recall that x^2-y^2 = (x-y)(x+y) so, substitute in x = a-b and y=c.
y = 3x-1 is probably the simplest, but it could also be y = 2x+7 y = 23 y = (1+∛x)^3 - 4 or lots of other possibilities. There are infinitely many relations which will take 8 to 23.
given the three sides, with c being the largest, you know that the triangle is a right triangle if c^2 = a^2+b^2 if c^2 < a^2+b^2, then it's acute, and obtuse otherwise. So, start checking.
(5+4-1)*3 = 24 (4-3+1)*12 = 24 (9-7)*(10+2) = 24
nope. You had to add 37x to both sides, so it should be y = 37x + 82
senior problem solving
x days from today, the number of ants n, is n = 350 * 2^(x/20) Now just plug in your values for n.
y" = 5cos(t) so, when is that zero?
Stop the Madness!
I think part of the problem is that when a new question is posted, there is no list of related questions available to check. That only appears later when people open up the posting. And the search engine does not appear to be very good at providing relevant matches. And, most ...
same as y": -cosx
I meant 336.7°
Assuming the normal 0° at North, and ignoring the misuse of the word "bearing", I'd say that the resultant heading of the plane is 246.7°
not sure about your notation, but such a reflection takes (x,y) -> (-x,y)
B Just add the components for X and Y
The hands of a clock in some tower are 4.5m and 2m in length. How fast is the distance between the tips of the hands changing at 9:00 at time t hours after 12:00, (at t=0) the minute hand is at 4.5 sin(2pi*t) the hour hand is at 2.0sin(2pi*t/12) at 9:00, the distance d is d^2 ...
Ahh. I see that I was interpreting 243^3/5 as (243^3)/5
No one is bothered by the fact that 5 does not divide powers of 2 and 3?
good except for this step: ln (243^3/5 *32^4/5) should be 1/5 ln (243^3 * 32^4)
If the balloon is between the stations h/x = tan45° h/(50-x) = tan60° h/tan45° = 50 - h/tan60° h = 50 - h/√3 h = 50/(1+1/√3) = 31.7 There is another answer if both stations are on the same side of the balloon.
C(x) = 2.00 for 0 < x <= 1 2.00 + .20(10x) for 1 < x < 2 since there are 10 charging units per mile.
First you need to figure how much helium is the balloons. Each balloon's lift is reduced by the weight of the helium inside. Once you know how much each balloon weighs, subtract that from the lifting power of the displaced air, which is 1.225kg * .01716 * 9.8N/kg, since ea...
this a line with slope -1/2 which goes through (0,-3) I think that should help some, eh?
Since velocity is the derivative of position, v(t) = -32t + 160 now just solve -32t+160 = 32 -32t+128 = 0 t = 4
yellow cards: 10 of 30 total so, P(y,y) = 10/30 * 9/29 = 3/29
r + .75(s-r) Just subtract r from s and add 3/4 of that to r.
If you mean (-8)^2 + (17 - 9)*4 + 7 = 64 + 8*4 + 7 = 103
yes. The reason the law of sines can give two triangles is because sin(x) is positive all the way from 0 to 180. cos(x) becomes negative for x>90, so the formula takes that into account, always leaving only one possible answer. I mean, think about it geometrically. If you k...
as Galileo showed, it is the same for all objects (ignoring air resistance). Since F = ma, and a 20kg object weighs twice as much as a 10kg object, 2F = (2m)a but a remains the same.
-1 13/14 = -1.9286 -1.9 = -1.9000 -1 9/11 = -1.8181 -1.95 = -1.9500 Now just order them, recalling that for negative numbers, greater means closer to zero.
But -3.363636 < -3.333333
x^2 + 4x + 4 = 0 (x+2)(x+2) = 0 the two solutions are -2 and -2, so their product is 4 In fact, for any quadratic ax^2 + bx + c = 0 the product of the solutions is c/a, so you don't even need to find what the roots are.
2√((8-2)^2+(3+1)^2) = 14.4
Ms. Sue please help
Looks like you need to review how to find the slope of a line between two points: change.in.y divided by change.in.x and the point-slope form of a line. point (h,k) with slope m is y-k = m(x-h)
Ms. Sue please help
(-8, -4) and B (1, -5). slope is (-5+4)/(1+8) = -1/9 You must have had either +8 or -1 for x. #2) FG has slope (-5-7)/(-4-3) = 12/7 HI has slope (6-0)/(4+5) = 2/3 neither #3) GH has slope 4 you want slope -1/4, so y-3 = -1/4 (x-2) y = -1/4 x + 7/2 (D)
Math please ASAP
what about it? If both points have the same y-coordinate, the line is horizontal. Does not matter what the x-coordinate is. ALL points on the horizontal line y=3 are of the form (x,3) no matter what x is.
Math please ASAP
Since both have y=3, the slope is zero
I guess they're just making sure you understand that all whole numbers are also integers.
ummm, 52 52 IS a whole number typo?
5.5a + 4(200-a) = 980 solve for a to get # adults tickets subtract that from 200 to get # child tickets
recall that sec^2 = 1+tan^2, so we have tan^2(x) + 3tan(x) - 4 = 0 (tan(x)+4)(tan(x)-1) = 0 tan(x) = 1 or -4 So, there will be 4 solutions, one in each quadrant.
differences: 5,10,15,20,25 so, the next difference ought to be 30, making the next term 111
math - great minds
what can I say?
sin 12πt has period 2π/12π = 1/6 cos 18πt has period 2π/18π = 1/9 So, the sum has period LCM(1/6,1/9) = 1/3 You can see this at http://www.wolframalpha.com/input/?i=10+sin%2812+pi+t%29%2B4+cos%2818+pi+t%29+
The room is 15x15 feet, making 225 sq ft. Each gallon covers 45 sq ft 5*45 = 225, so it will take 5 gallons. Or, as we solved it, 225/45 = 5
15*15/45 = 5
gotta work with moles. How many moles H2SO3 in 36.86g? Since each mole of H2SO3 requires 1 mole of SO2, you need the same number of moles. So, convert back to g of SO2
not sure just what you're after, but current in a conductor can cause it to heat, changing the resistance. Ohm's Law says I = E/R, and assumes a constant value for R. Heating may alter this, though at any instant the relation still holds, if the changing R is taken int...
(x,y) -> (x+8,y-2)
(A) is just PQR (B) is PQR reflected in the line y=0 (C) is PQR rotated 180° about (0,0) or, reflected twice, across both axes (D) is PQR reflected in the line x=0 Not sure what your notation is supposed to indicate.
well shucks -- us farm boys know that 1 sq mile is 640 acres.
so, we need to divide each tier into 1" square pieces. we need a^2+b^2+c^2 >= 300 6^2 = 36 8^2 = 64 10^2 = 100 12^2 = 144 14^2 = 196 16^2 = 256 a b c sum 6 8 10 200 6 8 12 244 6 8 14 296 6 8 16 356 8 10 12 308 That's it. Any other combination will be way over 308 s...
if you want integer dimensions, then since 28 = 2^2 x 7 the mold must be 2x2x7 or 1x1x28
Algebra - incomplete
unless you show the inequality, don't expect any help.
After winning 8 games (of 10) their % rose from 60% to 65%. So, .60n + 8 = .65(n+10) n = 30 So, they played 40 games altogether. check: 60% of 30 = 18 18+8=26 is 65% of 40
well 63/7 = 9 49/7 = 7, so 9.07
-1 - 1/√3
if the distance AP = x, the total distance flown is d = √(1+x^2) + (2-x) where x<2. If flying over land requires 1 unit of energy, then the energy cost is c = 10/9 √(1+x^2) + 1(2-x) Now just find the minimum of c.
I assume that C is the point on shore closest to B. So, if the pipeline leaves the shore at point P, between A and C, and the distance PC = x, the length p of the pipeline is p(x) = (7-x) + √(x^2+36) now we see that the cost c of the pipeline is c(x) = 400(7-x) + 500W...
6.0 N = (0.00,6.00) 4.0 NE = (2.83,2.83) 2.0 N25E = (0.85,1.81) 4.0 E = (4.00,0.00) Now just add 'em up and convert back to distance and heading
what figure has 3 lateral faces that are rectangales
I'd say yes, since a rectangle and a triangle can be combined to make a trapezoid. But not always; the triangle must meet certain conditions.
CAN A TRIANGULAR PRISM PLUS A RECTANGULAR PRISM MALKE A TRAPAZOIDAL PRISIM
can a trapezoiodal prism equal a triangular prism and rectangular prism combined
the solution set to an inequality is not just a number, but a whole range of numbers which satisfy the inequality. Any x between 2 and 4 is a solution to this problem, as well as in the other intervals. Graphed on the number line, we have http://www.wolframalpha.com/input/?i=s...
you need to solve two inequalities: x(40-2x)(25-2x) > 1512 x(40-2x)(25-2x) < 2176 x(40-2x)(25-2x) = 4x^3 - 130x^2 + 1000x So, you have to solve 4x^3 - 130x^2 + 1000x - 1512 > 0 4x^3 - 130x^2 + 1000x - 2176 < 0 a little synthetic division shows that the first has a ...
Calculus - good catch bob
Dang - forgot the restriction on the domain.
the number of apples is yield/tree * # trees. With x trees, yield per tree is 200 - 5(x-50) for x > 50 So, total crop is c(x) = x(200-5(x-50)) = x(450-5x) = 450x - 5x^2 c'(x) = 450-10x c' = 0 at x=45 So, the max yield is achieved with 45 trees
Grade 12 Calculus
if the sheet is x by y, and is rolled along the y axis, 2x+2y = 100 v = pi r^2 y where 2pi r = x, or r = x/(2pi), so v = pi (x/(2pi))^2 (100-2x)/2 = x^2(50-x)/(4pi) dv/dx = x(100-3x)/(4pi) dv/dx=0 when x = 100/3 at that point, max v is pi*(50/3)^3
What figure has exactly 4 lateral faces that are triangles?
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)