Tuesday

August 4, 2015

August 4, 2015

Total # Posts: 32,562

**Quadratic Functions**

Not sure what BiQuad is, but if you want to fit a parabola through those three points, you have, assuming that v(t) = at^2 + bt + c, 0a+0b+c = 10000 9a+3b+c = 2025 36a+6b+c = 6400 Somehow I suspect what you really wanted was v(3) = 1000-2025 = 7975, since most stocks don't...
*May 19, 2015*

**math**

well, shoot. Which is bigger: a ft^3 or a yd^3?
*May 19, 2015*

**math please help**

they are the same. It shows why 11^0 = 1 or why x^0 = 1 for any value of x. (except x=0)
*May 19, 2015*

**math help please!!**

3^3/3^6 = 3*3*3 ----------------- 3*3*3*3*3*3 Now you can see that three 3's cancel out in top and bottom, so you are left with 1/(3*3*3) = 1/27 This shows why x^n / x^m = x^(n-m)
*May 19, 2015*

**Math**

can you try that again in English?
*May 19, 2015*

**Math**

the cars are 360/16 = 22.5° apart. car 16 is 3 positions from car 13. So, a rotation of 67.5° will move car 16 to car 13's position.
*May 18, 2015*

**geometry**

in a triangle, angles are opposite sides. Two angles cannot be opposite each other. Using the tan function, the hypotenuse is not used. If the base 12 is opposite the 71° angle, then 12/d = sin 71° 12/c = tan 71° Better review your trig function definitions.
*May 18, 2015*

**Math**

you have a rectangle that is 100x(40+20) and a circle that has diameter 40+20=60 so, just add those two areas. If the area inside the circular ends but outside the rectangle does not count, then you have to subtract off the smaller circle of diameter 40.
*May 18, 2015*

**Math**

a^2 = 6 b^2 = 5 c^2 = a^2+b^2
*May 18, 2015*

**Math**

you want x minutes, where 12+.05(x-500) = 10+.04(x-300) x = 1100
*May 18, 2015*

**math**

each interval has width 2, so the midpoints are at 1,3,5,7,9
*May 18, 2015*

**math**

16-x^2 is concave down, so you want to use the right endpoints. Each interval has width 1, so you just have 2(f(1)+f(2)+f(3)) = 2(15+12+7) = 68 There are several good online Riemann Sum calculators. You can use them to verify your work.
*May 18, 2015*

**math**

see related questions below
*May 18, 2015*

**math**

just plug in the values for n: t1 = 7 t2 = 2t1-3(2) = 2(7)-3(2) = 8 t3 = 2(8)-3(3) = 7 ...
*May 18, 2015*

**math**

so, did you use the formula?
*May 18, 2015*

**algebra 2 and trig**

how many std's away is the 2.5% boundary? The mean is at (78+92)/2 = 85
*May 18, 2015*

**Calculus**

Assuming x is not zero, the fractions are zero when the numerators are zero. So, f'=0 when 8x+1 = 0, or x = -1/8 f"=0 when 4x-1 = 0, or x = 1/4
*May 18, 2015*

**Algebra 2**

revenue for x hats is y=15x cost is y=0.4(x-150)^2+9000 So, you want x when 15x >= 0.4(x-150)^2+9000 x >= 262.5 so, at least 263 hats
*May 18, 2015*

**Math**

if the chord length is 2x, then half the chord subtends half the angle: x/200 = sin 34°
*May 18, 2015*

**please help math**

it is that.
*May 18, 2015*

**math please help**

after n years, its value is 400(1.07)^n
*May 18, 2015*

**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*