Monday

April 27, 2015

April 27, 2015

Total # Posts: 30,837

**math**

I'll use h for height, because l for length looks like the number 1. the surface area is two semi-circular ends: 2*(πr^2/2) = πr^2 the rectangular cut surface: 2rh the curved surface: πrh Now just plug and chug
*March 7, 2015*

**math**

T = A+(B-A)/2 = (A+B)/2 B = 2T-A
*March 7, 2015*

**math**

If KN=x, and Z has speed b-20 Z flew for 3.5 hr B flew for 1.5 hr (b-20)(3.5) = 340+x b(1.5) = x (b-20)(3.5) = 340+b(1.5) 2b = 340+70 b = 205 check: K is 307.5 KM west of N So, K is 647.5 km west of G 647.5/185 = 3.5 307.5/205 = 1.5
*March 6, 2015*

**math**

clearly, 23 is not the largest. If it is the smallest, then 23 + 23+d + (23+d)+d = 180 3d + 69 = 180 3d = 111 d = 37 and the angles are 23,60,97
*March 6, 2015*

**math**

s(t)equals = 5cos(t) minus- sin(3t) ??? why all the words? Just write s(t) = 5cos(t) - sin(3t) v(t) = -5sin(t) - 3cos(3t) a(t) = -5cos(t) + 9sin(3t) a(π) = -5(-1) + 9(0) = 5 you are correct.
*March 6, 2015*

**SAT Math**

those two terms are 6 apart, so their difference is 6*4 = 24
*March 6, 2015*

**Math**

after a couple of others just like this, do you have some work of your own to show?
*March 6, 2015*

**Pre-Calc.**

just set y=0. x = ±3 So, the vertices are at (±3,0) You know that it has to be y=0, because if x=0, -y^2/4 = 1, meaning y^2 has to be negative.
*March 6, 2015*

**Math**

The domain is the set of first numbers of the pairs: {0,-2,2,1,-5} the range is the set of second numbers: {2,4,8,6,0} Looks like D to me.
*March 6, 2015*

**math, physics**

the initial velocity is 60 ft/s s = 60t - 1/2 at^2 So, a = 3 ft/s^2 60/3 = 20 seconds I wouldn't call that a sudden stop!
*March 6, 2015*

**MaTH**

meet 1: 7'46" meet 2: 7'46" - 42" = 7'04" next meet the reduction was 2*42 = 84 seconds meet 3: 7'04" - 84" = 5'40"
*March 6, 2015*

**math**

(i) 50(0.035) + 30(0.0475) = 3.175 (ii) 3.175/(50+30) = 0.03968 = 3.97% .035x + .0475(50-x) >= 0.04*50 x <= 30 That is, with 30kg from A and 20kg from B, the result is exactly 4% fat Since A's content is less than 4%, using any more of A will reduce the result below ...
*March 6, 2015*

**College Algebra**

y = x^2-13x+36 clearly the y-intercept is (0,36) y = x^2 - 13x + (13/2)^2 + 36 - (13/2)^2 y = (x - 13/2)^2 - 25/4 So, the vertex is at (13/2,-25/4) y = x^2-13x+36 y = (x-4)(x-9) The x-intercepts are at (0,4),(0,9) The coefficient of x^2 is positive, so it opens up. (C) No idea...
*March 6, 2015*

**algebra**

12t^2 + 17t - 40 = 0 (4t-5)(3t+8) = 0 t = 5/4 or -8/3
*March 6, 2015*

**Pre-Calc.**

you have to complete the squares. Rearrange things a bit and you have x^2+6x + 4y^2-8y = -9 x^2+6x + 4(y^2-2y) = -9 Now complete the squares, and be sure to make the same changes to both sides of the equation: x^2+6x+9 + 4(y^2-2y+1) = -9+9+4 (x+3)^2 + 4(y-1)^2 = 4 (x+3)^2/4...
*March 6, 2015*

**algebra**

F = Av^2/400 v^2 = 400F/A v = 20√(F/A)
*March 6, 2015*

**calculus**

if the base has width 2x, then the area is a = 2xy = 2x(27-x^2) = 54x - 2x^3 da/dx = 54 - 6x^2 = 6(9-x^2) da/dx=0 when x=3 So, the rectangle is 6 by 18 with area=108
*March 6, 2015*

**Math Check**

#1 C #2 B 4x + 2z = 10, so z = 5-2x That gives 2x + y - 3(5-2x) = 4 -2x + 3y - 13(5-2x) = -8 or 8x + y = 19 24x + 3y = 57 so, y = 19-8x That is, whatever x you choose, y and z can be found.
*March 6, 2015*

**Pre-Calc**

as you know, the hyperbola x^2/a^2 - y^2/b^2 = 1 has asymptotes of y = ± b/a x You have a=2, b=1, so the asymptotes have slope ±1/2. Also, your hyperbola is shifted, so its center is not at (0,0). Thus, your asymptotes are y-2 = 1/2 (x+1) y-2 = -1/2 (x+1) or, y...
*March 6, 2015*

**College Algebra**

4x+y-5=0 has slope -4 So, now you have a point and a slope: y-3 = -4(x-3)
*March 6, 2015*

**College Algebra**

just plug and chug. No mysteries here: f(g(2)) = f(1) = 0 g(f(-1)) = g(4) = 3 f(2)-g(3) = 1-2 = -1 f(-1)*g(-2) = 4(-3) = -12 f(x)/g(x) = (x-1)^2 / (x-1) = x-1 for x≠1
*March 6, 2015*

**Math**

sorry, bub. No diagrams here.
*March 6, 2015*

**Math**

the horizontal sections are the same shape as the base of the pyramid, but of different sizes. What do you think of the vertical sections? As with the cube, I assume the vertical sections are parallel to a side of the base.
*March 6, 2015*

**Maths**

come on, guy. For the first equation, pick any two values. Make them easy ones. y=-1, x=4 x=-3, y=4 So, plot (4,-1) and (-3,4) and draw the line. Do the same for the other one.
*March 6, 2015*

**Maths**

c'mon. Can you plot points? Plot two points for each equation. Connect the pairs into lines. See where the lines intersect.
*March 6, 2015*

**math**

see http://www.wolframalpha.com/input/?i=plot+x-y%2F3%3D2%2C+x-5y%3D0+ or use the graphing utility of your choice. See where the lines intersect.
*March 6, 2015*

**math**

ar^5 - ar^2 = 3(ar^6-ar^5) r^3-1 = 3r^4 - 3r^3 3r^4 - 4r^3 + 1 = 0 The only real root of this equation is r=1. Is there a typo? Can you find an error I made?
*March 6, 2015*

**Pre-calc/trig**

Both a and b are not truly periodic, since the oscillation gradually has lower and lower amplitude. But, all other things being equal, I'd say the pendulum keeps its amplitude much better than the ball, so I'd have to go with (a) Even (c), if the plane is always on ...
*March 6, 2015*

**Math**

using substitution, z = -3, so 3y-8(-3) = -9 3y = -33 y = -11 so, x-(-11)+2(-3) = 22 x+11-6 = 22 x = 17 So the solution is (17,-11,-3) The other is not quite so simple, but since z = 1-2x-4y, we have x-2y-3(1-2x-4y) = 2 x+y-(1-2x-4y) = -1 or, 7x+10y = 5 3x+5y = 0 So, since 5y...
*March 6, 2015*

**Pre-Calculus**

the vertices are at the ends of the major axis, and the covertices are on the minor axis. So, since 25 > 4, the vertices are on the y-axis. when x=0, y=5 when y=0, x=2 So, the vertices are at (0,±5) the covertices are at (±2,0) Looks like you need to review ...
*March 6, 2015*

**College Algebra**

the two-point form is y-3 = (-4-3)/(4-1) (x-1) and you can massage that into what you need.
*March 6, 2015*

**College Algebra**

just use the distance formula: d = √((-1-2)^2 + (-1-3)^2) = √(9+16) = 5
*March 6, 2015*

**College Algebra**

the domain is all reals except where the denominator is zero. see more information at http://www.wolframalpha.com/input/?i=%28x%2B4%29%2F%28x^2-1%29
*March 6, 2015*

**calculus**

#1 I think you have a typo here. Since the r = d/2, the radius always grows half as fast as the diameter. #2 If the base is x from the wall, and the top is y in height, then you have x^2+y^2 = 25^2 so, 2x dx/dt + 2y dy/dt = 0 or, getting rid of the 2's, x dx/dt + y dy/dt...
*March 6, 2015*

**College Algebra**

since body/snout is constant, body/3.5 = 7/2
*March 6, 2015*

**College Algebra**

680(1+.04/12)^(12*17)
*March 6, 2015*

**Business Math GBM 200**

cost is 300*.24 = 72.00 desired end revenue is 72*2 = 144.00 But only 79% = 237 lbs get sold $144/237lb = .6076 = $0.61/lb
*March 6, 2015*

**Physics**

how long does it take to fall 0.75m? 4.9t^2 = 0.75 the speed is 2/t m/s I don't think the above answer is correct.
*March 6, 2015*

**MATHS**

Looks to me as though there are in fact many possible solutions. Maybe the "correct" one is the smallest total. Nope. 1244 +125 ------- 1369
*March 6, 2015*

**o level mathematics**

Draw your Venn diagram. The intersection of all 3 circles is 2 The other 3 numbers are placed outside any intersections. So, that means we have accounted for 20 of the 35 students. The other 15 can play any combination of two of the games, or just one game. So, there could be ...
*March 6, 2015*

**AHOOOGAAH! HOMEWORK DUMP!**

...
*March 6, 2015*

**computer programming**

you're going to have to explain those lengths a bit better than that.
*March 5, 2015*

**Pre-Calc**

sinB/b = sinA/a so, sinB = 108*sin21°/76 = 0.50925 so, B = 30.6° or 149.4° Since A+B+C = 180, that means C is 128.4° or 9.6° So, since c/sinC = a/sinA there are two choices for c, as well.
*March 5, 2015*

**Math**

assume an infinite number of hops. In practice, of course, the ball stops earlier than that, but you have a geometric progression where a = 20 r = 4/5 So, the initial drop is 20, and each bounce is a round trip 4/5 as high as the one before, so the total distance after n hops ...
*March 5, 2015*

**maths**

12% off means 88% of the price. So, .88p = 1.21c p = 1.375c so, the marked price is 37.5% above cost
*March 5, 2015*

**math**

c+p+s = 100 .5c + 2p + 5s = 100 There are lots of answers, as long as there are an even number of chicks (so you have whole dollars for their cost) For example, getting the most sheep possible, you have 2 chicks, 4 pigs, 19 sheep Now, you can start converting each pig into 4 ...
*March 5, 2015*

**Directions**

Face North, and turn 70° eastward. Due east is N90°E NE is N45°E
*March 5, 2015*

**math**

the more negative the balance, the greater the debt.
*March 5, 2015*

**math**

3 1/3 = 10/3 1 1/4 = 5/4 10/3 * 5/4 = 50/12 = 25/6 = 4 1/6 miles
*March 5, 2015*

**Calculus**

dx/dt = (2-x)√(1-x) dx/(2-x)√(1-x) = dt Note that 2-x = 1 + 1-x = 1 + √(1-x)^2 2 arctan √(1-x) = t + c √(1-x) = tan(t/2 + c) = tan(t/2) + c 1-x = (tan(t/2) + c)^2 x = 1 - (tan(t/2)+c)^2
*March 5, 2015*

**Calculus**

what, have you forgotten your algebra I? y' = (y^2 + y^2 cosx)^2 y' = y^4 (1+cosx)^2 y'/y^4 = (1+cosx)^2 dy/y^4 = (1+cosx)^2 dx -1/3 y^-3 = 1/4 (6x+8sinx+sin2x) + c y = -∛(4/(3(6x+8sinx+sin2x) + c))
*March 5, 2015*

**algebra**

y = x^2-2x-8 = x^2-2x+1 - 9 = (x-1)^2 - 9
*March 5, 2015*

**math 5th grade**

8/(10+8) this assumes the socks feel identical, and they are stacked in a messy pile.
*March 5, 2015*

**math**

evidently you have not spent the intervening time studying the problems. #1 A flat rate is just that: a constant amount, say $25 A commission is a certain amount, say $50 for every sale. So, for x sales, he would earn 50x so, his total earnings are 25 + 50x #3 is just like #1...
*March 5, 2015*

**math**

If (x) = -2 and (y) = 2, then everything is false.
*March 5, 2015*

**math**

I assume (x) means |x| = absolute value? well, you have |x| = 2 |y| = 2 Looks like (c) is false. Can't tell with (a), since "1 (y)" eludes me.
*March 5, 2015*

**Math**

2x+y = 2400 so, y = 2400-2x a = xy = x(2400-2x) = 2400x-2x^2 That is just a parabola, with vertex at x=600 So, the field is 600 x 1200 As usual, these problems achieve maximum area when the perimeter is divided equally among lengths and widths.
*March 5, 2015*

**Math**

If the rectangle has width 2x, then the area is a = 2xy = 2x(12-x^2) da/dx = 24 - 6x^2 = 6(4-x^2) da/dx=0 when x=2 So, the largest area is 4(12-2^2) = 32
*March 5, 2015*

**Math**

how about 100 and 100 99*101 = 9999 100*100 = 10,000
*March 5, 2015*

**Acceleration**

SInce 1 mi = 1,6 km, 1 mi/hr/s = 1.61km/hr/s So, 23.3 mi/hr/s = 37.5 km/hr/s So, now you can compare them.
*March 5, 2015*

**math - typos**

Holy crap! Is ! the NOT operator? If so, we have !(!((5-10)*(4-(9/2)) > 60) has unbalanced parentheses (((5*7)/(4/(5+3)))=15)) has unbalanced parentheses In both cases there are not enough ('s Better go over it and make sure things balance out
*March 5, 2015*

**math: algebra**

so, who asked anything about the volume?
*March 5, 2015*

**math: algebra**

Drop an altitude from the vertex to the center of the base. Now draw a line from there to the center of a side of the base. Seen from the side, you now have a right triangle with legs 4 and 22. The hypotenuse is the slant height of the pyramid, and thus the altitude of one of ...
*March 5, 2015*

**Math**

there is only one statement, and it is true. why?
*March 5, 2015*

**Physics**

speed = wavelength*frequency So, if speed doubles you know nothing about the frequency. If the wavelength is unchanged, then the frequency doubles. But, if the wavelength doubles, the frequency remains unchanged.
*March 5, 2015*

**Math (Asap)**

aw, shucks... <scuff scuff>
*March 5, 2015*

**Math (Asap)**

If the width of the pool is w, then the length is 3w. So, the difference in area between the whole thing and just the pool is the area of the deck: 832 (w+8)(3w+8) - w(3w) = 832 w = 24 So, the pool is 24x72 feet Check: The pool+deck has area (24+8)(72+8) = 2560 The pool has ...
*March 5, 2015*

**Maths- Proof**

any even number is a multiple of 2. When you pick any old n, it might not be even. You have proven that any two numbers which differ by 2 have their squares differing by a multiple of 4. So, technically you have not done what was asked, but have proved an even stronger result...
*March 5, 2015*

**Chemistry**

Looks to me like Fe(NO3)3 + 3LiOH = Fe(OH)3 + 3LiNO3 I guess you can manage the ions and phases...
*March 5, 2015*

**math**

Assuming he earns an extra $10 for each carton of 30, we have 1.50x + 10(⌊x/30⌋) >= 450 x >= 246.7 check: 240 games is 8 carton, so he gets an extra $80 80 + 1.5*246 = 449.00 the 247th game earns him another $1.50, putting him at $450.50
*March 5, 2015*

**math**

To start, there were b boys and g girls. b = 3/7 g (b+g)-87 = b/2 + 4/5 g so, b + 7b/3 - 87 = b/2 + (4/5)(7/3)b b = 90 so, g = 210 check: after 87 students left, there were 45 boys and 168 girls. 300-87 = 213 = 45+168
*March 5, 2015*

**Math**

looks good to me
*March 5, 2015*

**calc**

y = √(8+x^3) dy/dt = (3x^2 / 2√(8+x^3)) dx/dt Now just plug in dy/dt=5 and x=2 to get dx/dt
*March 5, 2015*

**math**

The trailer went 300 km in t-1 hours, so its avg speed is 300/(t-1) Let the bus's speed be x, and the trailer's speed be y. The bus went 500 km. Its time spent on the trip before meeting the trailer was t = 400/x + 1 + 100/x = 1 + 500/x hours The trailer was on the ...
*March 5, 2015*

**Science**

#1 ok #2 no diagram, but a closed circuit would imply a flow of current, so I agree. #2 Nope. Mars has less mass, so its gravity is weaker. (C)
*March 5, 2015*

**Calculus, HELP!!**

If the paper is x by y inches, then the printed area is (x-2)(y-12) The paper has area a = xy = x(24/(x-2) + 12) = 12x^2/(x-2) da/dt = 12x(x-4)/(x-2)^2 da/dt=0 when x=4 So, the printed area is 4x6 and the page is 6x18 Now see what you can do with the other problem
*March 5, 2015*

**ALGEBRA**

the half-circle of radius 5 has area 12.5 pi = 39.27 The lower box has height 8cm and with equal to the diameter of the circle, or 10, so its area is 10x8 = 80 39.27+80 = 119.27 Maybe I misread the dimensions. I see no valid choice. Anyway, maybe you woked it out right.
*March 5, 2015*

**Pre-Algebra**

As you know, the diagonal of a 2-D rectangle is √(x^2+y^2) Similarly, the diagonal of a rectangular prism is √(x^2+y^2+z^2) So, since you have x,y,z just plug them in: √(144+16+81) = √241 = 15.52 Looks like B to me.
*March 5, 2015*

**math**

5.75 * 175 * 1.61 = 1620 km
*March 5, 2015*

**Geometry**

No way to know each base. The trapezoid could be a rectangle 32x16, with both bases 32. Or, it could be a triangle, with top base=0 and bottom base=64. Or anything in between. All we know is that (b+B)/2 = 32
*March 5, 2015*

**math**

If the distance each way is x, then since speed = distance/time, 2x/(x/24 + x/12) = 2x/(3x/24) = 16
*March 5, 2015*

**chemistry**

2Al + 6HCl = 2AlCl3 + 3H2 So, how many moles in 2.7g of Al? You get 3/2 times that many moles of H2
*March 5, 2015*

**Algebra**

P(1+0.0345/4)^(4*2.5) = 11000 P = 10094.75
*March 5, 2015*

**Calculus**

If the fenced side is y and the wall sides are x, then we have xy = 480 and the cost is c = 12*2x + 8y = 24x + 8(480/x) = 24x + 3840/x dc/dx = 24(x^2-160)/x^2 dc/dx = 0 when x = 4√10 So, the garden is 4√10 by 120/√10
*March 5, 2015*

**Calculus**

As with most such problems, I expect you will find that the maximum area occurs on a square page. But, let's check it out. If the dimensions are x and y, we have (x-4)(y-4) = 16 a = xy = x*(16/(x-4) + 4) da/dx = 4x(x-8)/(x-4)^2 da/dx = 0 when x=8 So, the page is 8x8, with ...
*March 5, 2015*

**delgado community college**

The seawater's mass is 1.025 g/ml * 1000ml = 1025g the bromine's mass is .000065 * 1025 = 0.066625 g
*March 5, 2015*

**math**

the new area is (7*2)(4*2) Note that the area grows by the square of the magnification ratio
*March 5, 2015*

**algebra**

each ball has volume 4/3 pi * (2.54/2)^3 = 8.58 cm^3 There are 15 of them, so add 15 times that to the volume of water, which is pi * 3^2 * 9 = 254.47 Note how they coyly avoided saying the balls were 1" in diameter. Saved you the step of doing the conversion to cm.
*March 5, 2015*

**Math**

B is the complement of A, so B/A = 69/21 =23/7 Surprise, surprise!
*March 4, 2015*

**MATH**

no, there are 2 halves in 1. There are 4/5 halves in 2/5.
*March 4, 2015*

**Calculus (Partial Derivatives)**

dH/dt = ∂H/∂x dx/dt + ∂H/∂y dy/dt ∂H/∂x = -13/(2√x) ∂H/∂y = 0.3√(0.1y+20) dx/dt = 300 dy/dt = 10/√y So, plug and chug, with t=2
*March 4, 2015*

**math**

divide the change in the dependent by the change in the independent. For #2, that is 2/1,2/1,2/1,... xo it's linear For #2, that's 9/2, 9/3, 9/3, 9/3, 9/3 They are all the same except for the first, so unless there's a typo, it's not linear.
*March 4, 2015*

**trig - eh?**

tan(40/90) ? That's an odd notation. Is there some reason why you did not say 4/9? Evan that seems unusual for such a problem.
*March 4, 2015*

**Algebra difference quotient**

f(x+h)-f(x) = 2(x+h)-3/4 - (2x-3/4) = 2h So, the quotient is 2h/h = 2 Next, f(x+h)-f(x) = 3(x+h-1)^2 - 3(x-1)^2 = 3(x^2 + 2(h-1)x + (h-1)^2) - 3(x^2-2x+1) = 3h(2x+h-2) Divide that by h, and you have 3(2x+h-2) = 6x+3h-6
*March 4, 2015*

**Geometry**

since the triangle are congruent, we know that EI=UA, so 3x-4 = 2x+3; x=7 Now you can find all the sides, and so on. For the other, you have FA = GR = 24 So, GR:RC:GC = 2:1:3 = 24:12:36 and now you can go on.
*March 4, 2015*

**math**

8sin2x + 2cosx - 5 = 0 16 sinx cosx + 2cosx - 5 = 0 16sinx cosx = 5 - 2cosx 256 sin^2x cos^2x = 25 - 20cosx + 4cos^2x 256cos^2x - 256cos^4x = 25 - 20cosx + 4cos^2x 256cos^4x - 252cos^2x - 20cosx + 25 = 0 I don't see any easy roots there, so a graphical approach may be your...
*March 4, 2015*

**trig equations**

cos^2x-5sinx+5=0 1-sin^2x - 5sinx + 5 = 0 sin^2x + 5sinx - 6 = 0 (sinx+6)(sinx-1) = 0 so, sinx = -1/6 or 1 Now you can find x. Watch for signs in quadrants.
*March 4, 2015*

**algebra**

did you check your earlier post? http://www.jiskha.com/display.cgi?id=1425519642 Was there something unclear there?
*March 4, 2015*

**algebra**

revenue = price * quantity, so if there are x price decreases, we have revenue is y = (6-.25x)(330+15x) = -3.75x^2 + 7.5x + 1980 Now, just find the vertex of that parabola to get the maximum revenue and its price. Recall that for ax^2+bx+c the vertex is at x = -b/2a
*March 4, 2015*

**Math (advice)**

A good rule of thumb is, that if you have a question, there are others who do also. Evidently they were also too shy to ask it, so you should. the teacher knows this, too, so he'd be much happier if someone (you!) asks the question, since it's easier to explain it to a...
*March 4, 2015*