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solve and give answer using interval notation

Solve and give answer using interval notation 2x2 – x  ≥  3

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Interval Notation 0 - Cool Math


This algebra lesson explains interval notation. ... This notation is my favorite for intervals. It's just a lot simpler! Let's look at the intervals we did with the ...
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Solving Linear Inequalities: Introduction and Formatting


Solving linear inequalities is almost exactly like solving ... "Interval notation" writes the ... along with the different answer formats: Solvex ...
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Solving Inequalities – Using Interval Notation – Cool


This algebra lesson explains interval notation. ... What about ? Here it is on the number line: x can't be -1, but it can be greater than -1...
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Virtual Math Lab/ Interval Notation - West Texas A&M University


Write the answer to an inequality using interval notation. Draw a graph to give ... When solving linear inequalities, ... Solve, write your answer in interval ...
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Solving Compound Inequalities & Interval Notation - YouTube


Jul 25, 2011 · Show several examples of how to solve compound inequalities and how to write the solution set in interval notation

How to Solve Inequalities With Interval Notation | Sciencing


To describe this infinite set of solutions, you would use interval notation, ... How to Solve Absolute Value Equations With a Number on the Outside.
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interval notation calculator - Solve-variable.com


Solve-variable.com gives useful advice on interval notation ... interval notation calculator? Can you give me more ... answer your math problems is by using a ...
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Solve and graph the inequality. Give answer in interval ...


Solve and graph the inequality. Give answer in interval notation. 7x ... Express the set -3 x-6 > 3 x+12 using interval notation Solve the ... Start here or give us ...
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How to Express Solutions for Inequalities with Interval Notation


the interval is called a closed interval, which you show on the graph with a filled-in circle at the point and by using square brackets in notation.
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Suggested Questions And Answer :


solve and give answer using interval notation

Let's check how's IQ is too high, give one answer of below question - He is joru ka gulam He has GPS chip in his ace He is note ka goti His name start from G and end on V
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solve inequality, give answer in interval notation |4-3x|+6≥7

|4 - 3x| + 6 >= 7 |4 - 3x| >= 1 4 - 3x <= -1 or 4 - 3x >= 1 3x - 4 >= 1 or 3x - 4 <= -1 3x >= 5 or 3x <= 3 x >= 5/3 or x <= 1 In interval notation, we have: (-infinity , 1] U [5/3, infinity)
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How do I write 13/2<x in interval notation?

wotz rong with x>6.5????? look fine tu me
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How do I solve -14<=-2x-2<=-8

Divide interval expression through by 2: -7<=-x-1<=-4. Now add 1 throughout: -6<=-x<=-3. Take this as two inequalities: -6<=-x and -x<=-3. In the first inequality add 6 throughout and in the second add 3: 0<=6-x and 3-x<=0. Now add x through both inequalities: x<=6 and 3<=x. Combining these: 3<=x<=6.
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4>8-3x/5>-2 I am having trouble solving this using interval notation.

Question: Solve 4>8-3x/5>-2. Split the above inequality into two inequalities. 4 > 8 - 3x/5 and 8 - 3x/5 > -2 Working both inequalities together, -4 > -3x/5    and   10 > 3x/5 -20 > -3x      and    50 > 3x 20/3 < x      and     x < 50/3 fitting the two inequalities together, 20/3 < x < 50/3 In set notation: x є  (20/3,50/3}
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Cos(x)=1/4

You did well by solving the quadratic, and you seem to have spotted that cos(x) can't be 2 because the cosine function can't give a result greater than 1 or less than -1. x=arccos(1/4) or x=cos^-1(1/4) is how the answer is found. This may be a function on your calculator. If you use tables you would look in the body of a cosine table to find what angle gives you a cosine of 0.25. The answer is 75.52 degrees approximately.
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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.
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how to solve 4X-5y=11 AND 7x+9y=3 USING ELIMINATION

how to solve 4X-5y=11,7x+9y=3 USING ELIMINATION The eqns are: 4x - 5y = 11 7x + 9y = 3 Let's eliminate the y's say, by multiplying the 1st eqn by 9 and the 2nd eqn by 5. This gives us, 36x - 45y = 99 35x + 45y = 15 Now simply add the two eqns together, to give 71x = 114 x = 114/71 Substituting for x = 114/71 in the 1st eqn, we get 4(114/71) - 5y = 11 456/71 - 11*(71/71) = 5y 456 - 781 = 71*5y -325 = 71*5y -65 = 71y y = -65/71 Answers: x = 114/71, y = -65/71
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Find the point(s) where the line through the origin with slope 6 intersects the unit circle.

The line passes through the origin so its equation is y=6x. The equation of the unit circle is x^2+y^2=1, which has centre (0,0) and radius 1. Substituting y=6x in the equation of the circle we have 37x^2=1 and x=+sqrt(1/37). Therefore the y values for the intersections are y=+6sqrt(1/37). The points of intersection are (-sqrt(1/37),-6sqrt(1/37)), (sqrt(1/37),6sqrt(1/37)).  If the unit circle has centre (h,k) the equation is (x-h)^2+(y-k)^2=1 and substituting y=6x we get (x-h)^2+(6x-k)^2=1, which gives the x value of the intersection. So x^2-2xh+h^2+36x^2-12xk+k^2=1; 37x^2-2x(h+6k)+h^2+k^2-1=0. There are only two factors of 37, which is prime, so to factorise rationally we must have (37x+a)(x+b)=0; 37x^2+x(37b+a)+ab=0. Using the quadratic formula: x=(h+6k+sqrt((h+6k)^2-37(h^2+k^2-1))/37. The square root can only be evaluated if the expression is positive, so (h+6k)^2>37(h^2+k^2-1). This requirement applies so that the line intersects the unit circle. When the expression is zero, the line is a tangent to the circle, so there is only one intersection point. h^2+36k^2+12hk>37h^2+37k^2-37; 36h^2-12hk+k^2<37; (6h-k)^2<37 and (6h-k)< +sqrt(37) (=+6.08). This connects the coordinates of the centre of the unit circle: k>6h+sqrt(37). On equality the line y=6x will be tangential to the circle. For example, if h=0 (centre of the circle is on the y axis), k=+sqrt(37) and the unit circle will lie above or below the axis with y=6x running tangentially on the right of the circle; or on the left touching the circle below the x axis. Ideally, we want the square root to be rational so 37-(6h-k)^2=a^2. If a=+1, 6h-k=6 and k=6(h-1); or 6h-k=-6, so k=6(h+1). That gives many possible values for h and k represented by pairs: (1,0), (2,6), (3,12), (0,-6), (-1,-12), (-2,-18), (0,6), (1,12), (-1,0), (-2,-6),  to mention but a few. Using (1,0) in the quadratic: 37x^2-2x(h+6k)+h^2+k^2-1=0 we have 37x^2-2x=0=x(37x-2) giving intersection points (0,0) and (2/37,12/37). The equation for the circle is (x-1)^2+y^2=1 or y^2=2x-x^2. Let's try (2,6). 37x^2-76x+39=0, (37x-39)(x-1)=0 giving intersection points (39/37,234/37) and (1,6). The equation of the circle is (x-2)^2+(y-6)^2=1 or x^2-4x+y^2-12x+39=0. If a=+6, 6h-k=1 or -1, so k=6h-1 or 6h+1. This generates more possible intersection points. There are clearly an infinite number of positions for the unit circle centre (h,k) and an infinite number of intersection points. However, the relationship between h and k so as to produce rational intersection points has been established. k=6(h+1), k=6h+1 are the equations linking the coordinates of the centre of the unit circle. With these equations in mind the quadratic determining the intersection points (x,6x) can be solved: 37x^2-2x(h+6k)+h^2+k^2-1=0. There are four variations of this quadratic because there are four equations linking h and k. Recap There are 2 values of a^2 where a^2=37-(6h-k)^2 and x=(h+6k+a)/37; a^2=1 or 36. When a^2=1, k=6(h+1). The equation of the circle is (x-h)^2+(y-6(h+1))^2=1 and x=(37h+36+1)/37. So the points of intersection are (h+1,6(h+1)), ((37h+35)/37,6(37h+35)/37), ((37h-35)/37,6(37h-35)/37), (h-1,6(h-1)). When a^2=36, k=6h+1. The equation of the circle is (x-h)^2+(y-6h+1)^2=1 and x=(37h+6+6)/37. The points of intersection are ((37h+12)/37,6(37h+12)/37), (h,6h), ((37h-12)/37,6(37h-12)/37). Note that (h,6h) is the result of (37h+6-6)/37 and (37h-6+6)/37. We can check the (h,k) values we used earlier. These were (1,0) and (2,6). We used the formula k=6(h-1) in each case (a=+1), so intersection points for h=1, k=6(h-1)=0, should be x=(h+6k+1)/37, giving (2/37,12/37) and x=(h+6k-1), giving (0,0). For (2,6) h=2 and k=6, giving intersection points x=(2+36+1)/37, giving (39/37,234/37) and x=(2+36-1)/37=1, giving (1,6). The values of h and k are not restricted to integers.
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write equation of the line with the given pt and parallel to the given line (8,9); x+5y=4

First thing:  Parallel to the given line.  That means the slope of your line should be the same as the the slope of x + 5y = 4.  We need to find the slope of x + 5y = 4. x + 5y = 4 5y = -x + 4 y = (-1/5)x + 4/5 The slope is -1/5 . Second thing:  Make the new line.  We know a point (8,9) and the slope (-1/5), so we can use point-slope form: y - y1 = m(x - x1) y - 9 = (-1/5)(x - 8) That is the equation of our line, but let's clean it up a bit: y - 9 = (-1/5)x + 8/5 y = (-1/5)x + 8/5 + 9 y = (-1/5)x + 8/5 + 45/5 y = (-1/5)x + 53/5 y + (1/5)x = 53/5 5y + x = 53 Answer:  x + 5y = 53 . Note:  Unless the problem specifically asks for a certain form (give the answer in whatever form), we want to give the answer in the form used in the problem.
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