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simplify the following to a function value of theta cos(540degrees + B)

amanda mchunu

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Find the values below for the angle theta with t... - OpenStudy


Find the values below for the angle theta with the following function value and contrait: function value: cos theta ... below for the angle theta with the following ...
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Simplify the trigonometric expression sec theta cos theta a ...


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Evaluate the 6 trigonometric functions of theta for 540 degrees


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Pre-AP Pre-Cal semester exam Flashcards | Quizlet


Pre-AP Pre-Cal semester exam. ... sin30= 1/2 and cos 30 = root 3/2, determine the following : ... 2. if sin theta = root 3/2, find the value of theta in degrees ...
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Trigonometric Equation Calculator - Symbolab


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Simplify a Trigonometric Expression - WebMath


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Suggested Questions And Answer :


simplify the following to a function value of theta cos(540degrees + B)

kosine(540+b)...540-360=180 Sine & koside go round in serkels evree 360 deg same as kosine(180+b) koside(180)=-1
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Interpolate the data set (1, 150), (3, 175), (4, 185), (6, 200), (8, 300) to estimate the amount of money Gracie may earn if she displays her items for 7 hours

Since 7 is halfway between 6 and 8, Gracie should earn an amount about halfway between 200 and 300, that is, 250 (triangular interpolation). This is the simplest interpolation, but see later. Interpolating for 2 and 5 we get 162.5 and 192.5, that is, respectively, a difference of 12.5 (162.5-150 or 175-162.5) and 7.5 (192.5-185 or 200-192.5), while 250 is a difference of 50 from 200 and 300. So the interpolated figures fluctuate. For a more sophisticated approach, we need to take the whole dataset and look for a formula that best fits. One way to do this is to fit a polynomial F(x)=ax^4+bx^3+cx^2+dx+e into the five given points. This polynomial has 5 unknown coefficients, so with 5 simultaneous equations we should be able to find them. The process can be simplified slightly by taking the lowest "x" coord and using that as the zero starting point. In this case the lowest coord is 1 (hour) so we subtract 1 from the first coord of each pair to get: (0,150), (2,175), etc. F(0)=e=150. So we have the constant 150. The next step is to subtract 150 from each of the other "y" coords so we arrive at the following set of equations: (1) F(2)=16a+8b+4c+2d=25 (2) F(3)=81a+27b+9c+3d=35 (3) F(5)=625a+125b+25c+5d=50 (4) F(7)=2401a+343b+49c+7d=150 and we already have F(0)=150=e. We can now eliminate d from (1) and (2): 2F(3)-3F(2): (162-48)a+(54-24)b+(18-12)c=70-75=-5; (5) 114a+30b+6c=-5. and we can eliminate d from (3) and (4): 5F(7)-7F(5): (12005-4375)a+(1715-875)b+(245-175)c=750-350; 7630a+840b+70c=400 which simplifies to (6) 763a+84b+7c=40 or 109a+12b+c=40/7 We can eliminate c between (5) and (6): 6(6)-(5): (654-114)a+(72-30)b=240/7+5=275/7; (7) 540a+42b=275/7 or 90a+7b=275/42. So b=(275/42-90a)/7. From (6) we have: 109a+12b+c=109a+12(275/42-90a)/7+c=40/7, so c=40/7-109a-12(275/42-90a)/7; c=40/7-550/7+(1080/7-109)a=-510/7+317a/7=(317a-510)/7. We now have b and c in terms of a. We can continue to find d in terms of a. From (1) d=(25-16a-8b-4c)/2=25-16a-8(275/42-90a)/7-4(317a-510)/7; d=25-1100/147+2040/7+(-16+720/7-1268/7)a= (3675-1100+42840)/147+(-112+720-1268)a/7; d=45415/147-660a/7. We have b, c and d in terms of a, so we can find a by substituting into an equation containing all four coefficients (but not (1), because we used it to find d). Let's pick (2) and hope we get a sensible result! 81a+27(275/42-90a)/7+9(317a-510)/7+3(45415/147-660a/7)=35. From this a=5143/2772=1.855. Therefore b=-22.919, c=78.510, d=-67.687, e=150. And F(x)=1.855x^4-22.919x^3+78.510x^2-67.687x+150. This results need to be checked before we use F to find an interpolated value. Unfortunately, this polynomial approach produces inconsistent results, and needs to be discarded. Lagrange's method seems the obvious choice, even if it is tedious to do. We have 5 x values which we'll symbolise as x0, x1, x2, x3, x4 and 5 function values f0, f1, f2, f3, f4. If the function we're looking for is f(x) then: f(x)=(x-x1)(x-x2)(x-x3)(x-x4)f0/((x0-x1)(x0-x2)(x0-x3)(x0-x4))+         (x-x0)(x-x2)(x-x3)(x-x4)f1/((x1-x0)(x1-x2)(x1-x3)(x1-x4))+         (x-x0)(x-x1)(x-x3)(x-x4)f2/((x2-x0)(x2-x1)(x2-x3)(x2-x4))+... x0=1, x1=3, x2=4, x3=6, x4=8; f0=150, f1=175, f2=185, f3=200, f4=300. We want x=7, so f(7) is given by: 4.3.1.-1.150/(-2.-3.-5.-7)+6.3.1.-1.175/(2.-1.-3.-5)+ 6.4.1.-1.185/(3.1.-2.-4)+6.4.3.-1.200/(5.3.2.-2)+ 6.4.3.1.300/(7.5.4.2) This comes to: -60/7+105-185+240+540/7=1600/7=228.57 (229 to the nearest whole number) compared with 250 from the simple interpolation. 
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Application of Derivatives =)) Help?

A. Determine where each of the following functions is increasing and where it is decreasing. 1) f(x) = x^2 - 6x + 19 f’(x) = 2x – 6 > 0 for x > 3 Function increases on x > 3 and decreases on x < 3 2) f(x) = 10x - x^2 f’(x) = 10 – 2x > 0 for x < 5 Function increases on x < 5 and decreases on x > 5 B. Determine the critical values of each of the following functions: 1) f(x) = x^2 - 16x f’(x) = 2x - 16 = 0 at x = 8 Critical value is f(8) = 8^2 - 16*8 = 64 – 128 = -64 Crit value: -64 2) f(x) = x^3 – 2 f’(x) = 3x^2 = 0 at x = 0 Critical value is f(0) = 0 - 2 = -2 Crit value: -2 C. Find all relative extreme points of each of the following functions: 1) f(x) = x^2 - 20x f’(x) = 2x – 20 = 0 at x = 10 f(10) = 10^2 – 20*10 = 100 – 200 = -100 Minimum point is at (10, -100) 2) f(x) = x^3 - 3x – 2 f’(x) = 3x^2 – 3 = 0 at x = 1 and at x = -1 f(1) = 1 – 3 – 2 = -4, f(-1) = -1 + 3 – 2 = 0 Minimum point is at (1, -4)       Maximum point is at (-1, 0) 3) f(x) = -x^3 - 3x^2 + 7 f’(x) = -3x^2 – 6x = 0 at x = 0 and at x = -2 f(0) = 0 – 0 + 7 = 7, f(-2) = -(-8) – 3(4) + 7 = 3 Minimum point is at (-2, 3)       Maximum point is at (0, 7) 4) f(x) = x^4 - 2x^2 + 3 f’(x) = 4x^3 – 4x = 0 at x = 0 and at x = 1 and at x = -1 f(0) = 0 – 0 + 3 = 3, f(1) = (1) – 2(1) + 3 = 2,  f(-1) = (1) – 2(1) + 3 = 2 Minimum points are at (-1, 2) and (1, 2)       Maximum point is at (0, 3)  
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How to use "if" & "and" functions in formula bar for calculating exact value of particular cell

This the Excel formula bar? General format IF({test statement},{value if true},{value if false}). IF statements can be nested so that more IF statements can be inserted where the values are. The AND function has the format AND({condition1},{condition2}...) and returns TRUE or FALSE. Combining these we have: =IF(AND(A1>=3,A1<=6),0.3,IF(AND(A1>=6,A1<=12),0.5,IF(AND(A1>=12,A1<=24),1,0))) as the formula in B1. EXPLANATION The three closing brackets or parentheses (")")are necessary to satisfy the "grammar" or syntax. The last one belongs to the first IF statement, the middle one to the second IF statement and the first belongs to the third IF statement. Excel will show a syntax error if any brackets are missing. A quick way of checking is to count the number of opening brackets and the number of closing brackets; they must be the same. In this case, there are 6 of each. Because AND returns TRUE or FALSE it's not necessary to follow it with "=TRUE", although you may do if you wish, but if you do it must come before the comma. The formula bar must start with "=" otherwise, Excel will assume you are writing a string of text. Excel will then help you through construction of the formula. Excel encounters "=" in the formula bar and knows a formula follows. It sees the IF statement and expects 3 parameters separated by commas. The first parameter is the AND function; the second is 0.3 and the third is another IF statement. If the AND result is TRUE, 0.3 goes on B1. Otherwise it moves on to the third parameter, the second IF statement and applies the same logic. The third parameter of the second IF statement is activated if the second AND function returns FALSE, and Excel encounters the third IF statement, applying the same logic. The third parameter of the third IF statement is zero, so B1=0 if the last AND function returns FALSE.
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Find an equation for the line that is tangent to the curve y=3x^3-3x at point (1,0).

y=f(x)=3x³-3x ··· Eq.1 Since the given curve is continuous function, Eq.1 is differetiable at every x and the first derivative f'(x), the slope (tangent) of the original curve, is found at every point (x,y) on the curve. The 1st derivative of Eq.1 is: f'(x)=9x²-3, so the slope of the tangent at x=1 is: f'(1)=9(1)²-3=6 While, the equation of a line that passes thru a point (x1,y1), and has a slope of m is: (y-y1)=m(x-x1) ··· Eq.2  (the point- slope form of linear function) Here, m=f'(x1)=6, x1=1,y1=0   Plug these values into Eq.2.   Eq.2 can be restated and simplified as follows: y-0=6(x-1) ⇒ y=6x-6 Therefore, the tangent to the given curve at point (1,0) is: y=6x-6 
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i need the answer for these questions

Part 1 Newton’s Method for Vector-Valued Functions Our system of equations is, f1(x,y,z) = 0 f2(x,y,z) = 0 f3(x,y,z) = 0 with, f1(x,y,z) = xyz – x^2 + y^2 – 1.34 f2(x,y,z) = xy –z^2 – 0.09 f3(x,y,z) = e^x + e^y + z – 0.41 we can think of (x,y,z) as a vector x and (f1,f2,f3) as a vector-valued function f. With this notation, we can write the system of equations as, f(x) = 0 i.e. we wish to find a vector x that makes the vector function f equal to the zero vector. Linear Approximation for Vector Functions In the single variable case, Newton’s method was derived by considering the linear approximation of the function f at the initial guess x0. From Calculus, the following is the linear approximation of f at x0, for vectors and vector-valued functions: f(x) ≈ f(x0) + Df(x0)(x − x0). Here Df(x0) is a 3 × 3 matrix whose entries are the various partial derivative of the components of f. Specifically,     ∂f1/ ∂x (x0) ∂f1/ ∂y (x0) ∂f1/ ∂z (x0) Df(x0) = ∂f2/ ∂x (x0) ∂f2/ ∂y (x0) ∂f2/ ∂z (x0)     ∂f3/ ∂x (x0) ∂f3/ ∂y (x0) ∂f3/ ∂z (x0)   Newton’s Method We wish to find x that makes f equal to the zero vector, so let’s choose x = x1 so that f(x0) + Df(x0)(x1 − x0) = f(x) =  0. Since Df(x0)) is a square matrix, we can solve this equation by x1 = x0 − (Df(x0))^(−1)f(x0), provided that the inverse exists. The formula is the vector equivalent of the Newton’s method formula for single variable functions. However, in practice we never use the inverse of a matrix for computations, so we cannot use this formula directly. Rather, we can do the following. First solve the equation Df(x0)∆x = −f(x0) Since Df(x0) is a known matrix and −f(x0) is a known vector, this equation is just a system of linear equations, which can be solved efficiently and accurately. Once we have the solution vector ∆x, we can obtain our improved estimate x1 by x1 = x0 + ∆x. For subsequent steps, we have the following process: • Solve Df(xi)∆x = −f(xi). • Let xi+1 = xi + ∆x
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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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Using the following values, find the approximate value of h'(x) at x=1.35

Question: Using the following values, find the approximate value of h'(x) at x=1.35. Where h(x) = f(g)(x). Here we use the chain rule. If h(x) = f(g) and g() is a function of x, then dh/dx = (df/dg).(dg/dx) To find h'(x) at x = 1.35, we need h'(x) = (df(g)/dg).(dg(x)/dx) at x = 1.35, g'(x=1.35) = ?? at x = 1.35, g(x=1.35) = 2.7 at g = 2.7, f'(g=2.7) = 3.97 We don't have information on the value(s) of g'(x) for various values of x, but we do have information on the variation of g(x) for different values of x, which shows that g(x) is a straight line function which works out as g(x) = 4x - 2.7 so, g'(x) = 4. i.e. g'(x=1.35) = 4. Now we have g'(x=1.35) = 4 and f'(g=2.7) = 3.97, So h'(x=1.35) = (df(g=2.7)/dg).(dg(x=1.35)/dx) = 3.97*4 = 15.88 Answer: 15.88
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(1, -20), (7, -20), (2, -40), (6, -40), (5, -50), (8, 10), (9, 46), (0, 7), (-1, 47), (-2, 89)

First put all the points (x,y) in order of the x value: -2 89 -1 47 0 7 1 -20 2 -40 5 -50 6 -40 7 -20 8 10 9 46   When x=0 y=7, so the constant term in the function is 7 (y intercept). There's a minimum (turning point) near x=5 because the values of y for x=2 and 6 (-40) are more positive than -50. The gradient (dy/dx) gets positively steeper as x increases after x=5, and negatively steeper between x=-2 and +2. The degree of the polynomial is even because the function is positive for larger magnitude positive and negative values of x. The zeroes are between x=0 and 1 and between x=7 and 8. The function can be split into two parts: (1) y for -2 Read More: ...

h(x)=f(g(x)) is a differentiable function

Question: Using the following values, find the approximate value of h'(x) at x=1.35. Where h(x) = f(g)(x). I'm assuming that you made a typo in your entry of the last line of data. You entered: if x=1.35, g(x)=1.40. It should be: if x=1.40, g(x)=2.9.   Here we use the chain rule. If h(x) = f(g) and g() is a function of x, then dh/dx = (df/dg).(dg/dx) To find h'(x) at x = 1.35, we need h'(x) = (df(g)/dg).(dg(x)/dx) at x = 1.35, g'(x=1.35) = ?? at x = 1.35, g(x=1.35) = 2.7 at g = 2.7, f'(g=2.7) = 3.97 We don't have information on the value(s) of g'(x) for various values of x, but we do have information on the variation of g(x) for different values of x, which shows that g(x) is a straight line function which works out as g(x) = 4x - 2.7 so, g'(x) = 4. i.e. g'(x=1.35) = 4. Now we have g'(x=1.35) = 4 and f'(g=2.7) = 3.97, So h'(x=1.35) = (df(g=2.7)/dg).(dg(x=1.35)/dx) = 3.97*4 = 15.88 Answer: 15.88
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