This means that the second number comes right after the first number in a number line. Ex. 1,2,3,4,5... The integers belong to the number 306: Factor 306.                    306                      ^                     2*153                          ^                         9*17                         ^                        3*3 Now look 3 * 3 * 2 = 18 The other factor 306 = 17 These two are consecutive numbers so these are your integers since 18 * 17 = 306. A more Algebraic approach would be to say: two consecutive intergers are multiplied together to make 306 x(x + 1) = 306 x^2 + x = 306 x^2 + x - 306 = 0 (x - 17)(x+18) = 0 x - 17 = 0; x = 17 x - 18 = 0; x = 18 There are your two consecutive integers. 
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Prove: for any two real numbers that are not equal, you can find a real number between them.

If the two numbers are a and b, and a0. Therefore x is equidistant from a and b. Since a0. x-a>0 and b-x>0 so x>a and b>x making a Read More: ...

10 vedic maths rules for class IX

2 instead of 5: 34/5 can be calculated by multiplying 34 by 2 instead of dividing by 5. 34*2=68. We just move the decimal point one place to the left: 34/5=6.8. 124/5=24.8 because 124*2=248. Move the decimal point: 248 becomes 24.8. 34*5 is the same as 34/2=17 but we add a zero to make 17 into 170. 73*5 is the same as 73/2=36.5 then move the decimal point one place to the right (or add zero): 36.5 becomes 365=73*5. So we only need to know how to multiply and divide by 2 to divide and multiply by 5. We just move the decimal point. Divisibility by 9 or remainder after dividing by 9. All multiples of 9 contain digits which added together give 9. As we add the digits together, each time the result goes over 9 we add the digits of the result together and use that result and continue in this way up to the last digit. Is 12345 divisible by 9? Add the digits together 1+2+3=6. When we add 4 we get 10, so we add 1 and zero=1 then we add 5 to get 6. The number is not exactly divisible by 9, but the remainder is 6. We can also ignore any 9's in the number. Now try 67959. We can ignore the two 9's. 6+7=13, and 1+3=4; 4+5=9, so 67959 is divisible by 9. Multiplying by 11. Example: 132435*11. We write down the first and last digits 1 ... 5. Now we add the digits in pairs from the left a digit step at a time. So 1+3=4; 3+2=5: 2+4=6; 4+3=7; 3+5=8. Write these new digits between 1 and 5 and we get 1456785=132435*11. But we had no carryovers here. Now try 864753*11. Write down the first and last digits: 8 ... 3. 8+6=14, so we cross out the 8 and replace it with 8+1=9, giving us 94 ... 3. Next pair: 6+4=10. Again we go over 10 so we cross out 4 and make it 5. Now we have 950 ... 3. 4+7=11, so we have 9511 ... 3. 7+5=12, giving us 95122 ... 3; 5+3=8, giving us the final result 9512283.  Divisibility by 11. We add alternate digits and then we add the digits we missed. Subtract one sum from the other and if the result is zero the original number was divisible by 11. Example: 1456785. 1 5 7 5 make up one set of alternate digits and the other set is 4 6 8. 1+5+7=13. We drop the ten and keep 3 in mind to add to 5 to give us 8. Now 4 6 8: 4+6=10, drop the ten and add 0 to 8 to give us 8 (or ignore the zero). 8-8=0 so 11 divides into 1456785. Now 9512283: set 1 is 9 1 2 3 and set 2 is 5 2 8; 9+1=0 (when we drop the ten); 2+3=5; set 1 result is 5; 5+2+8=5 after dropping the ten, and 5-5=0 so 9512283 is divisible by 11. Nines remainder for checking arithmetic. We can check the result of addition, subtraction, multiplication and (carefully) division. Using Method 2 above we can reduce operands to a single digit. Take the following piece of arithmetic: 17*56-19*45+27*84. We'll assume we have carried out this sum and arrived at an answer 2365. We reduce each number to a single digit using Method 2: 8*2-1*9+9*3. 9's have no effect so we can replace 9's by 0's: 8*2 is all that remains. 8*2=16 and 1+6=7. This tells us that the result must reduce to 7 when we apply Method 2: 2+3+6=11; 1+1=2 and 2+5=7. So, although we can't be sure we have the right answer we certainly don't have the wrong answer because we arrived at the number 7 for the operands and the result. For division we simply use the fact that a/b=c+r where c is the quotient and r is the remainder. We can write this as a=b*c+r and then apply Method 2, as long as we have an actual remainder and not a decimal or fraction. Divisibility by 3. This is similar to Method 2. We reduce a number to a single digit. If this digit is 3, 6 or 9 (in other words, divisible by 3) then the whole number is divisible by 3. Divisibility by 6. This is similar to Method 6 but we also need the last digit of the original number to be even (0, 2, 4, 6 or 8). Divisibility by 4. If 4 divides into the last two digits of a number then the whole number is divisible by 4. Using 4 or 2 times 2 instead of 25 for multiplication and division. 469/25=469*4/100=1876/100=18.76. 538*25=538*100/4=134.5*100=13450. We could also double twice: 469*2=938, 938*2=1876, then divide by 100 (shift the decimal point two places to the left). And we can divide by 2 twice: 538/2=269, 269/2=134.5 then multiply by 100 (shift the decimal point two places left or add zeroes). Divisibility by 8. If 8 divides into the last three digits of a number then the whole number is divisible by 8. Using 8 or 2 times 2 times 2 instead of 125 for multiplication and division. Similar to Method 9, using 125=1000/8. Using addition instead of subtraction. 457-178. Complement 178: 821 and add: 457+821=1278, now reduce the thousands digit by 1 and add it to the units: 278+1=279; 457-178=279. Example: 1792-897. First match the length of 897 to 1792 be prefixing a zero: 0897; complement this: 9102. 1792+9102=1894. Reduce the thousands digit by 1 and add to the result: 894+1=895. Example: 14703-2849. 2849 becomes 02849, then complements to 97150. 14703+97150=111853; reduce the ten-thousands digit by 1 and and add to the result: 11854. Squaring numbers ending in 5. Example: 75^2. Start by writing the last two digits, which are always 25. Take the 7 and multiply by 1 more than 7, which is 8, so we get 56. Place this before the 25: 5625 is the square of 75. The square of 25 is ...25, preceded by 2*3=6, so we get 625. All numbers ending in 0 or 5 are exactly divisible by 5 (see also Method 1). All numbers ending in zero are exactly divisible by 10. All numbers ending in 00, 25, 50 or 75 are divisible by 25. Divisibility by 7. Example: is 2401 divisible by 7? Starting from the left with a pair of digits we multiply the first digit by 3 and add the second to it: 24: 3*2+4=10; now we repeat the process because we have 2 digits: 3*1+0=3. We take this single digit and the one following 24, which is a zero: 3*3+0=9. When we get a single digit 7, 8 or 9 we simply subtract 7 from it: in this case we had 9 so 9-7=2 and the single digit is now 2. Finally in this example we bring in the last digit: 3*2+1=7, but 7 is reduced to 0. This tells us the remainder after dividing 2401 by 7 is zero, so 2401 is divisible by 7. Another example: 1378. 3*1+3=6; 3*6=18 before adding the next digit, 7 (we can reduce this to a single digit first): 3*1+8=3*1+1=4; now add the 7: 4+7=4+0=4;  3*4=12; 3*1+2+8=5+1=6, so 6 is the remainder after dividing 1378 by 7.  See also my solution to:
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how to find slope and y-intercept of y=3x-5

Any line equation which is in the form y=m*x + b is called slope-intercept form. What that gives you is the slope is the number which is multiplied by X (called the coefficient) while the slope intercept is the Y value when X=0. So if we take your first equation: y = 3x - 5 The slope (or m) = 3 The Y intercept = -5 Slope is defined as either "the change in y divided by the change in x" or "rise over run", so that 3, really can be considered as 3/1. Each change in X of 1 will change Y by 3. So slope = 3/1. Graphing any line can come from 2 methods. 1) create a table of values or 2) calculate a single point (x,y) and then apply the slope to find a second coordinate pair (x,y). For the equation y=3x - 5: y = 3x - 5 x y 0 -5 1 -2 Replacing the values for X into the original equation, will come out to the values for Y. So when X=0, y = 3*0 - 5, or simply -5.  When X = 1, Y = 3*1 -5 or simply -2. With these two coordinate pairs of points, you can plot a dot on your graph at each (0,-5) and (1, -2) then draw a straight line which goes through each point and continues straight in each direction, probably ending each end of this line with an arrow to show it continues. I do not have a way to include a picture here of a graph. The second way to graph a line is as follows. You need a starting point that will be on the line. Given the form of y=mx + b, you have a simple point which can be used at the y-intercept. The point is always in the form of (0, b), so in this case it is (0,-5). From that first point, you will apply your slope. The slope is 3 (or technically 3/1 which is a big help). From the initial point (0,-5) you will go UP 3 and RIGHT 1 and that will be the next point that is easy to find. Connect those two points and continue the line in each direction and that will be a graph of your line. Anytime your slope is positive, you will use it by going UP the top number (numerator of the slope) and going RIGHT the bottom number (denominator of the slope). But if your slope is negative (like your second problem is) you will use it by going DOWN the numerator and then RIGHT the denominator. The equation is y= -2/3x + 4 ( / = divided). I need to state the slope and y-intercept. I will not walk through the details on the second equation, but you should have enough information to get the answer from the above example.
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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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One number is 3 more than 2 times another. Their product is 27.

Question: One number is 3 more than 2 times another. Their product is 27. Find the numbers. Answer in the form of paired points with the lowest of the two numbers first. Let the two numbers be x and y. One number is 3 more than 2 times another. Therefore x = 2y + 3 Their product is 27. Therefore xy = 27 Substituting for x = 2y + 3 into the product equation, (2y + 3)y = 27 2y^2 + 3y - 27 = 0 (2y + 9)(y - 3) = 0 Hence y = -4.5, or y = 3, giving x = -6, or x = 9 As paired points, the two numbers are: (-4.5, -6) or (3, 9)
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what is each part of the equation y=mx+b mean and who do you find them using math vocabulary

The normal meaning for this standard linear equation is that x and y are coordinates in a rectangular arrangement of axes. The y axis is North-South while the x axis is East-West. Where they cross is called the origin with coordinates (0,0), that is, x and y are both zero. The equation y=mx+b defines a straight line. It slopes at a value given by m, the slope or gradient, and m is a number which can be a whole number, a fraction, or whatever, as long as it is constant so that the line remains straight. The slope, m, is also known as the tangent, and the tangent of the angle that the line makes with the x axis has a value of m. When the straight line is at an angle of 45 degrees to the x axis, its tangent is 1 so m=1. If the line slopes backwards at 45 degrees to the x axis, it's tangent is -1 and m=-1. Forward sloping lines have a positive m, while backward sloping lines have a negative m. The value of b is also called the y intercept, because it's the point on the y axis where the straight line crosses that axis. It can have a positive or negative value. b is a constant, just like m. mx is m times x. The x axis is divided by equally spaced numbers, 0, 1, 2, 3 etc to the right, and -1, -2, -3 etc to the left of the origin. The y axis is similarly divided, postitive numbers going up and negative numbers going down. By putting numbers in the equation you can work out where points go on the line. m will have a value, like 2, for example, and b a value, say, 3, so we have y=2x+3. If we put x=0 we get y=3 which is the y intercept. So we mark that point (0,3) by going up 3 divisions on the y axis. Now put x=1, then y=5. So we move to the point (1,5), which is right 1 and up 5. We can join that point to (0,3) and continue beyond these points. What we find is that, although we have only plotted two points, other values of x and y actually fit on the line. If we look at where x=3 and go up to meet the line, then go horizontally back to the y axis, we should find it meets the point 9 on the y axis. So the line represents the relationship between x and y as given by the equation for all points including points in between our whole number divisions, like, for example, 1.5 or one and a half, halfway between 1 and 2.
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