Guide :

find the value of k

Uk+Uk+1=33

Research, Knowledge and Information :


SOLUTION: find all values for k such that the equation 3x ...


SOLUTION: find all values for k such that the equation 3x squared - 2x + k = 0 has imaginary roots. Algebra -> Complex Numbers Imaginary Numbers Solvers and Lesson -> ...
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Find the value of k, (if any), for which the system below has ...


... {cases}$ I am looking to finding values of $k$, ... Find the value of k, (if any), for which the system below has unique, infinite or no solution.
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Find Values of K For Different Number of Solutions To ...


Aug 14, 2014 · ... of k for different number of solutions of given quadratic equation. ... Find Values of K For Different Number of Solutions To Quadratic Equation

How can one find the values of k for which the line 2x -k is ...


The simple way to do this is to clearly define what it means for tangent so that finding the k values is the easiest. Problem Specific Answer
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In the equation y=kx, find the value of k if y= 4 and x= 8 ...


In the equation y=kx, find the value of k if y= 4 and x= 8 1. Ask for details ; Follow; Report; by fooNelly6BlasandraV 01/14/2017. Log in to add a comment Answers ...
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Find value of $k$ for which the equation has ... - Stack Exchange


Find value of $k$ for which the equation has real roots. up vote 2 down vote favorite. ... Find the possible values of $p$ for which the equation has coincident roots. 2.
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Place Value of a Number - WebMath


Place Value of a Number. This selection will help you to find what the place value is of a particular digit in a number. Type your number here, ...
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Suggested Questions And Answer :


find the equation of the line containing the given pair of points

If the equation of a line is y=mx + b where m is the slope and b is the y intercept you can find the m value by finding the slope between these two points. Slope = the rise over the run or the change in y over the change in x take the y values first -6 to -4 is a rise of 2. take the x values next - 2 to -5 is a run of -3 so your m value is -2/3x The easiest way to find the y intercept is to plot points until this crosses the y-axis. You will find the y intercept is -22/3 which is your b value. The equation for the line is y=-2/3x -22/3
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How do I find thr greatest place value in a number for a fourth graders math assignment?

The greatest place value in a number is the place value of the digit on the far left. Example: 123.456 The digit on the far left is 1. The 1 is in the hundreds place. The greatest place value in 123.456 is the hundreds place. . Note:  Special Condition:  The greatest place value in a number is the place value of the non-zero digit on the far left. We don't usually write things with 0's on the left (00037 vs. 37), but if we do, the greatest place value in a number is the place value of the digit on the far left, ignoring any 0's to the left of the leftmost non-0 digit. Example: 0007.89 Ignore the 0's on the left 7.89 7 is on the far left 7 is in the ones place The ones place is the greatest place value in 0007.89 . Example: 00012300.456 We only remove the 0's on the far left 12300.456 The far left digit is 1 The 1 is in the ten thousands place The ten thousands place is the greatest place value in 00012300.456
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Describe the possible values of x on a triangle, if one side's value is 5, 7, and 2x-2.

If the triangle is right-angled, then we can find two solutions before continuing to find others. If 2x-2 is not the hypotenuse, then 7 must be, because it's the longer of the two known sides: 49=25+4(x-1)^2, so (x-1)^2=6 and x=1+sqrt(6). If 2x-2 is the hypotenuse, then: 4(x-1)^2=49+25=74, so x-1=sqrt(37/2) and x=1+sqrt(37/2). Now the general case. Drop a perpendicular from angle ABC on to AC at P. Let's refer to angle BAC as simply A. When A is known, the triangle ABC is determined and x can be calculated. So if we can find x in terms of A we know it can be defined for all values of A. AP=5cosA and PC=7-5cosA, BP=5sinA, so 4(x-1)^2=25sin^2A+(7-5cosA)^2 (BC^2=BP^2+PC^2, by Pythagoras), so  4x^2-8x+4=25sin^2A+49-70cosA+25cos^2A=74-70cosA. So 4x^2-8x-70(1-cosA)=0 or 2x^2-4x-35(1-cosA)=0. Using the quadratic formula we get x=(1+sqrt((37-35cosA)/2). The negative square root will be disallowed in most cases, because BC=2x-2 has to be strictly positive so x>1. (At x=0, A=0 and ABC ceases to be a triangle and B lies on AC.)  
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HOW TO SOLVE LINEAR EQUQTIONS WITH FIVE UNKNOWN

It's clear that the percentages exceed 100%, so I conclude that the dry matter content includes the crude ingredients, which are supplemented by another ingredient that is not specified. I also conclude that the dry matter content indicates that water is present, because none of the percentages for dry matter are 100%. First, remove water content and recalculate percentages: to do this we divide each percentage of the components by the percentage of dry matter. The result is expressed as a percentage. This enables us to mix the feed on a dry basis.  The adjusted percentages are: CORN: CP: 9.4909%, CF: 2.3812%, CFB: 1.1339% RICE: CP: 13.4427%, CF: 13.7776%, CFB: 8.3150% SBM: CP: 48.0682%, CF: 6.3636%, CFB: 5.1136% WHEAT: CP: 18.5698%, CF: 5.1903%, CFB: 8.5352% COPRA: CP: 20.7792%, CF: 13.2035%, CFB: 12.4459% We also have to adjust the percentages on the required feed: CP: 18.1818%, CF: 4.5455%, CFB: 13.6364% The last letter of each ingredient will be used to signify the fraction of that ingredient needed to make up the required feed (N, E, M, T, A). So we can write: CP REQUIREMENT: 9.4909N+13.4427E+48.0682M+18.5698T+20.7792A=18.1818 CF REQUIREMENT: 2.3812N+13.7776E+6.3636M+5.1903T+13.2035A=4.5455 CFB REQUIREMENT: 1.1339N+8.3150E+5.1136M+8.5352T+12.4459A=13.6364 N+E+M+T+A=1 (The ingredient fractions must add up to 1.) [To show these equations are valid, let's move away from percentages and consider actual amounts in the dry mix. For example, n grams of corn contains 9.4909n/100 grams (0.094909n grams) of crude protein. Similarly for the other ingredients: e grams of rice contains 13.4427e/100 grams (0.134427e grams) of crude protein. When we add together the crude protein contributions for all the ingredients we get the required amount, 18.1818x/100 (0.1818x grams), where x grams is the weight of the dry mix=n+e+m+t+a. So we can multiply through by 100 to leave the percentage numbers; if we divide through by x we get n/x, e/x, etc., and we can replace these by N, E, etc., where these are fractions of the total feed: N=n/x, E=e/x, ... X=x/x=1.] There are five variables but only four equations. The amount of copra is A=1-(N+E+M+T), so the equations can be reduced to omit A: CP REQUIREMENT: 11.2883N+7.3366E-27.2890M+2.2094T=2.5974 CF REQUIREMENT: 10.8222N-0.5741E+6.8398M+8.0132T=8.6580 CFB REQUIREMENT: 11.3120N+4.1308E+7.3323M+3.9107T=-1.1905 This last equation is suspicious because all the values on the left are positive but the value on the right is negative. Therefore at least one assumption in the logic is false. To resolve this difficulty we have to change the requirements to include inequalities. For example, the mix must contain at least a certain amount of an essential ingredient, or no more than a certain amount, as well as exactly a certain amount. We have no guide in the question to make any assumptions. Nevertheless, we'll continue with the solution and see where it leads us. The plan now is to treat T as an independent variable or constant, and to find each of N, E and M in terms of T. T is therefore a parameter from which the other fractions can be derived. We can eliminate M from CP and CF: 6.8398(11.2883N+7.3365E-27.2890M+2.2094T) + 27.2890(10.8223N-0.5714E+6.8398M+8.0132T)=6.8398*2.5974+27.2890*8.6580 77.2104N+50.1808E+15.1122T+295.3277N-15.6668E+218.6706T=17.7658+236.2681 372.5381N+34.5140E+233.7828T=254.0338 And we can similarly eliminate M from CF and CFB: 7.3323(10.8223N-0.5714E+6.8399M+8.0132T) - 6.8398(11.3120N+4.1308E+7.3323M+3.9107T)=7.3323*8.6580+6.8398*1.1905 79.3514N-4.2095E+58.7544T-77.3719N -28.2542E-26.7486T=63.4827+8.1427 1.9795N-32.4637E+32.0059T=71.6253. We now have two equations involving N, E and T, and we can eliminate E between them and so find N in terms of T: 32.4637(372.5381N+34.5140E+233.7828T) + 34.5140(1.9795N-32.4637E+32.0059T)=32.4637*254.0338+34.5140*71.6253, 12162.2997N+8694.1124T=10718.9626, 12162.2997N=10718.9626-8694.1124T, N=0.8813-0.7148T. If we substitute this value of N in 1.9795N-32.4637E+32.0059T=71.6253, we can find E in terms of T: 1.9795(0.8813-0.7148T)-32.4637E+32.0059T=71.6253; 1.7445-1.4149T-32.4637E+32.0059T=71.6253 -32.4637E+30.5910T=69.8808; E=(30.5910T-69.8808)/32.4637, E=0.9423T-2.1526. Substituting for E and N we can find M then A: using the CP requirement equation: 11.2883N+7.3366E-27.2890M+2.2094T=2.5974 11.2883(0.8813-0.7148T)+7.3366(0.9423T-2.1526)-27.2890M+2.2094T=2.5974, 9.9484-8.0689T+6.9133T-15.7928-27.2890M+2.2094T=2.5974, 1.0538T-27.2890M=8.4418, M=(1.0538T-8.4418)/27.2890=0.0386T-0.3093. We now have N, E and M in terms of T. We can find A from A=1-(N+E+M+T): A=1-(0.8813-0.7148T+0.9423T-2.1526+0.0386T-0.3093+T)=2.5806-1.2661T. All the ingredients have now been found in terms of T (wheat). These values are based on equality in the ingredient equations.  Summary N=0.8813-0.7148T, E=0.9423T-2.1526, M=0.0386T-0.3093, A=2.5806-1.2661T. These are all supposed to be positive fractions<1, and clearly they are not! This implies that the exact amounts required in the mix cannot be achieved. I've checked the arithmetic and logic fully and I can find no errors, so it does not appear to be possible to mix the desired quantities of the essential ingredients. It may be possible to get the right quantity of one particular ingredient at the expense of the others, or with an excess of at least one of the other ingredients. An excess could be as undesirable as a deficiency.
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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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Find the y-intercept of the line that goes through the points (5,175)&(8,190).

All lines, or linear functions, can be expressed as y=ax+b, where a is the slope or gradient and b the y intercept. To find out what the linear function is we use the values for x and y of two points on the line. So we have x=5 and y=175,  and x=8 and y=190. We can find a and b by solving two simultaneous equations: 175=5a+b and 190=8a+b. By subtracting one equation from the other we eliminate b and then we can find a: 15=3a so a=5.  Now substitute a=5 in one of the equations: 175=25+b, 150=b, so b=150. Check the other equation still balances: 190=40+150, which is true. So the equation is y=5x+150, and the y intercept is 150 (when x=0). Another way to solve is to find the gradient, a, by dividing the difference of the y values of the two points by the difference between the x values: a=(190-175)/(8-5)=15/3=5. We find the intercept using the equation b=y-ax and we simply substitute the coordinates of one point for x and y and put in the value of a we just found.
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What's a function of function?

If you just mean the purpose of a function then see immediately below. If you mean f of g where f and g are two functions, go to the end. The purpose of a function is like giving someone a to-do list. A function (normally shown as f(x)= meaning "function of x", or y=) will usually contain one variable, usually represented by x. Some functions may have more than one variable. If x isn't the variable, it will usually be a different letter like t or a or any letter. Let's say it is x. This variable will appear in a formula with numbers. The formula is the function and it contains instructions in symbols telling you what to do with the variable. For example, multiply the variable by 2 then add 3 and divide the result by 4. This would be written f(x)=(2x+3)/4 or y=(2x+3)/4. The equals sign means that the function is defined as the expression on the right. Just like someone might say to you, think of a number (that's x), but don't tell me what it is, then double it and add 3 and divide the result by 4. Functions can be plotted on a graph. The idea of this is that on the horizontal axis (x axis) you have markers for 0, 1, 2, etc. On the vertical axis (y or f(x)) you also have markers. The graph is usually a continuous line, and every point on the line is made by putting a different value for x and marking the point (on a rectangular grid f(x) units vertically and x units horizontally) for the result of working out the value of the function for different values of x. The points make up the graph. You can use the graph to find the value of x for any value of the function, and the value of the function for any value of x, provided the graph extends far enough. This is just a brief example of the purpose of a function. A practical example is the conversion of temperature in degrees Fahrenheit (F) to degrees Celsius: C=5(F-32)/9. This function tells you what do with the value of the temperature. You could plot this as a straight line graph and you can read off degrees C for any temperature in degrees F, and vice versa. Another example is d=30t where d=distance, speed=30mph, t=time. The function is 30t, and from it you can work out the distance a car moves in a particular time t when its speed is 30mph. Function of a function: this simply means you work out the value of applying one function and then feed this answer as the variable into the other function. An example could be that one function is used to find out how many miles a car travelling at an average speed of 30mph in a given time. The second function could be to find out how much fuel is used to travel that distance. So g(t)=30t is the first function. The second function is h(d)=d/24 where fuel consumption is 24 miles per gallon. h of g is h(g(t))=30t/24=5t/4.
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(-4,-2) (0,0) and (2,7) (k,5) what is k

(-4,-2) (0,0) and (2,7) (k,5) what is k i have to find the value of k, so that the two lines will be parallel. The first observation to make is that parallel lines have the same slope. We can determine the slope of the first line and use that to find the unknown x value for the second line. Let's give identities to the points so we can refer to them by name. P1 is (x1, y1)   (-4,-2) P2 is (x2, y2)   (0, 0) P3 is (x3, y3)   (2, 7) P4 is (x4, y4)   (k, 5) The x and y designations are so you can follow along with the general equation to calculate the slope, m. 1) m = (y2 - y1) / (x2 - y2) 2) m = (0 - (-2)) / (0 - (-4)) 3) m = (0 + 2) / (0 + 4) 4) m = 2 / 4 = 1/2 Now that we know the slope of the two lines, we can use the second set of points to find the value of k. 5) m = (y4 - y3) / (x4 - x3)   Of couse, x4 is k. 6) m = (5 - 7) / (k - 2) We know that m is 1/2, so we will now substitute that into the equation. 1/2 = (5 - 7) / (k - 2) 1/2 = -2 / (k - 2) We multiply both sides by (k - 2) 1/2 * (k - 2) = (-2 / (k - 2)) * (k - 2) 1/2 * (k - 2) = -2 Multiply both sides by 2 1/2 * (k - 2) * 2 = -2 * 2 (k - 2) = -4 Add 2 to both sides (k - 2) + 2 = -4 + 2 k = -2 Substitute that value into equation 6, for the slope. m = (5 - 7) / (k - 2) m = (5 - 7) / (-2 - 2) m = -2 / -4 m = 1/2 When k is -2, the second line segment has the same slope as the first line segment, therefore, the two lines are parallel.
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Flying bearing 325 degrees 800km,turns course 235 degrees flies 950km - find bearing from A.

1. Wendy leaves airport A, flying on a bearing of 325 degrees for 800km. She then turns on a course of 235 degrees and flies for 950km. Find Wendy's bearing from A. frown 2. Wendy decides to turn and fly straight back to A at a speed of 450 km/h. How long will it take her? There are two coordinate systems used to solve this problem. The first system, used in aeronautics, is a system in which angles are measured clockwise from a line that runs from south to north. The second, used to plot graphs, measures angles counter-clockwise from the +X axis, which runs from left to right. It is necessary to convert the angles stated in the problem to equivalent angles in the rectangular coordinate system. Bearing 325 degrees is 35 degrees to the left of north. North converts to the positive Y axis, so the angle we want is the complementary angle measured up from the negative X axis. 90 - 35 = 55 If we could show a graph, we would see a line extending up to the left, from the origin at (0,0) to a point 800Km (scaled for the graph) from the origin. What we want to know is the X and Y coordinates of that point. By dropping a perpendicular line down to the X axis, we form a right-triangle, with the flight path forming the hypotenuse. We'll call the 55 degree angle at the origin angle A, and the angle at the far end of the flight path angle B. Keep in mind that angle B is the complement of angle A, so it is 35 degrees. Because we will be working with more than one triangle, let's make sure we can differentiate between the x and y sides of the triangles by including numbers with the tags. Side y1 is opposite the 55 degree angle, so... y1 / 800 = sin 55 y1 = (sin 55) * 800 y1 = 0.8191 * 800 = 655.32    We'll round that down, to 655 Km Side x1 is opposite the 35 degree angle at the top, so... x1 / 800 = sin 35 x1 = (sin 35) * 800 x1 = 0.5736 * 800 = 458.86     We'll round that up, to 459 Km At this point, the aircraft turns to a heading of 235 degrees. Due west is 270 degrees, so the new heading is 35 degrees south of a line running right to left, which is the negative X axis. Temporarily, we move the origin of the rectangular coordinate system to the point where the turn was made, and proceed as before. We draw a line 950Km down to the left, at a 35 degree angle. From the far endpoint of that line, we drop a perpendicular line to the -X axis ("drop" is the term even though we can see that the axis is above the flight path). Side y2 is opposite this triangle's 35 degree angle at the adjusted origin, so... y2 / 950 = sin 35 y2 = (sin 35) * 950 y2 = 0.5736 * 950 = 544.89    We'll round that up, to 545 Km Side x2 is opposite this triangle's 55 degree angle, so... x2 / 950 = sin 55 x2 = (sin 55) * 950 x2 = 0.8191 * 950 = 778.19    We'll round that down, to 778 Km The problem asks for the bearing to that second endpoint from the beginning point, which is where we set the first origin. We now draw our third triangle with a hypotenuse from the origin to the second endpoint and its own x and y legs. Because the second flight continued going further out on the negative X axis, we can add the two x values we calculated above. x3 = x1 + x2 = 459 Km + 778 Km = 1237 Km The first leg of the flight was in a northerly direction, but the second leg was in a southerly direction, meaning that the final endpoint was closer to our -X axis. For that reason, it is necessary to subtract the second y value from the first y value to obtain the y coordinate for the triangle we are constructing. y3 = y1 - y2 = 655 Km - 545 Km = 110 Km Using x3 and y3, we can determine the angle (let's call it angle D) at the origin by finding the tangent. tan D = y3 / x3 = 110 / 1237 = 0.0889 Feeding that value into the inverse tangent function, we find the angle that it defines. tan^-1 0.0889 = 5.08 degrees The angle we found is based on the rectangular coordinate system. We need to convert that to the corresponding bearing that was asked for in the problem. We know that the second endpoint is still to the left of the origin. The -X axis represents due west, or 270 degrees. We know that the endpoint is above the -X axis, so we must increment the bearing by the size of the angle we calculated. Bearing = 270 + 5.08    approximately 275 degrees The second part of the problem asks how long it will take Wendy (the pilot) to fly straight back to the origin. Distance = sqrt (x3^2 + y3^2) = sqrt (778^2 + 1237^2) = sqrt (605284 + 1530169)              = sqrt (2135453) = 1461.319    We'll round that one, too   1461 Km Wendy will fly 1461 Km at 450 Km/hr. How long will that take? t = d / s = 1461Km / (450Km/hr) = 3.25 hours   << 3hrs 15 mins
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Max/min, doman, range, end behavior, interval of increase and decrease

y = 2x^2 + 3x - 2 = (2x^2 + 3x) - 2 = 2(x^2 + (3/2)x) - 2 = 2(x + (3/4))^2 - 9/8 - 2 = 2(x + (3/4))^2 - 25/8 Notice that 2(x + (3/4))^2 >= 0 for all real values of x. Thus, maximum value is infinite and minimum value is -25/8. domain is all real numbers, whereas range is [-25/8 , infinity) As x tends to infinity, y also tends to infinity. As x tends to negative infinity, y tends to infinity. The interval of increase will be (-25/8 , infinity) The interval of decrease will be (-infinity , -25/8) How do we know this? The answer is by looking at the end behaviour and the range.
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