Select the approximate values of x that are solutions to f(x) =0 where f(x)=-8x^2+4x+5?
This doesn't factor so use quadratic equation x =( -b ± √(b² - 4ac) )/ (2a) x =( -4 ± √(16 + 160) ) / (-16) x ≈ -0.579 x ≈ 1.079 m10162014
Select the approximate values of x that are solutions to f(x)=0 Where f(x)= -8x^2+7x+4?
You can use the intermediate value theorem to check the answers in the original equation, since synthetic division would take a while and wouldn't require a calculator. f(0) = 4 f(-1) = -8 - 7 + 4 = -11 There exists a zero between 0 and -1. Choices 1 and 3 f(1) = -8 + 7 + 4 = 3 f(2) = -32 + 14 + 4 = -14 There exist a zero between 1 and 2. Choice 1 There choice 1. {-0.39, 1.27} is the most likely answer.
Select the approximate values of x that are solutions to f(x) = 0, where...?
I hope you have a TI or any type of scientific calculator! Firstly the question is multiple choice so you can do it without much effort. Place any of the numbers as X in the equation. So in your calculator or by working it out, put any of the numbers (its better to choose positive numbers) like 7, Than replace with X, f (x)=8×(7)^2 + 7×7 + 2 = 443 Since f (x) is not 0 this isn't the right choice, so If one number in the pair is wrong, move to the next. Remember, they wanted an APPROXIMATE VALUE. So the value that comes nearest to Zero. Using a calculator I found out, using the values in C f (x)=0.75 so C is the nearest ( the approximate answer) to zero so C is the answer. :) Feel free to ask me if you are still confused. ;)
Select the approximate values of x that are solutions to f(x) = 0, where f(x) = -8x2 + 5x + 5.?
f(x) = -8x^2 + 5x + 5 f(1) = 2 > 0 f(2) = -1 < 0 So, f(x) = 0 somewhere on the interval (1, 2). f(1) - f(2) = 2 - (-1) = 3. So, 2/3 of the weight is applied to 2 and 1/3 of the weight is applied to 1. Therefore, x = 1/3 (1) + 2/3 (2) = 5/3. Similarly, f(0) = 5 > 0, and f(-1) = -8. So, another root lies between (-1, 0) -8 - 5 = -13. So, 8/13 weight is applied to 0 and 5/13 weight is applied to -1. x = 8/13 * 0 + 5/13 (-1) = -5/13
Select the approximate values of x?
f(x) = -6x2 + 4x + 3 and f(x) = 0 so -6x2 + 4x + 3 = 0 use quadratic formula a = -6 b = 4 c = 3 x = -b ± root(b² - 4ac) / 2a x = -4 ± root(88) / -12 x = -0.45 ----- or x = 1.12 so the answer is {–0.45, 1.12} Hope that helps! .
Approximate Value of sqrt{7}?
(1.) Write 7 as an extended decimal with pairs of zeroes. .........7 . 00 00 00 00 (for however far you want to go) (2.) Find the root of the largest square ≤ 7 ... {It is 2} ... put that above your 7 ... put the square below it .........2 .........7 . 00 00 00 00 .........4 ------------------- (3.) subtract and include the next pair of zeroes .........3 00 ........... (4.) Your partial answer { 2 so far }. Double it and tack on a zero { 40 } ................................... divide your "remainder" by that { gives you 7.5 } Use that digit in place of ................................... the zero and multiply by that digit { 47 × 7 = 329 } If that exceeds your remainder, ................................... decrement your digit and try again { 46 × 6 = 276 } (And again if necessary, etc) (5.) Put that digit in your "root" .........2 . 6 .........7 . 00 00 00 00 .........4 ------------------- .........3 00 46......2 76 -------------------------- (6.) essentially, go back to step (3.) and continue as long as you can stand it .............24 00 520.......{ 524 × 4 = 2096, so that works } The whole thing will look like this: .........2 . 6....4...5...7 .........7 . 00 00 00 00 .........4 ------------------- .........3 00 46......2 76 -------------------------- (6.) essentially, go back to step (3.) and continue as long as you can stand it .............24 00 524........20 96 ---------------------- ...............3 04 00 5285........2 64 25 ---------------------------- ..................39 75 00 52907.........37 03 49 --------------------------------- ....................2 71 51 And there is your root, to four places: 2.6457 { The real root is 2.645751311... } Incidentally, (2.6457)² is less than 7 by 0.00027151, the same as your final remainder. This works for any number, just arrange the number in pairs from the decimal point. For √(123.456) start with: ...01 23 . 45 60 (00 00 00 etc.)
Calculate the [H+] in a solution that is 0.10 M in NaF and 0.20 in HF. (Ka = 7.2 x 10-4)?
A buffer solution may be acidic or basic. Your question involves the acidic buffer. An acidic buffer solution contains a weak acid and its salt. In your question the weak acid is hydrofluoric acid; HF .... Ka = 7.2x10^-4 and its salt is sodium fluoride (NaF) As these formulas imply hyrdofluoric acid is a monoprotic acid. Concentrations of the substances; [HF] = 0.20 M [NaF] = 0.10 M (a) Dissociations: Weak acid ionizes only a small extent. HF(aq) <-----> H+(aq) + F-(aq) 0.20 - x M ...... x M ........ x M Ionic soluble salt ionizes completely. NaF(aq) -----> Na+(aq) + F-(aq) 0.10 M ........... 0.10 M ..... 0.10 M F- is the common ion, but the concentration of F- ion produced by the dissociation of the salt (NaF) will be very large compared to the concentration of F- ion produced by the dissociation of the acid, and hence x is neglected. Ka = [H+][F-] / [HF] This expression in some cases simplified as (since x is neglected) Ka = [H+][SALT] / [ACID] Substituting the values; 7.2x10^-4 = [H+] (0.10) / (0.20) [H+] = (7.2x10^-4 x 0.20) / 0.10 = 1.44x10^-3 Answer: C
Math Help for my College Final for Algebra MAT/117 .... 2 Algebra math questions I need answers for thank you !!!?
1. Select the approximate values of x that are solutions to f(x) = 0, where f(x) = -5x2 + 4x + 7. A. {–0.71, 0.57} B. {-5, 4} C. {–0.80, –1.40} D {–0.85, 1.65} 2. Select the approximate values of x that are solutions to f(x) = 0, where f(x) = -8x2 + 8x + 3. A. {1.29, –0.29} B. {-8, 8} C. {–1.00, –0.38} D. {–2.67, 2.67}