Saturday

May 28, 2016
Total # Posts: 40,861

**Brokport**

If the amounts invested are x,y,z then they have told us that x+y+z = 180000 .08x + .06y + .09z = 8520 .08x = 6 * .06y I suspect a typo, since even if the whole amount were invested at 6%, the interest would be 10,800. In other words, there's no way to solve these ...
*April 25, 2016*

**Algebra SOLVING LINEAR EQUATIONS AND INEQUALITIES**

#1 you can solve for 1/x and 1/y in the usual ways: elimination: double the 1st and subtract 6/x - 4/y = 28 6/x + 3/y = 7 ------------------- 7/y = -21 1/y = -3 then, 1/x = 8/3 or, y = -1/3 and x = 3/8 using substitution, 3/x = 14+2/y 2(14+2/y) + 3/y = 7 28 + 4/y + 3/y = 7 7/y...
*April 25, 2016*

**Math**

(2)(33)(19)[2(33)(19)-33-19] (2)(33)(19)[1254-33-19] (2)(33)(19)[1202] 1,507,308
*April 25, 2016*

**geometry**

Find the volume in terms of pi of a sphere with a surface area of 9 pi sq ft
*April 24, 2016*

**Math2**

f(x) = 3x^2 - 12x + 2 the y-intercept is clearly at 2. x-intercepts? where y=0. Use the quadratic formula to get x = 2±√(10/3)
*April 24, 2016*

**Math**

3 : 2 = 1 : 2/3 = 5 : 10/3
*April 24, 2016*

**Algebra**

slope = 15/9 = 5/3 y-4 = 5/3 (x-6) y = 5/3 x - 6 The lines are not perpendicular, because the slopes are not negative reciprocals. The line through the points is perpendicular to y = -3/5 x - 6 However, they do both have a y-intercept of -6.
*April 24, 2016*

**Math**

using the vertex form, y = a(x-3)^2 - 6 using the point given, 10 = a(-1-3)^2 - 6 10 = 16a-6 a = 1 y = (x-3)^2 - 6 = x^2-6x+3
*April 24, 2016*

**math**

clearly getting a 44 (lower than any of her scores so far) will not raise her grade!! Add up the total points she has so far. A "B" average for 7 tests needs at least 80*7 = 560 points. So, how many does she need to get there?
*April 24, 2016*

**Math**

The sequence is not geometric! 17111/19 = 900.5789 1539121/17111 = 89.9492 the ratio is not constant. Fix it, find the common ratio r, and then a6 = 19*r^5
*April 24, 2016*

**Calculus quick question pls**

huh? huh? How can you say that? x^2 is always positive, right? At least it is never negative. So, x^2+1 is always positive, and always at least 1. So, dividing by x^2+1 will never yield a vertical asymptote.
*April 24, 2016*

**Calculus quick question pls**

vertical asymptotes occur when you try to divide by zero. x^2+1 is never zero. You solved for x^2-1 = 0
*April 24, 2016*

**Math**

surely you can do the math. ∫[2,3] (3x^2)-(x^4-10x^2+36) dx = ∫[2,3] -x^4 + 13x^2 - 36 dx = -x^5/5 + 13/3 x^3 - 36x [2,3] = (-243/5 + 13*27/3 - 36*3)-(-32/5 + 13/3 * 8 - 36*2) = 62/15 double that for 124/15 Now you need to ask your teacher how 1436/15 can be right...
*April 24, 2016*

**Math**

what's the trouble? You are adding up a bunch of thin rectangles, with width dx and height the distance between the curves. The curves intersect at (±2,12) and (±3,27). Since both functions are even, we can use symmetry and use a = 2∫[2,3] (3x^2)-(x^4-...
*April 24, 2016*

**Absolute and relative change**

Nope. 1.07x = 16.05 the 16.05 includes the tax, right?
*April 24, 2016*

**math**

30P4 = 30*29*28*27
*April 24, 2016*

**math**

4! = 24
*April 24, 2016*

**Maths**

words, words, words. This is math, so use math 4x^2-5x+7 terms are separated by + and - signs, so there are three terms
*April 24, 2016*

**math(counting)**

if all you want is permutations of the letters, MAIRE: 5! = 120 OISEAU: 6! = 720 If you want your anagrams to be actual words, then of course there will be a lot fewer. Better get a good French dictionary.
*April 24, 2016*

**math(counting)**

what does "near" mean? There are only 5 kids, after all.
*April 24, 2016*

**Math**

If she started out with x, and was left with 15 more than half of what she started with, then x - x/8 - x/5 - x/10 = x/2 + 15 x = 200
*April 24, 2016*

**Ic**

I assume you can handle the java syntax, but the flow of logic is integer a,b,c,option,err print "Enter two integers: " read a,b print "Enter option: 1: add 2: subtract 3: multiply 4: divide " read option err=0 case option { 1: c=a+b 2: c=a-b 3: c=a*b 4: if...
*April 24, 2016*

**math**

233489
*April 24, 2016*

**precalc**

oops. tanϕ = -1/√15 tan(θ+ϕ) = (√8 - 1/√15)/(1-(√8)(-1/√15)) = (32√2 - 9√15)/7 = 1.48
*April 24, 2016*

**precalc**

cos(θ) = −1/3 QIII, so tanθ = √8 sin(ϕ) = 1/4 in QII, so tanϕ = -√15/4 tan(θ+ϕ) = (tanθ + tanϕ)/(1-tanθ tanϕ) = (√8 - √15/4)/(1-(√8)(-√15/4)) = (18√15 - 31√2)/52
*April 24, 2016*

**Absolute and relative change**

6%
*April 23, 2016*

**Precal/Math**

log(3-2x)-log(x+24)=0 log (3-2x)/(x+24) = 0 (3-2x)/(x+24) = 1 3-2x = x+24 x = -7
*April 23, 2016*

**geometry**

Thye area of a regular hexagon is 38 cm^2. What is the area of a regular hexagon with sides 4 times as long?
*April 23, 2016*

**Calculus!!**

dy/dx = xy/2 dy/y = x/2 dx ln y = 1/4 x^2 + c y = c e^(x^2/4)
*April 23, 2016*

**AP Calculus**

A yes, if f"=0 and f'≠0 B find g'(0) Then the tangent line is y-5 = g'(0) (x-0)
*April 23, 2016*

**Calculus**

The large cylinder has a cross-section 4 times that of the smaller one, so its water level rises 1/4 as fast as it falls in the smaller one.
*April 23, 2016*

**Calculus**

A a = ∫[0,5] 3/e^x dx = 3 - 3/e^5 B now you want c such that ∫[0,c] 3/e^x dx = ∫[c,5] 3/e^x dx 3 - 3/e^c = 3/e^c - 3/e^5 6/e^c = 3 + 3/e^5 e^c = 2/(1+e^-5) c = ln 2/(1+e^-5) C each semicircle has diameter equal to y. Adding up all those thin slices of ...
*April 23, 2016*

**math**

not sure what H= 10 inches from measuring the perimeter of the base is 2 inches means
*April 23, 2016*

**AP Calculus**

A (1+x)y^3 + (x^4)y - 85 = 0 y^3 + 3(1+x)y^2y' + 4x^3y + x^4y' = 0 y' = -(y^3+4x^3y)/(3(1+x)+x^4) = -(y^3+4x^3y)/(x^4+3x+3) B y'(3) = -109/93 So, the tangent line is y-1 = -109/93 (x-3) C y(3) = 1, so g(1) = 3 g'(1) = 1/y'(3) = -93/109
*April 23, 2016*

**GEOMETRY HELP PLEASE**

well, maybe not. It just might mean that AB is shorter than CD. In that case, h = 3√5 and x = -2, so CD=8 and AB=4, making the area (8+4)/2 (3√5) = 18√5
*April 23, 2016*

**GEOMETRY HELP PLEASE**

Drop altitudes CE and DF The trapezoid now can be seen to be rectangle CDEF and two right triangles of height h and base x. h^2+x^2 = 7^2 h^2 + (8+x)^2 = 9^2 Hmmm. I get x = -2 I guess I have drawn the figure incorrectly. Maybe you can use my ideas in your own drawing.
*April 23, 2016*

**math**

L = K∛M 15 = K∛125 15 = K*5 3 = K
*April 23, 2016*

**converting parabolic equations**

x = 8(y-1)^2-15 x = 8(y^2-2y+1)-15 x = 8y^2-16y-7
*April 23, 2016*

**physics**

So, if their ratio is 1.1, you have determined that Ve/Vb = 1.1 since distance = speed * time 3*10^-9 Ve = 1.1*3*10^-9 Vb = 3.3*10^9 Vb So, De-Db = 0.3*10^-9 Vb
*April 23, 2016*

**Calculus**

c'mon, you can do this. dm/dt = -0.22m dm/m = -0.22 dt ln(m) = -0.22t + c m = c e^(-0.22t) c is the initial amount, so m(t) = 20 e^(-0.22t) I'm sure you can find the half-life now, ok?
*April 23, 2016*

**Calculus**

we know the slope is dy/dx, so at (1,2), the slope is -3. The tangent line is thus y-2 = -3(x-1) y dy = -6x^2 dx 1/2 y^2 = -2x^3 + c y = √2 √(c-2x^3) you know that the domain of √ is non-negative numbers, so c-2x^3 >= 0 2x^3 <= c x <= ∛(c/2)
*April 23, 2016*

**Calculus**

recall that for parametric curves, the curvature is x'y" - x"y' ----------------- (x'^2 + y'^2)^(3/2) so, for your function, that is x' = 2t x" = 2 y' = 2t^2 y" = 4t (2t)(4t)-(2)(2t^2) ----------------------- = 4t^2/(8t^3(1+t^2)) = 1...
*April 23, 2016*

**Pre-Calc**

the domain of log(u) is u>0, regardless of the base of the logs. So, you need 3x-8 >0 x > 8/3 or, (8/3,∞)
*April 23, 2016*

**maths**

also, we have (x+y)(x-y) = 12 The pairs of factors of 12 are 1,12 2,6 3,4 x+y=6 x-y=2 x=4,y=2
*April 23, 2016*

**math**

is the light source between the center of the circle and the object's line, or on the far side of the line? That will affect where the shadow falls. Also, you have not indicated the speed of the object, dx/dt. In either case, label the light source L, the object P and the ...
*April 23, 2016*

**geometry (check answers)**

how can a rhombus be sometimes a square, but never a rectangle? a square is always a rhombus a rhombus is sometimes a square so, a rhombus is sometimes a rectangle, since a square is always a rectangle. as for trapezoids, some places define them as a quadrilateral with at ...
*April 22, 2016*

**Calculus I**

yes, so using symmetry, the area is a = 4∫[0,1] √(1-(x-1)^2) dx = π
*April 22, 2016*

**HELP ME MATH**

the perimeter is just the sum of the sides. So, that is large: 4x+2 + 7x+7 + 5x-4 = 16x+5 small: x+3 + 2x-5 + x+7 = 4x+5 large-small = (16x+5)-(4x+5) = 12x
*April 22, 2016*

**Precal**

but -2 is not a solution, since the domain of log(x) is x>0.
*April 22, 2016*

**Program**

assuming positive numbers, largest=0 for i=1 to 10 read val if (val > largest) largest=val end for print largest so, what change is needed if positive and negative numbers are allowed?
*April 22, 2016*

**Calculus I**

using shells of thickness dx, v = ∫[1,2] 2π(2-x)(x^3-1) dx using discs of thickness dy, v = ∫[1,8] π(2-∛y)^2 dy
*April 22, 2016*

**Calculus I**

each square of thickness dx has side 2y, so its area is 4y^2. Adding up all the thin squares, and using symmetry, v = 2∫[0,3] 4(9-x^2) dx
*April 22, 2016*

**Calculus I**

using shells of thickness dy, v = ∫[1,3]2πrh dy where r = y and h = x = (y-1)^2 v = ∫[1,3] 2πy(y-1)^2 dy = 40π/3 using discs (washers) of thickness dx, v = ∫[0,4]π(R^2-r^2) dx where R=3 and r=y=1+√x v = ∫[0,4]π(9-(1+√...
*April 22, 2016*

**math**

y = k/x^2 That is, x^2y = k is constant. So, you want y such that 19^2 * 9 = 17^2 * y
*April 22, 2016*

**Math**

come on. the angles of any quadrilateral sum to 360
*April 22, 2016*

**math**

what, can you not answer any of the guiding questions? I'll do #1 For x GB of data, the costs are Runfast: 25+10x BA&D: 18+15x now crank it out
*April 22, 2016*

**Algebra**

well, how many of the 52 cards are diamonds?
*April 22, 2016*

**Math**

since the opposite of "greater than or equal" is "less than", ~p = ~(x-2 >= 3) = x-2 < 3
*April 22, 2016*

**Trigonometry**

sinθ = tanθ cosθ, so tanθ - sinθ = tanθ (1-cosθ) squared that is tan^2θ (1-cosθ)^2 now we can add up the left side: tan^2θ(1-cosθ)^2 + (1-cosθ)^2 = (tan^2θ + 1)(1-cosθ)^2 = sec^2θ (1-cosθ)^2 on the...
*April 22, 2016*

**maths**

she walks on a heading, not a bearing. 50@25° = (50sin25°,50cos25°) = (21.13,45.32) add to that (200,0) and you end up at (221.13,45.32) the distance moved is thus 225.73m
*April 22, 2016*

**Math**

the sides are in the ratio 1:√3:2 multiply all those by 2.5 and you have a perimeter of (5/2)(3+√3)
*April 22, 2016*

**Math**

of the girls, brown: 4/5 long brown: 3/4 * 4/5 = 3/5
*April 22, 2016*

**College Algebra**

you know that an exponential function for the value v after t years looks like v = a e^(-kt) at t=0, v=26000, so v = 26000 e^(-kt) v(3) = 18000, so 26000 e^(-3k) = 18000 e^(-3k) = 9/13 -3k = ln(9/13) k = 0.1226 v(t) = 26000 e^(-0.1226t) That's all well and good, but what&#...
*April 21, 2016*

**Math**

depends on the size of the container, and how far the ball sank.
*April 21, 2016*

**Calculus**

if you draw the square at x, it has a side of length 2y, or 2√(64-x^2) So, the area of the square there is 4(64-x^2) The volume is the sum of all those squares of thickness dx, so v = ∫[0,8] 4(64-x^2) dx
*April 21, 2016*

**math**

it is just the base perimeter times the height: 150*5 = 750mm
*April 21, 2016*

**math**

the cost c for m messages is c(m) = 10.00 + 0.10m
*April 21, 2016*

**math**

175in:70yd = 175in:2520in = 1:14.4 175" is a pretty big drawing!
*April 21, 2016*

**Math**

x + 3x+4 = 12
*April 21, 2016*

**algebra HELP PLEASE**

72.8/280 = 13/50 = 0.26 pretty steep, since a 6% grade on a highway is considered a steep hill.
*April 21, 2016*

**Trigonometry**

sinθ = tanθ cosθ, so tanθ - sinθ = tanθ (1-cosθ) squared that is tan^2θ (1-cosθ)^2 now we can add up the left side: tan^2θ(1-cosθ)^2 + (1-cosθ)^2 = (tan^2θ + 1)(1-cosθ)^2 = sec^2θ (1-cosθ)^2 on the...
*April 21, 2016*

**Math**

correct
*April 21, 2016*

**math**

since (a-b)^2 = a^2-2ab+b^2, just divide the middle coefficient by 2 and square it. (9/2)^2 and 5^2
*April 21, 2016*

**geometry**

see the related questions. Same problem, different numbers.
*April 21, 2016*

**Math**

x+y = 19 (x+5)(y+5) = 208 now just solve for x and y.
*April 21, 2016*

**Math- PLEASE HELP**

using the fact that the acute angles of right triangles are complementary, and that vertical angles are equal, it's pretty easy to determine that HXZ is 31°
*April 21, 2016*

**math**

v = s^3 dv/dt = 3s^2 ds/dt so, 3s^2 ds/dt = 5/2 as you can see, the rate of change of the sides depends on the size of the cube.
*April 21, 2016*

**math**

v = 4π/3 r^3 dv/dt = 4π r^2 dr/dt when r=5, dv/dt = 4π * 25 * 1 = 100π cm^3/s
*April 21, 2016*

**Math**

those logs are just numbers, so x log19 - 5log19 = x log3 + 2log3 x(log19-log3) = 2log3 + 5log19 x = 2log3 + 5log19 -------------------- log19 - log3
*April 21, 2016*

**Math/Precal**

2(3^x) = 5 3^x = 5/2 x log3 = log(5/2) x = log(5/2) / log3
*April 21, 2016*

**math- pre calc**

or, in QII, x = -12/13
*April 21, 2016*

**Math**

http://www.wolframalpha.com/input/?i=y%3D3x%2B1 any function of the form y=mx+b where m and b are numbers, is just a straight line.
*April 21, 2016*

**Grammar**

#1 is better, but it's kind of useless. Of course usefulness is valuable, or it would not be useful. I can think of a context where #2 is syntactically correct, but it's kind of weird.
*April 21, 2016*

**Chemistry - Answer Check**

looks good to me.
*April 21, 2016*

**probability**

ok - now what?
*April 21, 2016*

**math**

EF = 3*BC = 18
*April 21, 2016*

**Math**

for that person and that speed, c(t) = 420t if the other factors are variable, then you need a more general function including weight and speed: c(w,s,t) no idea what that would look like, but c(154,15,t) = 420t
*April 21, 2016*

**math**

since y increases in steps of 3, the rule will be something like y = 3x + b
*April 21, 2016*

**Precalculs**

one way: Ax=B, so you want A^-1 so that x = A^-1 B A = (1 2 3) (-1 2 -1) (-1 1 -2) B = (7) (1) (-2) A^-1 = 1/2 * (3 -7 8) (1 -1 2) (-1 3 -4) Now if you supply the missing value in B, you can do the multiply and get (-1 1 2), the solution. http://www.wolframalpha.com/input/?i=...
*April 21, 2016*

**math**

15 <= 2.54W <= 17 9 <= 2.54L <= 11
*April 21, 2016*

**Precalculs**

one way: Ax=B, so you want A^-1 so that x = A^-1 B A = (1 2 3) (-1 2 -1) (-1 1 -2) B = (7) (?) (-2) A^-1 = 1/2 * (3 -7 8) (1 -1 2) (-1 3 -4) Now if you supply the missing value in B, you can do the multiply and get (x y z), the solution. Or, you can do row operations as in ...
*April 21, 2016*

**Algebra**

a+b+c = 54 c = a+b-4 3a = c+4 now start substituting to solve
*April 21, 2016*

**mathe**

that's a strange shaped Egyptian, but assuming he is pyramidal in shape, the height h is found using h^2 = (100/2)^2-50^2 huh? If the sides have a slant height of half the base width, the pyramid is totally flat!
*April 21, 2016*

**math**

V = H*W*L just multiply the dimensions. the result will be in units of m^3.
*April 21, 2016*

**Math**

I think C is farther out than D.
*April 21, 2016*

**Pre-Calc**

review your basic trig functions, and you will that the height h (in miles) can be found using h cotx - h coty = 3
*April 21, 2016*

**sequences and Series**

or check the related questions, where it has been solved.
*April 21, 2016*

**maths**

The amount donated starts at 10, then grows by 6 each week. So, the donated amounts are 10,16,22,28,... So, the donated amount for the nth week is an arithmetic progression with a = 10, d=6 The total donations in 20 weeks is thus S20 = 20/2 (2*10 + 19*6) = 1340 Add that to the...
*April 21, 2016*

**Ratios!!!!**

#8 2x+3x+4x = 63 x = 7 So, the actual quantities are 14:21:28 #9 3ab/27ab = 1/9 So, A,B,E are the equivalent ratios. D doesn't get rid of the a,b stuff. Odd how you could get B and E right, but miss A and pick D... #10 B and D only
*April 21, 2016*