Thursday

March 5, 2015

March 5, 2015

Total # Posts: 29,362

**Math**

5-2*(6-(4+2))=5 Still thinkin' on the other one
*January 22, 2015*

**help math algebra**

for addition and subtraction, just collect terms with the same power (7a^5-a^3)-(a^5-5a^3-1) 7a^5-a^3-a^5+5a^3+1) (7-1)a^5 + (-1+5)a^3 + 1 4a^5 + 4a^3 + 1 When multiplying, use the distributive property (5x-3)(x^3-5x+2) (5x)(x^3-5x+2) - 3(x^3-5x+2) 5x^4-25x^2+10x-3x^3+15x-6 5x...
*January 22, 2015*

**Math**

1.50n + 2.00+1.25n
*January 22, 2015*

**Math**

standard form does not involve an intercept. The standard form for a line is Ax + By = C You have done it correctly. And, the intercept was given. It is zero.
*January 22, 2015*

**geometry**

has 5 sides of length 9 ft each
*January 22, 2015*

**geometry**

KL/JL = tan 24° you know KL, so now just solve for JL.
*January 22, 2015*

**Algebra 1**

Thank you
*January 22, 2015*

**Algebra 1**

Does the table represent an expoential function? x 1 2 3 4 y -1 -8 -27 -64 yes or no my answer is no
*January 22, 2015*

**math**

T5/T2 = r^3, so r^3 = (x^4/8)/(4x) = x^3/32 r = x/∛32 I suspect a typo, but we'll go with what we have. T10 = T5 r^5 = x^4/8 * x^5/(256∛2) = x^9/(2048∛2)
*January 22, 2015*

**Math**

It's good. You don't really need the leading "-" if you say a is negative, but having it there does signal your intentions clearly.
*January 22, 2015*

**College Algebra**

read the "trajectory" article at wikipedia.
*January 22, 2015*

**Math**

#1 is always true. Draw a triangle. Extend one side so that you can see the interior and exterior angles. They add up to a straight line (the extended side). Thus they are supplementary. #2 sometimes #3 never Have you tried drawing the triangles to illustrate the ideas?
*January 22, 2015*

**hw dump**

I've done a couple. How abut you show some work of your own, rather than just doing a homework dump?
*January 22, 2015*

**math**

2x+(3x-(4x-5))=x/4-2 2x+(3x-4x+5)=x/4-2 2x+(-x+5) = x/4 - 2 x+5 = x/4 - 2 3x/4 = -7 x = -28/3
*January 22, 2015*

**math - bogus**

x^2 + y^2 = 20 xy = 192(39.37^2) = 297600 Clearly x and y are less than 20. No way could their product be so huge. I suspect a typo.
*January 22, 2015*

**math**

The nth row has 35-4(n-1) = 39-4n plants. So, what is the largest multiple of 4 less than 39? Subtract that from 39. Or, just divide 39 by 4. The remainder is how many plants are in the last row. It is the row that is too short to subtract 4 from it again.
*January 22, 2015*

**calculus**

v(t) = ds/dt = 50-4t find t when v=0
*January 22, 2015*

**math**

.99(1.05)
*January 22, 2015*

**Geometry**

If the longer ladder is x, and reaches up y feet on the wall, then 8.3^2 + y^2 = x^2 8.3^2 + (y-9.1)^2 = (x-7.2)^2 x = 17.9
*January 22, 2015*

**Geometry**

a general circle is (x-h)^2 + (y-k)^2 = r^2 since the center is on y=2x+1, k=2h+1 (x-h)^2 + (y-2h-1)^2 = r^2 Since the circle passes through (4,5), (4-h)^2 + (5-2h-1)^2 = r^2 Since the circle is tangent to the y-axis, h=r, so (4-r)^2 + (4-2r)^2 = r^2 r = 2 or 4 You can now ...
*January 22, 2015*

**Math-please help**

If there are x seats at $5, then income = 5x + 9(500-x) expenses = 4000 profit = income - expenses
*January 21, 2015*

**Math**

Since c = 2pi r, r = c/2pi a = pi r^2 = pi(c/2pi)^2 = c^2/4pi so, plug in your value for c.
*January 21, 2015*

**math**

2/3 = 16/x x = 24
*January 21, 2015*

**math**

The ratio is clearly 24:5. So, QR:10 = 24:5 45:FG = 24:5
*January 21, 2015*

**Math**

94+67+83+93 = 337 5*90 = 450 Looks like she needed a 113 on the final test. I suspect a typo. Fix it and redo the math.
*January 21, 2015*

**Math please help**

(1/3 mi)/(1/15 hr) = 5 mi/hr (1/5 mi)/(1/30 hr) = 6 mi/hr I suspect you can finish it off from here, eh?
*January 21, 2015*

**6th Grade Math**

Either way is right, but it looks like you're both wrong 6 3/8 + 2 1/16 = 8 7/16 8 7/16 + 4 3/4 = 13 3/16 He should have added the mixed numbers on the right, then subtracted the sum: 4 3/4 + 2 1/16 = 6 13/16 So, 6 3/8 = t - 6 13/16 6 3/8 + 6 13/16 = 13 3/16 Guess you'...
*January 21, 2015*

**Math**

.85*.87 + .15*.85 = .867 Sounds right, since the final score is just about the same as the average so far.
*January 21, 2015*

**Algebra 1**

My answer is 4.5 x 10^9
*January 21, 2015*

**Algebra 1**

What is the correct simplification of 5.4x 10^12/ 1.2x 10^3 written in scientific notation?
*January 21, 2015*

**math**

N = kA/P If A is doubled, so in N. So, (D)
*January 21, 2015*

**math**

f(x) = 15 + 4.24(x/120)
*January 21, 2015*

**Algebra 1**

thank you
*January 21, 2015*

**Algebra 1**

clearly, it's 36(6x^5)
*January 21, 2015*

**Algebra 1**

A fraction reduces to 36. If it's denominator is 6x^5,what is it's numerator?
*January 21, 2015*

**Math**

4^.5x = (4^.5)^x = 2^x This is easy to see, since log_4(x^2) = 2log_4(x) = log_2(x)
*January 21, 2015*

**Math**

36:(36+28+16) = 36:80 = ?
*January 21, 2015*

**precal**

5r + 3(r+4) = 52
*January 21, 2015*

**Physics**

when they act in the same direction.
*January 21, 2015*

**math**

it will travel 1km in 1km. If it takes 1000 hops, it will travel 2km It will take 500 hops to travel 1km.
*January 21, 2015*

**math**

(-2,12)
*January 21, 2015*

**Algebra II**

I don't think you really want a "z" floating around. I'll assume you meant a "2". The first two are just straightforward multiplication and division. You can surely do those. (g◦(h◦f))(-4) = g(h(f(-4)) = g(h(-2)) = g(9) = -33
*January 20, 2015*

**Calculus: Intergration**

This kind of stuff usually works best with a trig substitution. Note that 4x-x^2 = 4-(x-2)^2 So, let u = x-2 and you have du/(4-u^2)^(3/2) Now, let u = 2sinθ du = 2cosθ dθ 4-u^2 = 4-4sin^2θ = 4cos^2θ Now the integrand is 2cosθ dθ/8cos^3θ...
*January 20, 2015*

**Trig/Precalc #1**

If we call the two circles C1 and C2, respectively, then note that (2,1) lies on C2. So, the external tangents will be bisected by the line joining the two centers. Note that both circles are tangent to the x-axis. So, figure the slope of the line joining the two centers, and ...
*January 20, 2015*

**Pre-Calc**

This is just algebra I. You want x where -76x^3+4830x^2-320000 = 800000 I suggest a graphing utility, as this has no rational roots.
*January 20, 2015*

**Pre-Calc**

the circumference of the circle is 100-x. So, the area is a = (x/4)^2 + π((100-x)/2π)^2 This is just a parabola, so its vertex (minimum area) is easy to find. For the maximum area, compare a at the ends of the domain.
*January 20, 2015*

**Math**

2:3 = 18:x so, x=27, the no. of blue pens. Now, you have to decide the color(s) of the 3 pens removed to figure the new ratio.
*January 20, 2015*

**Pre-Calc**

the page is clearly 4" wider and higher than the print. That means that (x-4)(y-4) = 30 y-4 = 30/(x-4) y = 4+30/(x-4) The area A is xy, so A = x(4+30/(x-4)) = x(4(x-4)+30)/(x-4) = x(4x-16+30)/(x-4) = 2x(2x+7)/(x-4) I think the rest is fairly straightforward
*January 20, 2015*

**Math**

Piece of paper is 0.0075 inch thick. How many sheets of paper will be in a stack that is 2.25 inches high?
*January 20, 2015*

**Math**

Please help!!!!!!!!!!!!!
*January 20, 2015*

**Math**

First week sells 6 chairs for $80 each. Next week, if chair is not sold, it will sell for 0.85 times previous week's price. Store needs to sell 6 chairs for total of $270 to make a profit. what is the last week in which all 6 chairs could be sold so that the store makes a ...
*January 20, 2015*

**Science**

s = vt - 1/2 at^2 20t - 1/2 at^2 = 100 Also, v = Vo-at, so 20-at = 0 at = 20 so, 20t - 1/2(20)t = 100 t=10 So, a = 2 F = ma = 800*2 = 1600N
*January 20, 2015*

**Coll Algebra**

If the short side is x, then x + 2x + 2x = 30
*January 20, 2015*

**Math**

In a way. To say that the single equation on the left is the "same" as the two other equations is not technically true, but it does have the same solutions as the others. Actually, there is no value of x such that x=0 and x+6=0. x(x+6) = 0 is true if x=0 OR if x+6=0...
*January 20, 2015*

**physics**

F = ma You have m and a.
*January 20, 2015*

**algebra**

add the exponents of all the variables in each term. The degrees can be easily seen to be 4, 5, 5, 0 The degree of the polynomial is the highest degree among the terms. In this case, 5.
*January 20, 2015*

**Math**

No. In 1 minute, 1/20 of the tub fills, while 1/15 drains. So, the net rate of filling is 1/20 - 1/15 = -1/60 So, in 60 minutes the full tub drains. This assumes, of course, that the drain is opened when the tub is completely full.
*January 20, 2015*

**Math-Polynomials/Zeros**

if you're talking about quadratics with no constant term, then they look like y = ax^2+bx That's just y = x(ax+b) and it has two real roots (0 and -b/a). No complex roots are possible.
*January 20, 2015*

**Math**

what, no ideas? Check the discriminant. ax^2+bx+c has only complex roots if b^2-4ac < 0
*January 19, 2015*

**Math**

I gotta think you meant f(x) = x^4-4x^2+k we know that x^4-4x^2 = x^2(x^2-4) = x^2(x-2)(x+2) so it has 3 real roots, one repeated. x^4-4x^2+4 = (x^2-4)^2 = (x-2)^2(x+2)^2 so it has 4 real roots (2 distinct) x^4-4x^2+k has 4 distinct real roots if 0 < k < 4 x^4-4x^2+k has...
*January 19, 2015*

**math**

incorrect. |x| is never negative.
*January 19, 2015*

**Math**

f(x) = (x-1)(x-3) So, the vertex is at x=2. The restriction on the domain ensures that we only have the left branch of the parabola, so f is 1-to-1 and will have an inverse. x = y^2-4y+3 x = (y-2)^2 - 1 x+1 = (y-2)^2 y-2 = ±√(x+1) we only want the left branch, so ...
*January 19, 2015*

**math**

I assume "this region" means the green, less the hole. Since the area of a circle is pi r^2, just subtract the small area from the large area.
*January 19, 2015*

**Math**

1/20 - 1/15 = -1/60 So, 1/60 of the tub drains every minute. . . .
*January 19, 2015*

**Math**

f(x)=x^2+3x+2/x+1 = (x+1)(x+2)/(x+1) does not have a vertical asymptote. It just has a hole at x = -1 We want something with (x-3) on top, and (x+1) in the bottom, so f(x) = (x-3)/(x+1) has the zero and the vertical asymptote. Note that for large x, f(x) -> 1. So, we can ...
*January 19, 2015*

**calc**

nope. Just as d/dx x^n = n x^(n-1) ∫n x^(n-1) dx = x^n or, as it is ually written, ∫x^n dx = 1/(n+1) x^(n+1) So, ∫x^(1/2) dx = 2/3 x^(3/2) see when you take the derivative of that, that you get x^(1/2)
*January 19, 2015*

**calc**

just expand the expression (3+x)x^(1/2) = 3x^(1/2) + x^(3/2) then just use the power rule on each term.
*January 19, 2015*

**ap calc**

yes. so you are left with just (x^2-4) as usual, (√x)^2 = √(x^2) = x
*January 19, 2015*

**Algebra**

terms are separated by + or - signs just add exponents of all the factors. The degrees of the terms are thus easily seen to be 4,5,5,0
*January 19, 2015*

**arithmetic**

First, balance parens and brackets: 0.2[5.28-(0.7)] 0.2[5.28-0.7] 0.2[4.58] 0.916 Or, you can distribute the multiplication: 0.2[5.28-(0.7)] 0.2(5.28) - 0.2(0.7) 1.056 - 0.14 0.916
*January 19, 2015*

**math**

3 - 5 = -2 -2 is not in N
*January 19, 2015*

**calculus**

(If the base has side x and the box has height h, the area 2x^2 + 4xh = 600, h = (300-x^2)/2x So, the volume v = x^2 h = 150x - x^3/2 dv/dx = 150 - 3/2 x^2 So, dv/dx=0 when x^2 = 100 I think you can take it from there, ok?
*January 19, 2015*

**math**

just start writing down the facts as math. n = 2d+2 eight fewer quarters than twice the number of nickels q = 2n-8 value of the quarters was $1.60 more than four times the value of the nickels and dimes together 25q = 4(10d+5n)+160 Putting all that together, we have n = 2d+2 q...
*January 19, 2015*

**AP Calc**

well, it's clear that f(x) = x^2-3x+c and you know that f(2) = 5, so just find c.
*January 19, 2015*

**Geometery**

you just have to remember that the angles of a triangle add up to 180 degrees. So, with angles M+P+N = 180 x + 2x-1 + 2x+1 = 180 Now it's easy to find x, the evaluate the actual angle measures.
*January 19, 2015*

**Calculus**

dv/dt = pi r^2 dh/dt just plug in your numbers to find dh/dt
*January 19, 2015*

**grade 7 math**

by 3:00, the tug has traveled upstream 3*15 = 45 km, so it is now 20 km upstream from B. The boats' relative speed is 35 km/hr, so how long does it take the foil to make up the 20 km head start of the tug? Add that time to 3:00.
*January 19, 2015*

**grade 7 math**

Melvin travels 30 km/hr faster than Jessie. So, how long does it take to cover extra 15 km?
*January 19, 2015*

**Calculus**

first, it's "bacteria are" growing. You are told that the for a population p of these bacteria, dp/dt = kp so, that means that dp/p = k dt ln p = kt + c p(t) = c e^(kt) p(0) = c = 50 So, p(t) = 50e^(kt) p(1) = 200, so 50e^k = 200 k = ln 4 p(t) = 50e^(ln4 t) Or, ...
*January 19, 2015*

**math**

just plug the coordinates into your rotation matrix Rx: 1 0 0 0 cosθ -sinθ 0 sinθ cosθ A' = Rx * AT and so on
*January 19, 2015*

**math**

as usual, just recall that distance = speed * time Now just plug in your numbers...
*January 19, 2015*

**Math**

(2+5i)+(-2-4i) 2+5i-2-4i 2-2 + (5-4)i 0+i You can omit the 0+ and write is as just i, but if you want to make it clear that you are in fact working with a complex number, the standard form a+bi is used. Although, just the presence of "i" itself makes it evident that ...
*January 18, 2015*

**math**

You want 40/2 (2a+39d) where 12/2 (2a+11d) = 186 a+19d = 83 Just solve those last two for a and d, then plug them into the first formula.
*January 18, 2015*

**Algebra**

If y = x^3-2 y+2 = x^3 x = ∛(y+2) That means that ∛(x+2) is the inverse of x^3-2 To check, you know that f(f-1(x)) = x and f-1(f(x)) = x So, if g(x) = f-1(x), f(g) = g^3-2 = (∛(x+2))^3 - 2 = x+2-2 = x and the same for g(f).
*January 18, 2015*

**Math**

all look correct to me.
*January 18, 2015*

**Calc**

the distance d is found using d^2 = x^2 + (y+3)^2 Since x = -y^2, that means d^2 = y^4 + (y+3)^2 Now all you do is find dd/dy = 0 and that is where the distance is minimum. Then evaluate x = -y^2 and you have your point.
*January 18, 2015*

**maths**

2/5 * 1/2 = 1/5 1/3 * 3/4 = 1/4 1/4 is 25% more than 1/5
*January 18, 2015*

**Physical Science**

good question. Find the chemical formula for the sweetener, divide the mass of the package by the molecular mass of the formula, and multiply by Avogadro's Number. If that's all gobbledegook to you, then clearly you haven't been presented with the means to do the ...
*January 18, 2015*

**physics**

since v = at, clearly it is 1.1i + 0.56*4.8j
*January 18, 2015*

**Math**

No idea what a pattern of 1,3,5 means. You do know that y = a(x+4)^2 + 5 It opens up, so a > 0. Use that with the weird pattern data to finish it off.
*January 18, 2015*

**Math**

4+5+6+7+9 = 31 So, the new sum will be between 34 and 41 Divide both of those by 6 and match it up with the choices.
*January 18, 2015*

**Math**

I think you can easily fill in the first two. Simple multiplication and division, right? (g◦(h◦f))(-4) = g(h(f(-4))) = g(h(-2)) = g(9) = -33 Note that h(x) = (x-1)^2. So, f(h(a+4)) = f((a+4-1)^2)) = f((a+3)^2) = (a+3)^2+2 You can expand that if you want.
*January 18, 2015*

**Physics**

just plug your numbers into y = tanθ x - g/(2(v cosθ)^2) x^2 and solve for x when y = -125
*January 18, 2015*

**algebra**

64+35-14 = 86 So, two buses can hold the lot. Even chaperones! (A ∪ B) = {6,7,8,9,10,11} So, (A ∪ B) ∩ C = {8,10}
*January 18, 2015*

**algebra**

just recall the rule that total = one + other - both. So, counting just the ones taking something, you have 20+32-6 = 46 So, the other 4 took nothing.
*January 18, 2015*

**Mathematics arithmetic sequence**

a+d + a+5d = 4 a+2d = a+10d + 24 Collect terms and solve for a and d. Then write a,a+d,a+2d
*January 18, 2015*

**Algebra 1a**

just add the exponents. x^2 = xx x^3 = xxx x^4 = xxxx So, adding the exponents just means counting how many times you have x multiplied together.
*January 18, 2015*

**Algebra 1a**

(xy)^-3 = 1/(x^3y^3) (x^-5y)^3 = x^-15y^3 = y^3/x^15 So, you have 1/(x^3y^3) / (y^3/x^15) = 1/(x^3y^3) * x^15/y^3 = x^15/(x^3y^3y^3) = x^15/(x^3y^6) = x^12/y^6
*January 18, 2015*

**Physics**

#1 is correct
*January 18, 2015*

Pages: <<Prev | 4 | 5 | 6 | 7 | 8 | 9 | 10 | 11 | **12** | 13 | 14 | 15 | 16 | 17 | 18 | Next>>