Number of results: 6,971
A shopping center plot of land is in the shape of a parallelogram. Give two other facts about the plot or roads through it. I do not understand can you please explain? I know that A quadrilateral is a parallelogram if and only if it has two pairs of parallel sides. A ...
Friday, April 9, 2010 at 7:55pm by purple
wolframalpha . com can. Just enter plot (x+2)^2/4+(y+2)^2/15=1 Hyperbola with center at (-2,-2), eccentricity = √15/2
Wednesday, April 18, 2012 at 10:10am by Steve
5x-4y = 12 5x-12 = 4y 5/4 x - 3 = y None of the choices matches. Typo somewhere? For the graph, go to wolframalpha.com and enter plot 5x – 4y = 12
Tuesday, October 16, 2012 at 1:36pm by Steve
real-life step function
if you knew that gas went up 2 cents every week, you might use something like p = 3.299 + 2⌊x⌋ where x is the number of weeks and p is the price, which started at 3.299 To see some examples, you can go to wolframalpha.com and enter plot 3.299 + 2⌊x⌋ you...
Tuesday, October 16, 2012 at 10:16pm by Steve
to find dy/dx, 2x + y + xy' + 2yy' = 0 y' = -(2x+y)/(x+2y) y'=0 where 2x+y=0 x^2 + x(-2x) + (-2x)^2 = 7 x^2 - 2x^2 + 4x^2 = 7 3x^2 = 7 x = √(7/3) y = -2√*7/3) interchange the variables for dx/dy naturally, when dy/dx=0, dx/dy is undefined, and vice-versa. To see ...
Tuesday, October 23, 2012 at 2:47pm by Steve
hmm. let's see what we have, in more standard notation: x+y <= 2 x-2y <= 8 x - y/3 >= -2/3 You have a small triangular region where the vertices are the intersections of the lines. Go to wolframalpha.com and enter plot x+y <= 2, x-2y <= 8, x - y/3 >= -2/3
Monday, October 29, 2012 at 1:13am by Steve
visit rechneronline.de/function-graphs for excellent graphs. Set the limits for x and y and let 'er fly. The only inconvenience is that you need to express y as a function of x first. For more versatile graphs, but less control over ranges, try wolframalpha.com and enter plot ...
Friday, November 9, 2012 at 10:52pm by Steve
The given boundaries do not enclose a region. Do you mean x=2? visit wolframalpha.com and enter plot y=e^x and y=2 and y=0 to see what's wrong.
Thursday, November 15, 2012 at 9:58pm by Steve
the average value over [0,8] is k=(6π-16)/8 There are two x values where this occurs f(x) = k and g(x) = k to see where these are visit wolframalpha.com and enter plot y=2(x-4)+pi , y=arccos(x/2-3), y=(6*pi-16)/8, x=0..8 f(c) = (6π-16)/8 = 2(c-4)+π c = 3 - π...
Friday, December 21, 2012 at 1:35pm by Steve
You know that pi r^2 h = 1000, so h = 1000/(pi r^2) the surface area is a = 2pi r^2 + 2pi r h = 2pi r^2 + 2pi r * 1000/(pi r^2) = 2pi r^2 + 2000/r da/dr = 4pi r - 2000/r^2 = (4pi r^3 - 2000)/r^2 Since r > 0, da/dr=0 when 4pi r^3 = 2000 r^3 = 2000/4pi = 500/pi r = 5 ∛(...
Saturday, February 23, 2013 at 11:43pm by Steve
math pre calculus
surely the intercepts are easy f(x) is a polynomial of degree 4, so for large x it just looks like x^4. Since the graph just touches the x-axis at x=-5 and 2, It comes down from the upper left, touches at x=-5, goes back up and comes back down to touch at x=2, then heads on up...
Tuesday, March 26, 2013 at 11:00pm by Steve
Find the intercepts
intercepts: what is y when x=0? what is x when y=0? graph: visit wolframalpha.com and enter plot 3x+4y=12
Tuesday, April 9, 2013 at 10:02am by Steve
so, you have v = 2π∫(y^(2/3))y dy = 2π∫y^(5/3) dy = 2π (3/7) y^(7/3) but you need to specify the limits of integration, since the curve is open-ended. visit wolframalpha.com and enter plot y^2 = x^3
Friday, April 12, 2013 at 10:39am by Steve
(f◦g)(x) = f(g(x)) f(x) = 10^x, so f(g) = 10^g = 10^(logx) = x Similarly, g◦f = x Note that 10^x and logx are inverse functions domain for all polynomials is all reals g(x) = x^2 has range y>=0 g(x) = x^2-x+1 = (x-1/2)^2 + 3/4 has range y >= 3/4 since 10^x ...
Monday, September 2, 2013 at 4:08pm by Steve
-3(x+1) <= 5x-9 visit wolframalpha.com and enter plot y = -3(x+1) , y = 5x-9 You want the region where the blue line is below the red line: x >= 3/4
Sunday, September 8, 2013 at 10:41pm by Steve
The center of the circles is not at (1,2), so your first equation is incorrect. the line joining (1,2) to the centers of the circles will be perpendicular to the tangent line. So, since the tangent line has slope 1, the normal line will have slope -1. Now we have a point and a...
Thursday, September 12, 2013 at 4:53pm by Steve
the coefficient is there. It is 1. to see the graphs, go to wolframalpha.com and enter plot y=|x+2|+4 and y=|x+5|+3 you want the area below the blue line and above the purple line.
Thursday, September 12, 2013 at 8:05pm by Steve
(x+1)(x-4) >= (x-2)^2 x^2-3x-4 >= x^2-4x+4 x >= 8 for the graphs, go to wolframalpha.com and enter plot y=(x+1)(x-4) , y=(x-2)^2, x=6..10 or, enter (x+1)(x-4)-(x-2)^2 and see where it's positive
Monday, September 23, 2013 at 2:57pm by Steve
visit wolframalpha.com and enter plot y=4x+6 , y=x+8 The easiest points to plot on a line are the intercepts, where x=0 or y=0. In this case, they would be (0,3) and (5,0) plot those two points and draw a line through them.
Monday, September 23, 2013 at 3:00pm by Steve