Guide :

Explain why you cannot use substitution to determine the limit.

Why can't you use substitution. Find the limit if it exists. limit (x--> -2): square root of x-2

Research, Knowledge and Information :


L. 2.1 Solutions - Webs


Determine the limit by substitution. ... explain why you cannot use substitution to determine ... use the graph to estimate the limits and value of the
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Please Explain The Problem Step By Step And Give ... - chegg.com


... Please explain the problem step by step and give me the final answer to it. Explain why you cannot use substitution ... use substitution to determine the limit.
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Find Limits of Functions in Calculus - analyzemath.com


Find Limits of Functions in Calculus. ... Use limit properties and theorems to rewrite the above limit as the product of two limits and a constant.
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calc hw 2.1a (1).pdf - Calc AB HW 2.1a In Exercises 1 and 2 ...


... or explain why the limits do not exist. 1. a ( ) ... Determine the limit by substitution. ... Explain why you cannot use substitution to determine the limit.
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www.northallegheny.org


In Exercises 7—14, determine the limit by substitution. Support graph- ically. ... In Exercises 15—1 8, explain why you cannot use substitution to deter-
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11 Limits and an Introduction to Calculus - Cengage Learning


750 Chapter 11 Limits and an Introduction to Calculus The Limit ... direct substitution to evaluate limits. Why you should ... a limit, remember that you cannot
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Limits (An Introduction)


Limits (An Introduction) ... But we can use the special "−" or "+" signs (as shown) to define one sided limits: ... When you see "limit", ...
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www.northallegheny.org


determine the limit by substitution. ... explain why you cannot use substitution to determine the limit. ... or explain why the limits do not exist.
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Calculus I - Substitution Rule for Indefinite Integrals


Substitution Rule for Indefinite Integrals ... it cannot be stressed ... The idea that we used in the last three parts to determine the substitution is ...
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Suggested Questions And Answer :


Explain why you cannot use substitution to determine the limit.

thats same as sqrt(x) as x gotu 0...anser=0 ????????? substitute wot ????????? me assume yu kum from rite side so x alwaes > 0, so take root av + number
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i have a venn diagram that has a large yellow triangle in the middle. what can i lable the other two loops

Until the table is provided, only a general answer can be given. Let's suppose that one loop is the set of objects with property A, and the second loop that with property B. Forgetting about the triangle for a moment, the intersection of the loops is the set of objects with properties A and B. Now, introduce the triangle. By itself it represents the set of objects with property C. But because it's within the intersection of the loops, C is a subset of the set of objects with properties A and B. As a subset it can only contain objects with properties A and B and it does not contain all such objects. In fact, the area outside the triangle but within the intersection of the two loops represents the set of objects with properties A and B but excludes those contained by the triangle. The Venn diagram contains 4 regions: A objects only B objects only A and B but not C A, B and C objects Let's use an example. A objects are red; B objects are square; C objects are striped. The 4 regions are: Red objects that are neither square nor striped Square objects that are neither red nor striped Square red objects that are not striped Square striped red objects The loops can be labelled RED and SQUARE and the triangle can be labelled STRIPED. By substituting other properties for red, square, striped, the diagram can be used to solve sorting problems. Let's substitute LARGE for striped. The triangle represents large objects. Both loops contain some large objects. The 4 regions become: Small red objects that are not square Small square objects that are not red Small red square objects Large red square objects. The loops can be labelled RED and SQUARE, and the triangle can be labelled LARGE.
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How to find c and y so that each quadrilateral is a parallelogram

The opposite internal angles of a parallelogram are equal, and adjacent angles are supplementary, but which of the given angles are opposite and which are adjacent? We know that all angles must be positive, so 7x-11>0 and 5y-9>0 so x>11/7 or 1.57 and y>1.8. We also know that all the angles of a parallelogram can be determined if just one is known, because of the relationships. This means that if we take any pair of angles we know that they are (a) equal or (b) supplementary. Take the first pair: 5x+29 and 5y-9: if (a), 5x+29=5y-9, so 5(y-x)=38 and y=(38+5x)/5; or if (b), 5(y+x)=160, or y+x=32 and y=32-x. Also, the remaining pair 3y+15 and 7x-11: if (a), 7x-3y=26, y=(7x-26)/3; or if (b), 7x+3y=176 and y=(176-7x)/3. We can see that if (a) is applied to the first pair at least one of x or y will contain a fraction. If (a) is applied to the second pair, which can be written 2x-8+(x-2)/3, x must be x=5, 8, 11, ..., 3n+2 for y to be an integer (where n is a positive integer) and y is 3, 10, 17, ..., and if (b), which can be written 58-2x-(x-2)/3, x must be 3n+2 for y to be an integer (where n is an integer 0 Read More: ...

evaluate tha definite intergral top 3 and bottom 1 (3x^2+2x)

My previous answer:  integral((3x^2+2x)dx)=integral(3x^2dx)+integral(2xdx) for 1 Read More: ...

how many 1/2" candies will fit in a 41/2" x 4" bowl?

Assuming that 4" is the height of the bowl, we can conclude that it is ellipsoidal, like a rugby football, with the shorter axis (x axis) being 4.5" in diameter and its longer axis (y axis) needing to be determined from the given figures. Two domes are cut off each end of the ellipsoid leaving a 3.5" diameter circular top and a 3" diameter circular base. The general equation of the ellipse that will be used to model the bowl is x^2/2.25^2+y^2/a^2 (2.25" being the "radius" of the x axis, and a the radius of the y axis). The ellipse has its centre at the x-y origin. To find a we work out in terms of a where the base of the top dome has a width of 3.5" and where the base of the bottom dome has a width of 3". Putting x=1.75 (half of 3.5) in the equation of the ellipse, we get y=sqrt(a^2(1-1.75^2/2.25^2))=4asqrt(2)/9. Similarly for the base dome: y=asqrt(1-1.5^2/2.25^2)=-asqrt(5)/3, negative because it's below the centre or origin. The difference between these two y values is the height of the bowl, 4", so 4asqrt(2)/9+asqrt(5)/3=4, and a=36/(4sqrt(2)+3sqrt(5))=2.9114 approx.  The volume of the bowl is found by considering thin discs of radius x and thickness dy so that we can use the integral of (pi)x^2dy (the volume of a disc) between the y limits imposed by the heights of the two domes. x^2=5.0625(1-y^2/a^2). Because a is a complicated expression, we'll just use the symbol for it. We can also write 5.0625 as 81/16. The integrand becomes (81(pi)/16)(1-y^2/a^2)dy, with limits y=-asqrt(5)/3 and 4asqrt(2)/9. The integrand is simply the sum of the volumes of the discs between the bases of the two domes. Integration gives us (81(pi)/16)(y-y^3/(3a^2)) between the limits, which I calculated to be 53.39 cu in approx. How many candies fit into this volume? We know the length of the candies is 1/2", but we don't know any other dimensions, so we don't know the volume of each candy. Also, the alignment of candies will improve the number of candies fitting into the bowl, but if this cannot be arranged, we have to assume they will be randomly orientated. We could make the assumption that they are cone-shaped with height 0.5" and base diameter 0.25" (radius 0.125" or 1/8") giving them a volume of (1/3)(pi)0.125^2*0.5=0.0082 cu in each approximately. So divide this into 53.39 and we get about 6,526, with very little space between the candies. A more realistic figure can be obtained by finding out how many candies fill a cubic inch when tightly packed. Two will fit lengthwise into an inch. Think of the candies as cuboids 1/4" square and 1/2" long (volume=1/4*1/4*1/2=1/32 cu in). That gives you 32 in a cubic inch, over 1,700 in a bowl. The cuboid is a sort of container that will hold just one candy and allow it some lengthwise freedom of movement. At the other extreme think of the candies as 1/2" cubes (volume=1/8 cu in) then only 8 will fit into a cubic inch and only about 427 will fit into the bowl. But the candies have greater freedom of movement in a cubic container. Let's say you can get 100 into one cubic inch packed tightly, then you would get about 5,340 in the bowl. Get a small box and pack in as many candies as you can. Measure the volume of the box (length*width*height). If N is the number of candies, then each has an average effective volume of (box volume)/N. Use this number to divide into the volume of the bowl.  
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area under a curve at the right of a circle

Remember that the area under a curve has to be bounded, so limits need to be applied to define the boundaries. Usually it's the area between the curve and one axis or both. Occasionally it's the area between two curves or the area produced by the intersection of two curves. Sometimes, as in the case of a circle or ellipse which already encloses an area, it's the whole or part of an area inside the curve. The "formula" is based on the area of very thin rectangles, infinitely thin, in fact, which are laid side by side to fill the area. An integral is applied which sums the areas of the rectangles over the whole region specified to get the area of that region. The most common formula is integral(ydx) where y=f(x) defines the curve. Limits of x are then given to enclose the area, so that definite integration is applied. The best way to approach any problem involving finding areas related to a curve is to sketch the curve and draw the area that needs to be found. Imagine a large curve had been drawn on the ground and you had a roll of sticky tape. You work out where the area is and cut strips of tape and stick them down so that you fill the required area. You can only lay the strips side by side, no criss-crossing and no laying the tape in a different direction. You can only cut the strips into rectangles using a cut straight across, not at an angle. You end up with not quite filling, or slightly over-filling the space because the tape will overlap the curve slightly. But when you step back it will look like the area has been properly covered by tape. If you used narrower tape the area would be even better filled. That's the principle on which the integration is based, because the area of each rectangle is the length of the strip y times the width we call dx (the width of the strips of tape). The area outside a circle must be bounded. You need the equation of the circle so that you can relate x and y. For example, the general equation of a circle is (x-h)^2/r^2 + (y-k)^2/r^2=1, where r is the radius and (h,k) is the centre. So y=k+sqrt(r^2-(x-h)^2) is half the circle, because the other half is k-sqrt(r^2-(x-h)^2).  If you need the area between the circle and the x axis between a and b, you need the lower part of the curve given by the second expression; if you need the upper part of the curve you need the first expression. If the circle is coloured red and the outside of the circle is blue, the area between the lower circle and the x axis will be entirely blue, whereas the area between the upper circle and the x axis will be red and blue. Get the picture? Once you've decided what you want, you compose the integral: integral(ydx)=integral((k-sqrt(r^2-(x-h)^2))dx) for b Read More: ...

(2x^2+x-6)/(x+2)=

2x^2+x-6 factorises: (x+2)(2x-3) so we have (x+2)(2x-3)/(x+2). There is a common factor in the denominator and numerator: x+2, so divide them by this common factor: 2x-3 is the answer. There's just one thing: you can only divide through by a common factor if the factor isn't zero, in other words x+2 cannot be zero, so x cannot be -2. Any other value except x=-2 is OK. This is because we can't divide by zero. How did I know how to factorise the top? Well, I guessed that the denominator was likely to be a factor, so I used the "solution" of x+2=0, subtracting 2 from each side: x=-2, then I substituted this value of x into 2x^2+x-6: 2*(-2)*(-2)-2-6=8-2-6=0, which tells me that x+2 divides into the numerator. To find the other factor making up the quadratic, I asked myself what do I need to multiply x by to get 2x^2, and the answer is 2x^2/x=2x; then I asked, what do I need to multiply 2 by to get 6? 6/2=3. Finally, what sign + or - is needed to go in front of 3? We already have +2 in the factor x+2, so we need a minus to make -6. So the other factor of the quadratic must be 2x-3.
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use a graphical method to solve this simultaneous equation: y=x^2 and y=x+6

Plot y=x^2 and y=x+6 on the same graph. The first is a U-shaped curve sitting on the origin (0,0), the second is a straight line crossing the x axis (y=0) at x=-6 and crossing the y axis (x=0) at y=6. Where the line cuts through the U curve represents the solution to the equation x^2-x-6=0 (x^2=x+6). The solution to this quadratic is x=-2, +3. The intersection points are therefore (-2, 4) and (3, 9). The remaing part of your question involves straight lines only, linear equations. For (a) draw a straight line joining y=-4 on the y axis to x=2 on the x axis. Also draw a line joining y=3 1/3 to x=-1 2/3. These points on the axes are where x=0 and y=0 for the two functions. The lines don't stop at the axes, so just continue them after they cut the axes. What may have been confusing to you is that the first equation starts y=..., but the second equation has x and y together in an expression. However, you can move things around in an equation and, if you want, you can make the equation look like y=... or x=... or just combine x and y in an expression. It doesn't matter. What you'll find is that the lines are parallel (have the same slope) so that means they never cross and that means there's no solution. In (b) the two lines have different slopes so we would expect them to intersect. The first equation means joining (0,3) to (9/2,0) and (0,11/2) to (11/3,0). Again, extend the lines to beyond where they cut the axes. Note that the second equation simplifies to 3x+2y=11. The solution to the equations solved simultaneously or by substitution give a single intersection point at (3,1).
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When solving a system of equations, how do you determine which method to use?

Unless you're told what method to use, you can use any of those methods. They all have to give the same solution. I use the one that I think will be quickest. A clue can sometimes be seen in the coefficients, suggesting perhaps elimination as the best method. Non-linear systems suggest substitution. Graphing can show inconsistency or multiple solutions.
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Research and explain how it creates parallel rays of light, name 4 uses for retroreflectors and why were they placed on moon's surface.

The three mirrors are arranged to form the corner of a cuboid, or more specifically a cube with the reflecting surfaces facing inwards. An incident ray of light can be considered to have an x, y and z component, where x, y and z are Cartesian components of the ray vector. The ray will strike one mirror first and then be reflected on to another, which in turn reflects on to the third mirror. With each reflection one component of the original ray is reversed. So (x,y,z)→(-x,y,z)→(-x,-y,z)→(-x,-y,-z). In the last reflection all three components of the ray have been reversed, meaning that the ray is reflected back parallel to the path it arrived. A retroreflector on the moon, for example, would bounce a laser ray back the way it came and by timing the journey from source back to source using accurate timing devices, the distance of the moon to the earth can be calculated because the speed of light in a vacuum is known. Using satellites instead of terrestrial sources gives better results because there is no atmosphere to cause variation in the speed of light. Retroreflectors on the moon are usually in arrays so that they capture and reflect light over a wide area. Retroreflectors can be used similarly for navigation systems to provide great accuracy in determining location. So they can be installed in satellites for just such a purpose. They can also be used to determine the speed of light in various media where distances between points are known and the timing is measured. They're used in reflectors on the road surface and in vehicle reflectors. They can also be used in hi-viz clothing.
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