Wednesday

May 25, 2016
Total # Posts: 40,793

**algebra**

good as far as you went, but 10/√27 10/(3√3)
*May 9, 2016*

**math**

sinA = 2 sinA/2 cos A/2 similarly for B and C Now add them up.
*May 9, 2016*

**math,science**

a = -22m/s / 13s = -22/13 m/s^2
*May 9, 2016*

**geometry**

you can easily see that the region is a rectangle. Rotate a rectangle around one of its sides and you get a cylinder. Then it's easy. See my work at http://www.jiskha.com/display.cgi?id=1462581546
*May 9, 2016*

**math**

This may help: http://mathforlove.com/2012/02/midpoints-of-a-quadrilateral-form-a-parallelogram/
*May 9, 2016*

**math**

the interest is the same for each year: 13000 * 0.08 clearly you need to review why the "normal" method works as it does.
*May 9, 2016*

**Algebra II**

well, π radians = 180°
*May 9, 2016*

**Algebra II**

use the law of sines
*May 9, 2016*

**geometry**

If they have the same base area, then the volumes are in the ratio of the heights, cubed. That is 250/2 = (H/h)^3
*May 9, 2016*

**Algebra II**

did you draw a diagram? If so, you can see that the distance x is found using the law of cosines: x^2 = 3.6^2 + 2.3^2 - 2*3.6*2.3*cos66˚
*May 9, 2016*

**math**

There are 20 greens in 70 gumballs The chance of drawing zero green ones is 50/70 * 49/69
*May 9, 2016*

**math**

the two terms are 9d apart. So, d = -2 Now you have d, and can easily find a, so S15 = 15/2 (2a + 14d)
*May 9, 2016*

**math**

huh?
*May 9, 2016*

**Trigonometry**

the height in feet can be found using h/5280 = tan6°
*May 8, 2016*

**Geometry**

V = 18*w*12 where the width is w.
*May 8, 2016*

**math: pre-calculus**

by definition, e^(ln x) = x ln(e^x) = x ln and e^ are inverse functions, just like x^2 and √x Assuming no typos, I see ln(e^3y)+ln(ey)-ln(e^4) = ln(e^3y * ey / e^4) = ln(y^2) = 2ln(y)
*May 8, 2016*

**math: pre-calculus**

you want p = a e^(kt) clearly p(0) = 53 p = 53 e^(kt) p(3) = 90, so 53 e^(3k) = 90 e^(3k) = 90/53 3k = ln(90/53) k = ln(90/53)/3 = 0.1765 so, we have p(t) = 53 e^(.1765t) Now you try the others, lest you be accused of homework dumping...
*May 8, 2016*

**geometry (please help)**

The area of the triangular base is (1/2)(9)(10) = 45 There are two of those, for a total of 90 There are three rectangular faces of area 9*10+9*9+9*13.5, for a total of 292.5 Total area is thus 90+292.5 = 382.5 in^2
*May 8, 2016*

**math: pre-calculus**

note that this is just log(x) shifted left 3 and down 8. So, now you can relate its properties to those of h(x)
*May 8, 2016*

**math**

just plug in the data and solve for w. Then you can get the length. w(3w-2) = 65 Of course, it also helps to know that 5*13 = 65 ... just sayin'
*May 8, 2016*

**algerbra**

since distance/speed = time, (1600/400) + (1600/320) = ?
*May 8, 2016*

**Math Algebra**

since distance/speed = time, 200/(x+3) = 200/(x-3) - 1
*May 8, 2016*

**math**

290-50 = 240 240/60 = 4 so, in 4 more hours
*May 8, 2016*

**Calculus**

V(0) = 500,000 V(5) = 405,000 (V(5)-V(0))/(5-0) = -19,000 ft^3/min
*May 8, 2016*

**math , steve or damon pls**

right on both of them.
*May 8, 2016*

**math**

pi, not pie! 2pi(r^2+rh) = 2pi(25+75) = 200pi Now just multiply that are by the cost, which is .0450/cm^2
*May 8, 2016*

**calculus**

let u = 2x you know lim sin(u)/u, right?
*May 8, 2016*

**damonnnn math help**

since multiplication is commutative, 6t^3 * 6t^3 = 6*6 * t^3*t^3 = 36t^6 t^3*t^3 = t*t*t * t*t*t = t^6, not t^9 add exponents when multiplying powers
*May 8, 2016*

**Geometric series**

a = -1 r = 3 Sn = (-1)(3^n-1)/(3-1) = (1-3^n)/2 = -3280 1-3^n = -6560 3^n = 6561 n = 8
*May 8, 2016*

**calculus**

what, you forgot your Algebra I? (x+1/x)^2 = x^2 + 2 + 1/x^2 Surely you can handle each of those terms...
*May 8, 2016*

**math**

figure the volume in ft^3, then multiply by how many gal/ft^3. That will give you how many gallons will fit. I have no idea how many gallons are actually in the tank...
*May 8, 2016*

**calculas**

let the distance from 3rd base be x. Then the distance z from home is z^2 = 90^2 + x^2 so, to find dz/dt, 2z dz/dt = 2x dx/dt So, find z when x=20, then plug in dx/dt and x.
*May 8, 2016*

**Maths - eh?**

If B is a point, how can a line be perpendicular to it?
*May 8, 2016*

**mathematics-permutation and combination**

The two boys are separated by 10 others on each side, right? Now you can easily figure the solution.
*May 8, 2016*

**math**

Just draw a diagram. All those right angles help a lot. XZ/2800 = tan(90-63)° 2800/XP = cos(90-25)°
*May 8, 2016*

**math**

x/180 = cot 27°
*May 8, 2016*

**Trigonometry and Right Triangles!**

the sides are in the ratio 1:√3:2 so if the hypotenuse is 16, then the sides are 8,8√3,16 and the perimeter is 24+8√3
*May 8, 2016*

**Math**

2x+5x = 4.27 x = .61 So, they have 1.22 and 3.05 lbs. Now you can decide how much has to be given so they have the same amount.
*May 8, 2016*

**math**

180/x = tan 27°
*May 8, 2016*

**math**

with A at (0,0) we just add the x- and y-displacements to get the final location: <4,0>+<0,-3>+<-3.06,-2.57> = <0.94,-5.57> Now you can find the distance and bearing.
*May 8, 2016*

**Maths**

x+y=120 0.9x + 1.2y = 120 80+40 = 120 72+48 = 120
*May 8, 2016*

**Math**

If the diamond is just a rotated square, then its diagonal is 3, so its area is 9/2. So, the rectangle has width 3 and area 27, so its height is 9.
*May 8, 2016*

**math**

Draw a diagram. It will be clear that using the law of cosines, the distance d is found using d^2 = 300^2 + 200^2 - 2*300*200 cos33°
*May 7, 2016*

**MATH**

1/14 (196x+17-10)= (14/20)(20x+12x) has no solution since (14/20)(20x+12x) = 14x + 42/5 and 1/14 (196x+17-10) = 14x + 1/2 and there is no value of x such that 14x+42/5 = 14x + 1/2
*May 7, 2016*

**MATH**

The equation 1/14 (196x+17-10)=______ (20x+12x) has no solution. The equation (4x+24/122)=_____ (x+16/122)
*May 7, 2016*

**MATH!**

not sure what question is answered by the 4, but you are correct in assuming that the equation is an identity.
*May 7, 2016*

**MATH!**

The equation 1/14 (196x+17-10)=______ (20x+12x) has no solution. The equation (4x+24/122)=_____ (x+16/122) Im sure the answer for the second one is 4, (4x+24/122) = 4 (x+6/122)
*May 7, 2016*

**math**

The answer is as follows: 9-3÷1/3+1 9-3*3/1+1 9-9+1 0+1 Answer = 1
*May 7, 2016*

**math**

other interpretations of 9 minus 3 divided by 1/3 plus 1 are (9-3)/(1/3) + 1 = 6*3+1 = 19 (9-3)/(1/3 + 1) = 6/(4/3) = 9/2 9 - 3/(1/3) + 1 = 9-9+1 = 1 9 - 3/(1/3 + 1) = 9 - 9/4 = 27/4 As you can see, using math symbols is a lot less ambiguous than just words, and grouping ...
*May 7, 2016*

**math**

you have correctly factored the left side. What has that to do with the right side? Or is that 16/22 a typo?
*May 7, 2016*

**math**

The equation 1/14 (196x+17-10)=______ (20x+12x) has no solution. The equation (4x+24/122)=_____ (x+16/122) Im sure the answer for the second one is 4, (4x+24/122) = 4 (x+6/122)
*May 7, 2016*

**math**

The equation 1/14 (196x+17-10)=______ (20x+12x) has no solution. The equation (4x+24/122)=_____ (x+16/122) Im sure the answer for the second one is 4, (4x+24/122) = 4 (x+16/122)
*May 7, 2016*

**math**

R = R1 + R2 = 10 + 15 = 25Ω
*May 7, 2016*

**math**

clearly a permutations problem, since the order of award matters. So, there are 50P3 ways to arrange the selections.
*May 7, 2016*

**math**

the desired point will be the center of the circle passing through the three points. So, if that is (h,k), we have (x-1)^2 + (y-1)^2 = r^2 (x-7)^2 + (y-1)^2 = r^2 (x-1)^2 + (y-9)^2 = r^2 Now, you can solve those to find (h,k) and r, or you can note that the three points form a...
*May 7, 2016*

**calculas 1**

f(0) = 0 f(9) = 9^9 (wow!) (f(9)-f(0))/9 = 9^8 f'(x) = 9x^8 so, you want c such that f'(c) = 9^8 9c^8 = 9^8 c^8 = 9^7 c = 9^(7/8) ≈ 6.84 ∊ [0,9]
*May 7, 2016*

**math**

The equation 1/14 (196x+17-10)=______ (20x+12x) has no solution. The equation (4x+24/122)=_____ (x+16/122) Im sure the answer for the second one is 4, (4x+24/122) = 4 (x+16/122)
*May 7, 2016*

**math**

if f(x) = x^2+bx+c, we have (x^2+bx+c) + (x^2-bx+c) = 2x^2 2x^2+c = 2x^2 so, c=0 and f(x) = x^2+bx f(2016) = 0, so 2016^2 + 2016b = 0 b = -2016 f(x) = x^2 - 2016x = x(2016-x) f(10) = 10(2016-10) = 10(2006) = 20060
*May 7, 2016*

**trig**

not quite sure what "double angle of tan" means, but if you want tan(u) where u is in QI and sec(u) = 7/2, then draw the triangle. clearly tan(u) = √45/2 = 3√5/2. So, tan(2u) = 2tan(u)/(1-tan^2(u)) = 2(3√5/2)/(1-45/4) = -12√5/41
*May 7, 2016*

**maths-factor**

1/(x-3) < -1 what do you mean factors? You want a solution set If x-3 > 0, then 1 < 3-x x < 2 Not possible if x>3 If x-3 < 0 or x < 3, then 1 > 3-x x > 2 So, the only solutions are 2<x<3 verify here: http://www.wolframalpha.com/input/?i=1%2F%...
*May 7, 2016*

**geometry**

the distance from (-6,-8) to (0,0) is 10. So, (x+6)^2 + (y+8)^2 = 10^2
*May 7, 2016*

**geometry**

Write the standard equation for the circle center (-6,-8), that passes through (0,0). I know that the formula would leave the 1qstr part as (x+6)+(y+8) I'm not sure about the radius when it goes through (0,0)
*May 7, 2016*

**integral calculus**

no idea what box you're talking about. Probably an integral sign which does not render in your font. Judging from the answer produced by wolframalpha.com http://www.wolframalpha.com/input/?i=integral+dx%2F%283x^3-5%29^3 you have some major tricks to use.
*May 7, 2016*

**algebra**

the max height is where t = -b/2a = 3 h(3) = 256 ft h=0 when t=7
*May 7, 2016*

**calculas 1**

You give no limits on the area.
*May 7, 2016*

**calculas**

clearly the rectangles are bounded at x = 0, π/6, π/3, π/2, 2π/3, 5π/6, π So, evaluate g(x) at the boundaries and add up the areas. What do you get? There are plenty of online calculators for this to check your work.
*May 7, 2016*

**calculas 1**

Have you tried laying out a few rows? I'm sure you will see a pattern. You will be adding up sequences of even or odd numbers...
*May 7, 2016*

**Geometry**

Since s=rθ, the circumference of the base of the cone is 15 * 4π/3 = 20π So, the radius of the base of the cone is 10. Now you can find the height of the cone, and thus the volume.
*May 7, 2016*

**math**

recall that x = vcosθ t y = vsinθ t - g/2 t^2 dy/dx = dy/dt / dx/dt = (vcosθ - gt)/(-vsinθ) -cotθ + g/(vsinθ) t we want dy/dx = tan(θ/2) = (1-cosθ)/sinθ = 1 - cotθ -cotθ + g/(vsinθ) t = 1 - cotθ t = vsinθ/g
*May 7, 2016*

**math**

just use your equation of motion, and find when its slope (derivative) is tan(alpha/2).
*May 7, 2016*

**Trig**

If we call the lengths OA and OB x and y, then we have, using the law of cosines, x^2+y^2-√3 xy = 1 That is just an ellipse with its extreme values at x=2 or y=2. So, the maximum value of OB is 2.
*May 7, 2016*

**Calculus**

well, the two curves intersect at (±√(10/3),40/3), so the area is 2*integral[0,√(10/3)] (x^2+10)-(4x^2) dx = (40/3)√(10/3)
*May 6, 2016*

**Calculus**

work = force * distance, so ∫[0,25] 30-.1x dx
*May 6, 2016*

**Math**

Come on, man. The area is just a 6x4 rectangle. Rotating it around the y-axis jives you a cylinder with r=6 and h=4, so its volume is 144π Or, if you must use calculus, v = ∫[0,6] 2πx*4 dx = 144π Or, using the Theorem of Pappus, it is the area of the ...
*May 6, 2016*

**math**

yes it is
*May 6, 2016*

**Math Help Please**

anything that ends in .9... is an integer. So, 0.999999... = 1 1.999999... = 2 4.999999... = 5 and so on
*May 6, 2016*

**Programming**

Oops. I dislayed the total, not the average. Change that to avg /= n Display "The average score for Student #",s," is ",avg
*May 6, 2016*

**Programming**

how about something like this: minSubjects = 8 maxSubjects = 12 Display "Enter the number of students: " enter students for s = 1 to students Display "Enter the scores for student #",s enter scores n = length(scores) &...
*May 6, 2016*

**MATH Please Help Please**

f(n) is a sequence of integers, and the domain is all integers. f(x) is a function whose domain is all real numbers, not just integers.
*May 6, 2016*

**Pre-Cal (Trig) Help?**

try arccos(.957269)
*May 6, 2016*

**Pre-Cal (Trig) Help?**

all you need is the sum formula for cosines. in other words, cos(A+B) = .957269 A+B = 16.81°
*May 6, 2016*

**maths-word problem help**

A&B together can do 1/12 + 1/18 = 5/36 of the job in a day So, it will take them 36/5 * 2 = 72/5 days. or 14 2/5 days After 14 days, they have each worked 7 days, so 7 * 5/36 = 35/36 of the job is done. On the 15th day A works 1/3 of a day to finish up the last 1/36 of the task.
*May 6, 2016*

**Math (Ms. Sue)**

why all the duplications? I only see H1 H2 H3 H4 TT TH Those are the only outcomes if only two events are to be considered.
*May 6, 2016*

**math**

In better notation, M(x) = 50 * 2^x change the 50 to 450
*May 6, 2016*

**math**

x((1+r)^2-1) = 440 x((1+r)^3-1) = 728 ((1+r)^3-1)/((1+r)^2-1) = 728/440 r = 1/5 = 0.2 = 20%
*May 6, 2016*

**calculus**

x^3 = at^2 3x^2 x' = 2at x' = 2at/3x^2 x" = (2a)(3x^2) - (2at)(6xx') ------------------------------- 9x^4 2ax - 4at(2at/3x^2) ----------------------------- 3x^3 (6ax^3 - 8a^2t^2)/(9x...
*May 6, 2016*

**college algebra**

ln(310*3x) ln(930x) ... Hmmm I suspect you mean log base 3 instead of ln. ln is the natural log, using base e=2.71828... So, since all your logs are base 3, let's just write log, since there is no confusion. log(10)+log(x) = log(10x) log(10-x) cannot be simplified log(10/x...
*May 6, 2016*

**college algebra**

Since it starts in 2000, and 2001 is the next year, I'd pick 2,419,774 Not sure how the 5 values represent a population over 8 years.
*May 6, 2016*

**Math**

r = 95 b = 5g r+g = b-189 95+g = 5g-189 4g = 284 g = 71 so, b = 355 95+71+355 = 521
*May 6, 2016*

**math**

xy=42 x+y=3 There are no real solutions to this. However, if we suppose there is a typo, note that 6*7 = 42 6+7 = 13
*May 6, 2016*

**Geometry**

π/3 * 2^2 * 4 + 2π/3 * 2^3 = 32/3 π
*May 6, 2016*

**math**

3/10 * 2/9
*May 6, 2016*

**5th grade math**

y < 2x y < 2*15 y < 30
*May 5, 2016*

**math**

(6.4+12.3)/2 h = 37.4 9.35h = 37.4 h = 4
*May 5, 2016*

**math**

better use Heron's formula, if all you have is the sides.
*May 5, 2016*

**Pre calculus**

sinθ = √((1-cos2θ)/2) = √(1/8) since 2θ is in QIV, θ is in QII
*May 5, 2016*

**Algebra**

both ok
*May 5, 2016*

**mason**

There are 7C2 = 21 ways to pick the 1st two horses, leaving 5 to arrange in the other places. So, there are 21 * 5! ways to finish the race.
*May 5, 2016*

**Math**

oops - 3/10 of 2/3 = 2/10 = 1/5 1/3 + 1/5 = 8/15 so, 7/15 x = 35 x = 75 check: on Monday read 1/3, or 25 pages, leaving 50 unread On Tuesday read 3/10 of 50 = 15 pages, leaving 35 unread
*May 5, 2016*

**calculus help me plz**

dx/((x-1)sqrt(x^2-2)) OK. First step is a trig substitution. x = √2 sec(u) x^2-2 = 2tan^2(u) dx = √2 sec(u) tan(u) du now the integral is sec(u)/(√2 sec(u)-1) du see what you can do with that.
*May 5, 2016*