Tuesday

June 30, 2015

June 30, 2015

Total # Posts: 32,041

**Calculus**

it reverses direction when its velocity changes sign. That is, when v(t) = 0. So, find t when v(t) = s'(t) = 3t^2 - 18t + 24 = 0 then plug that value into s(t) and a(t)
*May 18, 2015*

**Math**

he used 15.75-10.5 = 5.25 gallons of paint in 1.5 hours. That is 3.5 gal/hr Thus, using the point-slope form of the line, if y is the amount of paint left at x hours after 12:00, then y-10.5 = -3.5(x-2) So, find y when x=0. Or, knowing that he used 3.5 gal/hr, and knowing that...
*May 18, 2015*

**math please**

when dividing numbers expressed as powers, you subtract the exponents. Think about it 10^7/10^3 = 10*10*10*10*10*10*10 --------------------------- 10*10*10 Three of the factors cancel, and you are left with 7-3 = 4 multiples of 10. That is 10^(7-3) = 10^4 So, you want (C)
*May 18, 2015*

**Math**

216 = 6*6^2 96 = 6*4^2 so, you wind up with 6√6t + 4√6t = 10√6t = √600t
*May 18, 2015*

**Algebra**

a = w(w+4) . . .
*May 18, 2015*

**MATH if work is correct**

78125 = -5^7 So, starting with 5^-2, the 10th term is -5^7 Note that the 3rd term is 1=5^0. So, you have to multiply by -5 another 7 times to get to -5^7.
*May 18, 2015*

**Math**

the full volume is 60*60*45 2/3 of that is still empty. So, divide that volume by 6 L/min (6000 cm^3/min) to get the time needed.
*May 18, 2015*

**Math**

In how many different orders can you read 4 books
*May 18, 2015*

**science**

a body of mass 5kg is thrown to a height of 100m.what is its potential energy at the maximum height.?
*May 18, 2015*

**math Help!!!!**

There are 6 ways you can draw the two marbles to include one red: RB,RY,RW,BR,YR,WR There are 4P2 = 12 ways to draw any two marbles So, P(one red) = 6/12 = 1/2
*May 18, 2015*

**math...Please help!!!!**

I think B, but the language is garbled. The SD will not change. Read up on how the SD is figured, and you should see why.
*May 18, 2015*

**Geometry**

h^2 + 7^2 = 28^2
*May 18, 2015*

**Math**

as you recall, if the axes are rotated through θ, tan 2θ = B/(A-C) so, tan 2θ = 24/7 and cosθ = 3/5 sinθ = 4/5 Now just plug those values into your rotation matrix and crank it out.
*May 18, 2015*

**Math**

tan(2θ) = -10/0 2θ = -pi/2 θ = -pi/4 x' = (1/√2)x + (1/√2)y y' = (-1/√2)x + (1/√2)y 13((1/√2)x + (1/√2)y)^2 - 10((1/√2)x + (1/√2)y)((-1/√2)x + (1/√2)y) + 13((-1/√2)x + (1/√2)y)^2 = ...
*May 18, 2015*

**grade (5) math**

If the pen has dimensions x and y, with n gates, and assuming there are posts only at the corners and on both sides of the gate(s) and that the gate(s) do not share a corner post, then using up the whole budget, (2x+2y-3n)+2(4+2n)+5n = 50 x+y+3n = 25 and, since the gates are 3...
*May 18, 2015*

**Calculus**

since h is not continuous at x=0, h'(0) is not defined.
*May 18, 2015*

**Calculus**

cannot say what the limit is. If f(x) is continuous, though, the limit is 7.
*May 18, 2015*

**Calculus**

this is true only if f(x) is continuous. Otherwise, all bets are off.
*May 18, 2015*

**Math!!! Please help!!!**

x^7 is the 5th term in the expansion, so it is 11C4 x^7 (-3)^4 = 26730x^7 The coefficients are 1 5 10 10 5 1, so (2y-3x)^5 = (2y)^5 + 5(2y)^4(-3x) + 10(2y)^3(-3x)^2 + 10(2y)^2(-3x)^3 + 5(2y)(-3x)^4 + (-3y)^5 Now just expand all those values nCr = n!/[(n-r)!r!] nC(n-r) = n!/[(n...
*May 17, 2015*

**math**

compare the sums of the two geometric sequences: (2^30 - 1)/(2-1) vs (3^20 - 1)/(3-1) There's a difference, but either one will set you up for life!!
*May 17, 2015*

**math**

not quite. The area is base times height, divided by 2. So, (B)
*May 17, 2015*

**Math**

consult your various conversion formulae. One way would be 6.58559 sin(x+0.73169) There are, of course, ways to express that using cosine and other phase shifts.
*May 17, 2015*

**Calculus**

See the related questions below. They should help. If not enough, some back and let us know how far you got.
*May 17, 2015*

**Math**

for the cards, does the way you arrange your cards make any difference? If not, it's a combination. For the actors, of course it makes a difference who plays which part. So, a permutation. But, if you just want to get four guys, it's a combination.
*May 17, 2015*

**Math**

3!
*May 17, 2015*

**math**

compare the sums of the two geometric sequences: (2^30 - 1)/(2-1) vs (3^20 - 1)/(3-1) There's a difference, but either one will set you up for life!!
*May 17, 2015*

**Maths**

after n years, the interest will be 800(0.02n)
*May 17, 2015*

**math for college readiness**

since y=kx, just find k: -30 = k*6 k = -5 ...
*May 17, 2015*

**trigonometry**

use your half-angle formula to find sin(15°) Now use your sum formula to find sin(270°+15°)
*May 17, 2015*

**Math**

one possible answer: 48
*May 17, 2015*

**math**

if the sides are in ratio r, the areas are in the ratio r^2. So, the sides are in the ratio 12:9
*May 17, 2015*

**math**

for an angle θ, the chord is 2r sin(θ/2) the arc length is rθ the area of a sector is 1/2 r^2 θ Now use that to find your answers
*May 17, 2015*

**Math**

4 * 6 - 2 * 3/2 = 21
*May 17, 2015*

**Math**

you give no dimensions of the key's rectangle. Its sides must be in the ratio of 94:50 to be similar to the court. All circles are similar to each other.
*May 17, 2015*

**Math**

Well, the standard form is (y-k)^2/b^2 - (x-h)^2/a^2 = 1 and for an hyperbola, c^2 = a^2+b^2
*May 16, 2015*

**algebra**

Well, you have the formula. What's the trouble?
*May 16, 2015*

**Maths**

f^-1(x) = (x-1)/2 (g◦f)(-2) = g(f(-2)) = g(-3) = 23 (f◦g)(x) = f(g) = 2g+1 = 2(3x^2-4)+1 = 6x^2-7
*May 16, 2015*

**Math**

I'd say either can be true, but not all shapes can be inscribed in a rectangle. That is, so that all their vertices touch the sides of the rectangle.
*May 16, 2015*

**Math**

thousands of years ago it was learned that the ratio of the circumference to the diameter of a circle is pi = 3.14159265... This is true for all circles: c = pi * d or, c = 2pi * r
*May 16, 2015*

**math**

clearly T16 = 1464-1290 = 174 S16 = 16/2 (a+174) = 1464 a = 9 T16 = 9+15d = 174 d=11 S20 = 20/2 (2*9+11*19) = 2270
*May 16, 2015*

**calculus**

find y' at the given point (h,k). That is the slope of the tangent line. And then just plug in that slope (call it m) into the point-slope form of the line: y-k = m(x-h) Don't forget your Algebra I now that your'e doing calculus. So, what do you get for y=√(x...
*May 16, 2015*

**calculus**

y = x/(x-1) = 1 + 1/(x-1) y' = -1/(x-1)^2 y'(2) = -1 so, the tangent line is y-2 = -1(x-2) see the graphs at http://www.wolframalpha.com/input/?i=ploy+y%3Dx%2F%28x-1%29%2C+y%3D-%28x-2%29%2B2
*May 16, 2015*

**math**

clearly T16 = 1464-1290 = 174 16/2 (a+174) = 1464 a = 9 T16 = 9+15d = 174 d=11 S20 = 20/2 (2*9+11*19) = 2270
*May 16, 2015*

**Math**

6x4 - (1/2)x2
*May 15, 2015*

**Math**

you have 2 complete circles of 12' diameter. Subtract from that area the area of the small circle of radius 2. Now divide the result by 110
*May 15, 2015*

**math**

oops. forgot about r. dv/dt = a dr/dt da/dt = 8πr dr/dt So, find r and a when 36π and plug them in to find da/dt
*May 15, 2015*

**math**

v = 4/3 πr^3 a = 4πr^2 dv/dt = 4πr^2 da/dt = a da/dt So, find a when v = 36π and plug it in to find da/dt.
*May 15, 2015*

**math**

you have two triangular bases, and three rectangular faces. I expect that you know how to find the area of triangles and rectangles ...
*May 15, 2015*

**Math**

you need to provide some dimensions of the court. All circles are similar to all other circles.
*May 15, 2015*

**MATH**

you have 1/2 of a circle of radius 12 and a rectangle that is 12x19 You just add up those areas.
*May 15, 2015*

**math**

If the perimeter is p, the area is A = 1/2 ap. So, since p = ns, what is the side s of an n-gon? The central angle of each of the n isosceles triangles is θ=360/n, so the side is s = 20 tan(θ/2) Thus, the area is A = 1/2 a (ns) = 1/2 (10) (n*20 tan 180/n) = 100n tan(...
*May 15, 2015*

**Calculus**

well, we have y' = -1/2 e^(-2t) + 1/ln10 * 1/4 10^(4t) + c y'(0)=0 means -1/2 + 1/(4 ln10) + c = 0, so c = 1/2 - 1/(4 ln10) Now follow the same logic to get y(t)
*May 15, 2015*

**Calculus**

T'(6) could be approximated by the value (T(8)-T(6))/2 = 5/2 T'(6) means how fast T is changing at t=6, in °F/hr.
*May 15, 2015*

**geometry**

there is no angle opposite the right angle. However, the sides in this triangle are in the ratio 1:√3:2 So, if d is opposite the 60° angle, it is the long side, and the sides are 2, 2√3, 4 If d is opposite the 30° angle, it is the short side, so the sides ...
*May 15, 2015*

**Caclulus**

Hmm. I see an extra sin, not cos: 9x sin(6x^2) sin(3x^2)
*May 15, 2015*

**Algebra 1**

Z=2 means 2 std above the mean. So, what's 80+2*3?
*May 10, 2015*

**Geometry**

one complete circle of diameter 12 and 2 side lengths of 106
*May 10, 2015*

**Trigonometry**

Newton's math is correct, but there is certainly no reason to convert to degrees.
*May 10, 2015*

**Maths**

b boys, 80-b girls Now add up the total points scored: 75(b) + 65(80-b) = 69(80)
*May 10, 2015*

**maths**

avg = totalcost/totalpeople = (2*84 + 6*62)/(2+6)
*May 10, 2015*

**Math**

the line through the two given points is y-4 = -7(x-1) Now use that to find the desired values.
*May 10, 2015*

**maths**

25^2 = 625 so, figure 26^2, and subtract 630 from that. or, find √630 = 25.099 so, since 25^2 < 630 and 26^2 > 630 as above...
*May 10, 2015*

**maths**

well, 1000^2 = 1,000,000, which is the smallest 7-digit number. 10 times that will be 8 digits. √10 = 3.1622, so 3162^2 = 9,998,244 3163^2 = 10,004,569
*May 10, 2015*

**Maths - eh?**

apparently the sum of all the shaded areas is less than the shaded area of the triangle. I don't see how that can be.
*May 10, 2015*

**algebra**

that will be when 1.05^t = 2 Now add that t to 2000
*May 10, 2015*

**Algebraic Geometry**

If the 2nd is x, then we have (x-5) + x + 2(x-5) = 33 x = 12 Now figure the legs.
*May 10, 2015*

**Math**

A+B+C = 26 (B+3)+B+(2B-1) = 26 Now you find B, and thus A and C.
*May 10, 2015*

**matrices**

well, B^2 = 0 0 1 1 0 0 0 1 0 Now do it again.
*May 10, 2015*

**Math too hard please help**

n: C=10n+26 ------------------ 27: 10(27)+36 = 270+36 = 306 and so on with the other values
*May 10, 2015*

**trig**

find the angle between the pole and the wire using the law of sines: 35/sin30 = 25/sinθ Now use the law of sines again to find x, since you can now figure the angle opposite x.
*May 10, 2015*

**Trig**

or, having found B, you can use the law of cosines: x^2 = 35.4^2 + 10.2^2 - 2(35.4)(10.2)cosB
*May 10, 2015*

**Trig**

If the ladder makes an angle A with the ground and B with the embankment, then using the law of sines, 35.4/sin 117.5 = 10.2/sinB = x/sinA Now you can find B, and A = 180-(117.5+B) and you can find x.
*May 10, 2015*

**binomial expansion**

since the powers add up to 12, that would be the 10th term: 12C9 (√2)^9y^3 = 12C3 (16√2)y^3 = 3520√2 y^3
*May 9, 2015*

**binomial theorem**

That would be 22C19 a^3 b^19 = 22C3 a^3 b^19 = 1540a^3b^19
*May 9, 2015*

**trig**

note that h cot38° - h cot75° = 65
*May 9, 2015*

**Pre-Calculus**

cos x cot^2 x= cos x cos x cot^2 x - cos x = 0 cos x (cot^2 x - 1) = 0 cosx = 0 cotx = ±1 Now you can just find x, right?
*May 9, 2015*

**math**

4 pink out of 14 total, so P(pink) = 4/14 = 2/7
*May 9, 2015*

**Common Core Math II A-CR**

recall that the height h is given by h(t) = 128t - 16t^2 That's just your good old friend, the parabola. Find its vertex -- that is the maximum height.
*May 9, 2015*

**precalculus**

suppose the terms were -3,-2,-1,0,... would that be hard tto figure out? Well, it's the same here: (t-2)-(t-3) = 1 (t-1)-(t-2) = 1 The common ratios for the 2nd one are found by dividing each term by the previous one: 1,-3/2,2,-5/2 (-3/2)/(1) = -3/2 2/(-3/2) = -4/3 (-5/2...
*May 9, 2015*

**precalculus**

see related questions below
*May 9, 2015*

**Trig**

all you need here is the perpendicular distance from Nancy's plane to a N-S line. If that distance is x, then x/400 = sin 23° x = 156.29 you are correct. Nasty how they sneak in unrelated problems, eh?
*May 9, 2015*

**algebra**

1.22x = 3200000 x = 2622950.82
*May 9, 2015*

**math**

5x+7 = 7x-21
*May 9, 2015*

**math**

well, A+B=180, so solve for x. Also, A+D = 180
*May 9, 2015*

**math**

2*15/√3
*May 9, 2015*

**math**

the two angles are equal, so just set them equal and solve for x Then figure MRQ
*May 9, 2015*

**Math**

one statement is by definition independent.
*May 9, 2015*

**algebra**

If x at 40%, the rest (20-x) is 85%. So, .40x + .85(20-x) = .60(20)
*May 9, 2015*

**Maths**

2560/58.0 = x/(9.50*8.75)
*May 9, 2015*

**maths**

clearly a = √7 r = √3 An = √7*√3^(n-1)
*May 9, 2015*

**english**

1st step is learning to write it. As in Can you help me in my debate topic, 'Knowing English well is equal to having a college degree'?
*May 9, 2015*

**physics**

consider that T^2 = kr^3 That means that T^2/r^3 is constant T^2/r^3 = (2T)^2/(xr)^3 = 4T^2/x^3r^3 So, 4/x^3 = 1, and x = ∛4 That is, our small planet's orbital radius is ∛4 times earth's.
*May 9, 2015*

**maths**

d = (32-11)/(12-5) = 3 a = 11-4*3 = -1 now you can fill in the rest.
*May 9, 2015*

**math**

nth from front is a + (n-1)d nth from l is l - (n-1)d their sum is a+l Not sure what you meant by "a l"
*May 9, 2015*

**precalculus**

the coefficients are 1,3,3,1 so you have 1(2x)^3(-3y)^0 + 3(2x)^2(-3y)^1 + 3(2x)^1(-3y)^2 + 1(2x)^0(-3y)^3 = 8x^3 - 36x^2y + 54xy^2 - 27y^3
*May 9, 2015*

**extreme math help please**

change in x: 2 change in y: 3 y = 3/2 x - 5
*May 9, 2015*

**college precalculus**

check for common difference or ratio: #1: d = 1 #2: r = 2/3 #3: differences: -5/2, +7/2, -9/2, ... ratios: -3/2, +4/3, -5/4, ... neither d nor r is constant, so the sequence is not AP or GP
*May 9, 2015*

**algebra**

y grows by 5 when x grows by 1. So, since 6-1=5, when x=6, y=4+5*5 = 29 or, using the point-slope form, y-4 = 5(6-1)
*May 8, 2015*

**Calculus**

(a) ok (b) ok (c) g" = (x^2-4x+16)/(x-2)^2 Since g" is never negative, g is never concave down (d)correct, but since you know g(3)=4, y-4 = -9(x-3) (e) since the graph is always concave up, any tangent lines must lie below the graph. Doodle around some, and you will ...
*May 8, 2015*

**Physics**

distance = speed * time, so that wold be 3.00*10^5 km/s * 1.28s = 3.84*10^5 km
*May 8, 2015*