Can someone help me on algebra?
The superhero’s height is given by:[math]f(x) = -16x^2 + 200x[/math]From this, we note that he is starting on the ground. Also, we will ignore the superhero’s own height.Suppose we want to know whether his height can ever be 612 ft high.We set his height function equal to 612 and then see if we can solve it. We will use a discriminant to check what kind of solutions it has.[math]612 = -16x^2 + 200x[/math][math]-16x^2 + 200x - 612 = 0[/math][math]-4x^2 + 50x - 153 = 0[/math]The discriminant of this polynomial is [math]50^2 - 4*(-4)(153) = 4948[/math]. Because this discriminant is positive, this polynomial has two real solutions. This means the expression, at some time [math]x[/math], [math]f(x) = 612 ft[/math].Note:The question does not ask for the specific time when [math]f(x) = 612[/math].It asks a yes-no question, and asks you to explain.Using a discriminant, you can say—for certain—whether his height reaches 612 ft. without knowing when it happens.However, some teachers may expect a graph or even a solution.Going a step further:If one were using calculus, you can find when the superhero reaches his maximum height.For a function [math]f(x)[/math], the maximum or minimum value is obtained when [math]f’(x) = 0[/math].Where [math]f(x) = -16x^2 + 200x[/math], then[math]f' = -32x + 200[/math][math]0 = -32x + 200[/math][math]x = 200/32 = 6.25[/math]So, the maximum height is attained after 6.25 seconds.That height is [math]f(6.25) = -16(6.25)^2 + 200(6.25) = 625 ft.[/math]Because the maximum value is 625 ft., the superhero can clear a building that is 612 ft.I only include this last part to stoke your curiosity.
Can someone help me with my Algebra 2 homework.?
Ok, my teacher explains things too fast. And I do not understand how to do this particular problen / step. Question: Write and equation in slope intercept form for the line that contains the given point and is perpendicular to the given line. problem : (3,-1), 12x + 4y = 8 If you guys can give a step by step instructions to how to do it I would gladly appreciate it :]
Can someone help me with plato algebra 2?
well im doing this thing called plato and its a credit recovery and i need this in order to graduate but ive been really trying and its very difficult and i have noone to help me . its algebra 2 if someone can please leave there email to help me out or know a website i can go to for assistance please help. :,(
Can someone help me with Algebra?
When you write problems in Y!A, everything is on single lines, so you need to add parentheses to make the problem clear if it contains fractions. For problem 1, I am assuming that the equation is (2/11)x = 14 and that we can manipulate it as follows: 2x = 154 [multiplying each side by 11] x = 77 [dividing each side by 2] If, on the other hand, it was 2 / (11x) = 14 then we'd manipulate it as follows: 2 = 154x [multiplying each side by 11x] x = 2/154 = 1/77 [dividing each side by 154 and simplifying the fraction] Choose whichever answer actually fits the problem you have. Problem 2: An angle and its supplement add up to 180 degrees, so x + 19x = 180 degrees 20x = 180 degrees x = 9 degrees
Can someone help me with 10th grade algebra 2? (binomial theorem)?
I have a quiz tomorrow and do not understand a problem in the homework. Directions: Find the specified term of each binomial expansion. Problem: Fourth term of (x+2)^5 the answer in the back of the book says 80x^2 how do i go about solving this? please explain. also, we use a TI84 in this course so any help on that would also be appreciated. (i know how to access the menus and such so no need to go into details)
Can someone help me with these Algebra II problems?
1) is this written coreectly? As it is it is unsolveable. start by dividing by 3 > x^2 = -27 then get square root --however there is no square root of a negative number 2) multiply both sides by 6 > 5q^2 - 2q^2 = 432 simplify q terms > 3q^2 = 432 divide both sides by 3 > q^2 = 144 find sq root > q = 12 3) start by getting the x terms on the same side and putting the number terms on the other 5x/12 - x/6 = -1/2 - 1/4 multiply by 12 to get rid of denominators 5x - 2x = -6 - 3 simplify 3x = -9 divide by 3 x = -3 hope that helps :)
Can someone help me with these ALGEBRA 2 problems?? PLEASE?
All of these are solved in the same way. For any equation of the form ax² + bx + c = 0 the two solutions are ------------------ x = ( -b ± √ ( b² - 4ac ) ) / 2a Express the given problems in standard form; extract the constants a, b, and c; plug them in to the solution and there are your answers. Complex roots will have negative values under the square root.
Can someone help me with my algebra 2 homework ?
Before I start, the angle variable "feta" is really spelled "theta." To calculate the amplitude of a sinusoidal function, look at the coefficient (the multiplier) outside of the function, if the function is sine or cosine. For example, the amplitude of y = 3 sin (theta) is 3. This is not the case with tangent, since it has a vertical asymptote at x = pi/2. So tangent has infinite amplitude. To find the period, look at what is inside the trigonometric function (inside the parenthesis). Divide 2*pi by the coefficient inside the parenthesis, and you will get the period. Remember, if there is no number inside the parenthesis, then the coefficient is one. Thus, the period for y = 3 sin (theta) is 2 pi (coefficient is 1). Also, the period for y = 6 sin (4*theta) is 0.5 pi (2 pi / 4). To find the phase shift, you have to look inside the parenthesis again. If you are adding to the angle, you will shift to the left. If you are subtracting from the angle, you will shift the function right. So for y = sin(theta + 90 degrees) + 2, the sine function will be shifted to the left 90 degrees. The vertical shift is found outside the trigonometric function. If you are adding to the function, you will move it up. If you are subtracting, you will move it down. This makes sense since you are adding or subtracting directly to the y value. For your example y = sin(theta + 90 degrees) + 2, you will move the sine function up 2 units. When you describe a transformation, you will need to apply all of the above characteristics: amplitude, period, phase shift, and vertical translation. Here's the first example y = 3 sin (3*theta) - 1 Amplitude: Increases by factor of 3 Period: Shrinks by a factor of 3 Phase Shift: None Vertical Translation: Down 1 unit Things get really tricky when you have an equation like this: y = 3 * (sin (3*(theta-180)) + 2) Is the vertical translation up 2 units? Not so fast. Multiply out the 3 so the equation looks like this: y = 3 * sin (3*(theta-180)) + 6 Now you can see the vertical displacement is really 6. You should do the same with the stuff inside the trigonometric function.
Algebra 2 help, please?
Easier way to look at this... y=23.45x-5.02 a. Slope: 23.45 <-- it's always the number in front of the x b. Anything that has the same slope c. The new slope of the perpendicular line has to be 1 over original slope. In this case an equation could be y=(1/23.45)x + 45