3 ft = 3 X 12 36 inches 1/2 of 36 = 18 B
as you know the vertex of a parabola is at x = -b/2a, so here the minimum is at x = 60/0.2 = 300 So, figure C(300)
assuming you meant arcsin(14v), think of the triangle with leg 4v and hypotenuse 1. The other leg is √(1-16v^2), so tan arcsin(4v) = 4v/√(1-16v^2)
we want 2πr = πr^2 r = 2
translations or rotations or scaling alone are unaffected by order. Toss in a mix of those, and the order makes a big difference. Reflections in general also depend on order.
c'mon, guy. cos π/3 = 1/2, so cos 2π/3 = cos 4π/3 = -1/2 x = 2π/3 or 4π/3
sec x = -2, so cos x = -1/2 solutions in QII and QIII
s = 1/2 at^2, so a = 2s/t^2 v = at = (2s/t^2)t = 2s/t = 73.3 m/s
since the x values steadily increase, and the y-values decrease and then increase, I'd say quadratic.
oldv = pi r^2 h newv = pi (2/5 r)^2 (4h) = pi * 4/25 r^2 * 4h = 16/25 pi r^2 h = (16/25) * oldv
u <= -1 or u >= 5 draw a dot at -1 and shade to the left draw a dot at 5 and shade to the right you can see the graph at http://www.wolframalpha.com/input/?i=solve+|4u-8|+%3E%3D+12+
think of the 5-12-13 right triangle. In QIV, y is negative So, what do you think?
assuming the ribbon is tied on top, and holds on the lid, its length will be 2h+2d + 10 = 130 h+2r = 60 So, since v = pi r^2 h, v = pi r^2 (60-2r) = 60pi r^2 - 2pi r^3 dv/dr = 120pi r - 6pi r^2 = 6pir(20-r) dv/dr=0 at r=20 So, the box has radius = 20 height = 20
At 0 Hz, the capacitor has high impedance At high Hz, the inductor has high impedance The circuit's impedance reaches a minimum at the resonant frequency of the LC portion. The resistor keeps it from ever being zero, to protect the voltage source.
15/36 = 5/12
you can play around with Z table stuff here: http://davidmlane.com/hyperstat/z_table.html
max if they point in the same direction. Just add them min if they point in opposite directions. Just subtract them.
Convert fraction decimal and percent
missed 5, got 20. 20/25 = 0.80 = 80%
well, since 1 atm = 760 mm Hg, what do you think? Be careful with the units.
v = 4π/3 r^3 v' = 4π r^2 at r=50, v = 500π/3 and v' = 100π So, the tangent line at r=50 is v - 500π/3 = 100π(x-50)
don't forget your algebra I just because you're doing trig now. 2cos2θ = √3 cos2θ = √3/2 Now it's time to recall where cos(x) is positive: QI and QIV 2θ = π/6 or -π/6 Now, cos(x) has period 2π, so cos(2x) has period π,...
you have 3z and you want 1z. So, divide both sides by 3. 3z/3 = 15/3 z = 5
The pull apart at 66+82 = 150 mi/hr. So, it only takes 111/150 = 0.74 hours to get 111 miles apart. That is 44'24", so it will happen at 3:25:00 + 0:44:24 = 4:09:24 You sure do make it hard to parse your text, what with no punctuation and garbled syntax!
the cutoff frequency is r/(2pi L) = 15/(2pi*5.6 * 10^-3) = 426 Hz
That's what I get. Good work.
h/3.5 = sin 21°
h' = (x^2-2x-9)/(x-1)^2 h" = 20/(x-1)^3 Clearly there are no inflection points, since f" is never zero h' is zero at two places, since the numerator is. So, there will be two extrema, one on each side of x=1. So, one will be a max, the other a min, depending ...
x + 4xy = y^2 1 + 4y + 4xy' = 2yy' y' = (1+4y)/(2y-4x)
so, plot those points and draw the graph. Should look something like this: http://www.wolframalpha.com/input/?i=plot+y%3D2sin%28x-2pi%2F3%29+and+y%3D2sin%28x%29 Since there's that pesky 2pi/3 in there, I'd pick points where you expect a min/max/zero. Since sin(x) has t...
Each post is connected to 4 others. Since each streamer connects in both directions, you don't want to count them twice, so you have 5*4/2 = 10 streamers
each long side has 6 candles, counting the corners. with no corners to worry about, each short side only needs 2 more candles. Total is thus 6+6 + 2+2 = 16 Or, you can consider the entire perimeter as one long line, of length 32. So, it will need 16 candles to decorate it.
tan^2(5x)cos^4(5x) = 1/8 - 1/8cos(20x) since tan = sin/cos, you have sin^2(5x)cos^2(5x) = 1/8 - 1/8cos(20x) Now use the double-angle formula on the left, and the half-angle formula on the right to get 1/4 sin^2(10x) = 1/4 sin^2(10x) QED
I'd help, but can't figure out (tx At any rate, use the identity tan^1+1 = sec^2 to reduce everything to a polynomial in cos(x). Solve for cos(x) and then for the values of x.
6/10 see a pattern here?
625/1000 = 5/8 (divide top and bottom by 125) by the way, .125 = 1/8
Have you considered a tutor? A few sessions with some good person-to-person coaching may be all you need. Invest a few bucks and it may be worth it. Too bad you don't live in Austin, TX, or I'd volunteer to do the job, as I do math tutoring all the time.
sec x is negative in QII,QIII So, your angles will be between 90 and 270 degrees.
12kL * 1000L/kL * 1000mL/L = 12 million mL
math - incomplete
better provide a function.
Oh, well, why didn't you say so? I gave you the answer above. Read things carefully, ok? Stop sittin' around waiting for someone to hold your hand.
Try converting to fractions. Then you have 11/8 14/8 17/8 23/8 I think the last one should be 2 1/2, since you added 3/8 the first two times. If it really is 2 7/8, ya got me. Or, it could be that you just missed a term of 2 1/2 in between 2 1/8 and 2 7/8.
(459.45)/(1-0.51)(1+.24)(1+.17)(1-.10) = 1224.31
Give Writeacher her due, ok? Her wisdom is as dew on the lily. Or, if you're a Scottish laddie, defend her honor with your sgian dubh!
I knew that. It's not new to me.
Did you try checking to see whether the equation balances with that value? As it happens, it does.
Depends on the method. For substitution, it could be 2025a + 45(1.81-324a)/18 = =1.39 For elimination, it could be 4050a+90b = -2.78 1620a+90b = 9.05 For any method it could be to reduce the size of the coefficients: 45a+b = -1.39/45 18a+b = 18.1/18
z1 z2 = (4*3)cis(7π/6+π/3) = 12cis(3π/2)
Calc Help :(((((
Hmmm. Are you sure you're reading it right? Remember that log(1/3) = -log(3)
same as theta=45°, or y=x Actually, it's y=x rotated by 180°, but it's still the same line. Recall that tan 225° = 1 = y/x
in QIV, cosθ = 3/5 sinθ = -4/5 and you know that sin2θ = 2sinθ cosθ So, plug and chug
you're kidding, right? The definition of a square specifies that all 4 sides are equal. That's what equilateral means. In a rectangle, such as a book, a desk top, a door, etc. the sides are not all equal. So, no for rectangles.
well, you have the formula sin(u)+sin(v) = 2 sin (u+v)/2 cos (u-v)/2 So plug in u=3x and v=x
since time = distance/speed, she has been running 5.6km*1000m/km / 3.4m/s = 1647 seconds or 27:27
Math - PreCalc (12th Grade)
since sec^2 = 1+tan^2, (x/.2)^2 = (y/-.25)^2 + 1 25x^2 - 16y^2 = 1 Looks like A to me
Math - PreCalc (12th Grade)
C(7,5) = C(7,2) = 7*6 / 1*2 = 21
That would be the number of points where the denominator is zero. The denominator is (x + 2)(x − 5) so, what do you think?
Math - PreCalc (12th Grade)
or, to see the foci plotted, http://www.wolframalpha.com/input/?i=foci+of+%28y-2%29^2%2F9+-+%28x-2%29^2%2F16+%3D+1
Math - PreCalc (12th Grade)
there are lots of them, depending on the eccentricity. Recall that an hyperbola with equation y^2/a^2 - x^2/b^2 = 1 has vertices at (0,±a) and foci at (0,±c) where c = ae and e>1 is the eccentricity. Also, recall that c^2 = a^2 + b^2 Here, we have the center o...
assuming no amputees, there are 12 too many legs to be all ducks. So, 6 of the animals had 2 extra legs each. 8 ducks and 6 cows or, more algebraically, if there are d ducks and c cows, d+c = 14 2d+4c = 40 now solve for d and c.
correct 6*11 + 7*11 = (6+7)*11 = 143
the central 80% of the Z table lies within µ ± 1.282σ So, plug in your numbers You can play around with Z table stuff at this excellent site: http://davidmlane.com/hyperstat/z_table.html
Trig - sum and difference formulas
the important identity here is cos(A+B) = cosAcosB - sinAsinB So, it looks like you have cos(6x+x) = cos 7x You also know that sin(A+B) = sinAcosB + cosAsinB so, sin(90+x) = sin90cosx + cos90sinx = 1*cosx = 0*sinx = cosx Do lots of these, since they will come in handy later on.
(x+3)(x+6)-2(x+1) x(x+6) + 3(x+6) - 2(x+1) x^2+6x + 3x+18 - 2x-2 x^2 + 7x + 16
-x(x-1) - (-4)(x-1) + 5(3x)(2x+1) - 5(1)(2x+1) -x^2+x + 4x-4 + 30x^2+15x - 10x-5 29x^2 + 10x - 9
0 < -9x / x(1-x^2) x=0 is not in the domain. Excluding that, we have 0 < -9/(1-x) If the quotient is positive, 1-x < 0 x > 1
dunno. Is 6^2 - 4(-7)(3) positive? The discriminant will always tell you.
Gr 10 math - finding slopes in shapes
correct. -3/-4 = (-1)(3) / (-1)(4) = 3/4 The -1's cancel out
Assuming an initial position of zero, s(t) = 5/2 t^2 for 0<=t<1 so, at t=1, s = 5/2 Now, using the 2nd function, s(t) = 5/2 + 4t^(3/2) - log(t) solve that for s(t) = 4
use implicit differentiation: x/2 + y/8 y' = 0 y' = -4x/y
The second equation is y = 2(x^2-1)^2 so, -2(x^2-1) = 2(x^2-1)^2 so, x^2-1=0 is a solution, meaning x = ±1 Otherwise, divide by 2(x^2-1) to get -1 = (x^2-1) x = 0 So, the solutions are -1,0,1 View the graphs at http://www.wolframalpha.com/input/?i=solve+y%3D-2%28x^2-1%2...
man, ya gotta check your answer. It does not fit the first equation. y = x-3, so x + 3(x-3) = 6 4x = 15 x = 15/4 y = 3/4
If you rearrange things a bit, you get x - 6y = 2 3x - 18y = 4 which is the same as x - 6y = 2 x - 6y = 4/3 The lines are parallel, so there is no intersection, and no solution to the equations. Your answer fits the 2nd equation, but not the first.
subtract the equations to get 2x^2 - x = 1 2x^2-x-1 = 0 (2x+1)(x-1) = 0 x = -1/2 or 1 so, y^2 = 11/2 or 4
sinx = 2cosx tanx = 2 tanx > 0 in QI or QIII, so there are two values for x. Verify this at http://www.wolframalpha.com/input/?i=solve+sinx+%3D+2cosx
Math -Quadratics Help!
#1 I assume you can find the intersection of two lines. In this case, that is at (1,1/4). The line 5x+7y+3=0 has slope -5/7, so now we have a point and a slope, and the desired solution is y - 1/4 = -5/7 (x-1) #2 y=3x-5 has slope 3, so the angle it makes with the x-axis is arc...
2x + (90-x) = 96
I hope the selections include 3/16 (1±√33)
math word problems
a = b/2 b = 4c a+c = 648 so, b/2 + b/4 = 648 3/4 b = 648 b = 648 * 4/3 = 864
(84-8)/x = tan 60° Now just solve for x
probability math question
10 marbles 8 not blue what do you think?
minimize 2(xy + xy/4 + y^2/4) subject to xy(y/4) >= 231 x=5.29 y=13.22 h=3.30
Hmm. I got v = 5.29 x 13.22 x 3.30 = 230.78 a = 262.12
the reflection takes (x,y) -> (x,-y) so, g(x) = -f(x)
velocity changed by 20m/s in 10s. So that means an average acceleration of (20m/s)/(10s) = (20/10) m/s^2 = 2 m/s^2 That is, every second, the velocity increased by 2 m/s, thus going from 0 to 20 in 10 seconds.
PreCalc PLEASE HELP
a8 = a1*r^7, so r^7 = 100/(25/32) = 3200/25 an amusing little sequence, no?
PreCalc PLEASE HELP
62500*1.05^6 = 83755.98
well, you've done the words. CuCO3 -> CuO + CO2 or CuCO3(s) -> CuO(s) + CO2(g) You can put a ∆ over the arrow to show addition of heat
looks like 32 g of O2, no?
99% = µ±2.576 std
Chemistry - incomplete?
Wow. Since none of the elements in the product are included among the reactants, there's a lot of room to play around. I think there's something missing here.
generally this site works to answer specific questions. There is a wealth of online resources at your google fingertips. Or at your local used bookstore or library.
3,500,000 + 48,800 = 3,548,800
AB has direction (-3,4,1) BC has direction (0,-2,6) CD has direction (3,-4,-1) DA has direction (0,-2,6) So, AB || CD and BC || AD since the sides are parallel in pairs, they must have the same lengths in pairs (I trust you can verify this). the area is just |(b-a)x(d-a)| wher...
just substitute and evaluate g(x) = x^2-x, so g(g) = g^2-g = (x^2-x)^2 - (x^2-x) g(f) = f^2-f = (2x-1)^2 - (2x-1) f(x) = 2x-1, so f(g) = 2g-1 = 2(x^2-x)-1 f(f) = 2f-1 = 2(2x-1)-1 expand those out if you desire.
what's the trouble? Just substitute and evaluate: g(x) = x^2-1, so g(g) = g^2-1 = (x^2-1)^2-1 g(f) = f^2-1 = (1-3x)^2-1