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

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

amanda mchunu

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### Pre-AP Pre-Cal semester exam Flashcards | Quizlet

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### 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 (

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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

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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

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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,

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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

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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 no

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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 =

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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

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### 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

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