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If M=√2÷√(×+3)(×-2) determine the value of × for which M is real

if M=√2×÷√(×+3)(×-2) determine the value of x which M is real

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In the following equation, m is a real number. z^2 - (3 + i ...

... m is a real number. z^2 – (3 + i) z + m + 2 i = 0 Calculate the values of m such that the equation has a real root. Calculate the second root.
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Chapter I Introduction Solutions and Initial Value Problems ...

Section 1.2 Solutions and Initial Value Problems Example 1. (a) Determine for which values of m the function ϕ(x) = xm, ... and m2 = −3.
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if the equation below is to have 1 double real root ...

if the equation below is to have 1 double real root, determine the value for m. ... So, for this parabola to have one real solution, m must be equal to 4.
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#4x^2-6mx+(2m+3)(m-3)# -

... (m - 3) = 0 we obtain {(x_1=1/2(m-3 ... **Determine all the values of the parameter m for which the operation: ** #4x^2-6mx+(2m+3)(m-3)# has TWO different real ...
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Shortlisted Problems with Solutions

Shortlisted Problems with Solutions ... 3 white and C 2 red. Find the maximum value T can attain. C4. ... Determine all integers m ≥ 2 such that every n with m 3
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Linear Functions Part A: Basics: Slope and Intercept

Part A: Basics: Slope and Intercept ... Notice from the table that the value of y increases by m = 3 for every ... say (x 1, y 1) and (x 2, y 2), determine a line in ...
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mid ans dvi - Northwestern University

(2) Determine the truth value of each of these statements if the universe ... (3) (a) TRUE (√ 2 is a real number.) (b) ... mid_ans_dvi Created Date: Wed May ...
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Chem Final All of the Problems Flashcards | Quizlet

Chem Final All of the Problems. STUDY. PLAY. ... Calculate the value of q (kJ) in this exothermic reaction when 1.70 g of methane is combusted at constant pressure.
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Section 2.1: BASIC PROOFS WITH QUANTIFIERS Existence ... by the Intermediate Value Theorem, there is a real number r ... Example 2: Prove that for every real number x ...
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Suggested Questions And Answer :

minimun of sqrt(x^4-3x^2+4)+sqrt(x^4-3x^2-8x+20)

f = sqrt(x^4-3x^2+4)+sqrt(x^4-3x^2-8x+20) To find the turning points, differentiate f and set that value to zero. df/dx = (1/2)*(4*x^3-6*x)/sqrt(x^4-3*x^2+4)+(1/2)*(4*x^3-6*x-8)/sqrt(x^4-3*x^2-8*x+20) = 0 (4*x^3-6*x)/sqrt(x^4-3*x^2+4) = -(4*x^3-6*x-8)/sqrt(x^4-3*x^2-8*x+20) cross-multiplying and squaring, (4*x^3-6*x)^2*(x^4-3*x^2-8*x+20) = (4*x^3-6*x-8)^2*(x^4-3*x^2+4) expanding this gives, 16*x^10-96*x^8-128*x^7+500*x^6+384*x^5-1068*x^4-288*x^3+720*x^2 = 16*x^10-96*x^8+244*x^6-236*x^4-64*x^7+288*x^5-544*x^3-48*x^2+384*x+256 which simplifies as, 64*x^7-256*x^6-96*x^5 +832*x^4-256*x^3-768*x^2+384*x+256 = 0 2x^7 - 8x^6 - 3x^5 + 26x^4 - 8x^3 - 24x^2 + 12x + 8 = 0 which factorises as, (x^2 - 2)(2x^5 - 8x^4 + x^3 + 10x^2 - 6x - 4) = 0 Now the (x^2 - 2) term gives you the two roots, +/- sqrt(2) The quintic equation has 3 real roots and two complex roots. The real roots are at (approx) x = -1, x = -0.5, x = 3.5. (I know where the roots are because I graphed the quintic and read from the graph.) If you wish to get these roots more accurately, then use the Newton-Raphson method with the above root values as strarting values. The fastest way of finding the minimum of the function, f, is to plug in all the above root values and determine the minimum value from that. A faster method would be to graph the curve of f(x), then say, "From the graph of f(x) it can be seen that the smallest local minimum is at where x lies between x = 1 and x = 2. From the solution, x^2 - 2 = 0, we can say that the mimimum occurs at x = sqrt(2)." Then include the minimum value. f(sqrt(2)) = sqrt(4 - 6 + 4) + sqrt(4 - 6 - 8sqrt(2) + 20) sqrt(2) + sqrt(18 - 8sqrt(2)) sqrt(2) + sqrt(16 - 8sqrt(2) + 2) sqrt(2) + sqrt{(4 - sqrt(2))^2} sqrt(2) + 4 - sqrt(2) f(sqrt(2)) = 4 I'm afraid that I couldn't find all the real roots analytically, so this solution is a bit incomplete, however you may still find it useful.
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Basic functions

1. What is the range of k(x)=-|x| Asking for the range is another way of asking for valid Y values (or in this case k(x) values). So the range is (-infinity, 0]. 2.Is y=2 a constant function? Yes. Constant functions have no variation in the output for any input. 3. x=-3 a function? No it is not. In order to be a function, there are several properties which must be satisfied, one of which is that for any input, there must be a single output. Often we simplify this by asking "does the function pass a vertical line test?" So if you are to draw a vertical line, would it only cross the function in one place for all values of X? For X=-3, it is a vertical line, so it fails this property of functions because there are an infinite number of solutions (Y values) for this single X value (-3). 4. What is the minimum value of y=x^2-4 The minimum value of parabola's will occur either at the vertex (where its slope is 0) or at its domain boundaries. There are no stated domain restrictions for this function and the slope is 0 at X=0. The first derivative is y'=2x and y'=0 at x=0. The second derivative is y''=2, indicating it is concave for all values X. So the minimum is x=0 and y=-4. For a function with a higher power, there may be more than one minimum (each called local minimum) and any place the second derivative is concave up, both to the left and right segments where the first derivative is 0 will be one of these local minimums. You will take your inputs for each relative minimum and any domain limits (for functions which do not have all real numbers for their domain) and compare the Y values of the original function to determine the absolute minimum for that function.
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If M=√2÷√(×+3)(×-2) determine the value of × for which M is real

This is easier than it looks. The square root term including x is only meaningful in this context when the quantity under the square root is strictly positive, so (x+3)(x-2)>0.  So we have just 2 conditions when this is positive: x+3 and x-2 are both positive; and x+3 and x-2 are both negative. Both positive means x+3>0 and x-2>0, so x>-3 and x>2. Since 2>-3, x>2 satisfies both inequalities.  Now when are they both negative? x+3<0 and x-2<0, so x<-3 and x<2. Since -3<2 then x<-3 satisfies both inequalities. So there you have it: M is real when x>2 or x<-3. But if √2x is the numerator rather than √2 and the expression is √(2x) rather than (√2)x, we must eliminate x<-3 leaving x>2 as the solution.
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find the real root of x^3-5x+1 by regula-falsi method

 find the real root of x^3-5x+1 by regula-falsi method. f(x) = x^2 - 5x + 1 Test for a root f(0) = 1, f(1) = -3. There is a change of sign between f(0) and f(1) , so there is a root between x = 0 and x = 1. The first (smallest) false root is denoted as a, the second (greatest) false root is denoted by b, the interpolated (false) root is denoted by c and c is given by, c = (a.f(b) - b.f(a)) / (f(b) - f(a)) After c is determined, it will replace either a or b for the next iteration. The sign of f(c)  is found, and c will replace a if f(c) has same sign as f(a), otherwise it will replace b. a b f(a) f(b) c=(a.f(b) - b.f(a)) / (f(b) - f(a)) sign(f(c)) Regula-falsi method 0 1 1 -3 0.25 -ve 0 0.25 1 -.234375 0.2025668264 -ve 0 0.2025668264 1 -0.4522143e-2 0.2016336225 +ve 0.2016336225 0.2025668264 0.0000295283 -0.4522143e-2 0.2016396263 +ve 0.2016396263   2.406*10^(-7)       When c reaches the value 0.2016396263, f(c) = 2.406*10^(-7), which is small enough to allow c to be taken as a real root of f(x). Root of f(x) is: c = 0.2016396263    
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solve for y=3x+4, and show both ordered sets on graph

The equation is linear in two variables, x and y. It can't be solved, as such, for particular values of x and y, but a straight line graph represents the dependency of one variable on the other. The graph is a mapping of the function in which the range of values of x and y is the set of all real numbers, so the graph maps x values onto y and y onto x in a unique way: there is only one value of x for a particular value of y, and only one value of y for a particular value of x. There is a real value of x for every real value of y and vice versa, subject to the dependency function y=3x+4, and no real values are excluded. The ordered pair (x,y) forms the set of real numbers satisfying the equation. Every point on the line (and there are an infinite number of them) is an ordered pair, and forms an infinite set. Each ordered pair is effectively a solution to the equation, so there are an infinite number of solutions.
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determine consecutive values of x between which each real zero is locatedf(x)=11x^4-3x^3-10x^2-9x+18

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Mathmatical sequences

i have a sequence that is: 60,57,51,42 what are the next three numbers
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If 2 + 22 = 15 & 4 + 16 = 16 & 6 + 10 = 17 & 8 + 4 = 18 what does 10 + 10 =

2 + 22 = 15, 4 + 16 = 16, 6 + 10 = 17 and 8 + 4 = 18   There must be certain common multiples, x and y, that make the real value of left-hand side(LHS) in each expression above equal to the real value of its right-hand side(RHS): x for the 1st terms of each LHS, { 2, 4, 6, 8 }, and y for the 2nd terms of each LHS, { 22, 16, 10, 4 }. So that, the expressions above can be rewritten in their real values, assuming the number of each RHS shows the real value of its LHS: 2x + 22y = 15 ··· Eq.1,  4x + 16y = 16 ··· Eq.2,  6x + 10y = 17 ··· Eq.3, 8x + 4y = 18 ··· Eq.4.    From Eq.1 and Eq.2, we have: x = 2 and y = 0.5. CK: Plug these values of x and y into each LHS of Eq.1, Eq.2, Eq.3 and Eq.4.   2·(2) + 22·(0.5) = 15,    4·(2) + 16·(0.5) = 16,   6·(2)+10·(0.5) = 17 and 8·(2) + 4·(0.5) = 18   CKD. Thus, in each given expression, the 1st term of LHS indicates 2 times of its number, the 2nd term of LHS indicates one half of its number, and the number of RHS expresses the real value of LHS.  Therefore, the answer is 10 + 10 = 25  {=10·(2)+10·(0.5)} 
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explain complex number and vectors

Start with real numbers. A number line is often used to represent all real numbers. It has infinite length and somewhere we can mark zero, dividing positive numbers (on the right of zero) and negative numbers (on the left of zero). The line is continuous so no real number can be left out. A line is 1-dimensional. A complex number has two components: one real, the other imaginary. A complex number can be represented by a plane, and it's 2-dimensional. So what does imaginary mean? The basis of imaginary numbers is the square root of minus one (sqrt(-1)) and traditionally it is given the symbol i. The square root of any negative number can be expressed using i. So, for example, the square root of minus 4 is 2i, because -4=4*-1 and the square root of 4*-1 is 2sqrt(-1)=2i. A complex number, z, can be written a+ib, where a and b are real numbers. But, more importantly perhaps, they can be represented as a point in 2-dimensional space as the point (a,b) plotted on a graph using the familiar x-y coordinate system. So, just like the number line represented all real numbers like an x axis, so all complex numbers can be represented by a plane, an infinite x-y plane. Now we come to vectors. There is a commonality between complex numbers and vectors. A straight road with cars , houses, people, etc., on it make the road like a number line. Any position on the road can be related to a fixed point on the road we'll call "home". Objects to the right, or eastward, could be in front of home and those to the left, or westward, behind home. The position of an object is the distance from home. This is a 1-dimensional vector field, where all objects have position. If the objects move their speed will have direction, towards the right or towards the left and we can say that a speed to the right is positive and a speed to the left is negative. Now we introduce another straight road at right angles to the first road. Now the picture is 2-dimensional. The position of an object is defined by two values: position east or west and position north or south. North can be positive and south negative. This is equivalent to the complex plane, which represents all complex numbers. The 2-dimensional plane represents all 2-dimensional vectors, whether it's position or speed. But the word "velocity" is used instead of "speed", because velocity includes direction, but speed is just a number, or magnitude, of the velocity. A vector has an east-west (EW) component (x value) and a north-south (NS) component (y value) and a vector r=xi+yj, where i is called a unit vector in the NS direction and j in the EW direction, so the point (x,y) fixes the positional vector r. Vectors are usually written in bold type, so you won't confuse i with i. The magnitude of a vector is sqrt(x^2+y^2) so it is represented by the hypotenuse of a right-angled triangle whose other sides are x and y. Pythagoras' theorem is used to work out the value. The magnitude of a vector is sometimes written |r| and is called a "scalar" quantity, so it doesn't have a direction or a sign (positive or negative), because the sign is a property of the direction of the vector.  A vector is not limited to a 2-dimensional plane. It can have as many dimensions as necessary. An aeroplane's positional vector and velocity would involve another dimension: height. A submarine's positional vector and velocity would involve depth. Height and depth are perpendicular to the EWNS plane and together form 3-dimensional space. The unit vector for height and depth is k, and height would be positive while depth would be negative, and the letter z is used with x and y so that a point in 3-space is (x, y, z). The magnitude |r| is sqrt(x^2+y^2+z^2). When working with vectors, addition and subtraction requires adding and subtracting the x, y, z components separately. When adding or subtracting complex numbers, the same applies to the x and y, real and complex, components. Multiplication and division are a special topic beyond the scope of this introductory explanation.  
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