Monday

March 30, 2015

March 30, 2015

Total # Posts: 30,183

**factorization**

3*5*7 = 105 now any multiple of 105 has them as prime factors.
*February 6, 2015*

**math**

y(6) = 2, so y = -2(x-6)+2 = 14-2x for 0 <= x <= 7
*February 6, 2015*

**math**

His increments each year were 0,x,2x,3x,4x, summing to 10x So, 10x = (32400-6000)/5
*February 6, 2015*

**algebra**

7x + 3x + 90 = 180 10x = 90 x = 9
*February 6, 2015*

**math**

correct does the answer i convert right? yes, but your English needs some work.
*February 6, 2015*

**math -hw help**

√3 s^2 = 3/2 s^2 = √3/2 clearly, s is not rational √3s^2 = 4√3 s^2 = 4 s = 2, a whole number. Now you try the other.
*February 5, 2015*

**geometry**

using similar triangles, 1.5/8 = FG/(32+8)
*February 5, 2015*

**algebra**

(2x+5)/(x^2+3x+2) - (x+4)/(x^2+3x+2) = (2x+5-x-4)/(x^2+3x+2) = (x+1) / (x+1)(x+2) = 1/(x+2)
*February 5, 2015*

**calculus**

y = 18/√x = 18x^(1/2) y = √(18/x) = 3√2 x^(1/2) Hard to tell what you meant, with that trailing √ sign. Do you write 2x+ to mean 2+x as well?
*February 5, 2015*

**Trig**

sinx/cosx(sinx+cosx/sinx×cosx) As written, it's sinx/cosx (sinx + (cosx/sinx) * cosx) which I doubt is what you meant. I expect you wanted (sinx/cosx)*((sinx+cosx)/(sinx*cosx)) = sec^2x (sinx+cosx) which is not secx. so, what do you really mean?
*February 5, 2015*

**Trig**

try tossing in some parentheses to make it clear just what is in the denominators...
*February 5, 2015*

**Calculus**

∫3√x (2x^2-4) dx If we let u = √x du = 1/2√x dx x = u^2 and we have ∫3√x (2x^2-4) dx ∫6x (2x^2-4) dx/2√x ∫6u^2(2u^4-4) du and that's just power stuff.
*February 5, 2015*

**math**

assuming positive integer n, n even and x negative cannot be evaluated in real numbers. As in ∜(-3) or √(-17) If we allow any real values for n, then things get more complicated. Of course, any negative power of 0 is undefined.
*February 5, 2015*

**Calculus**

the anti-derivative of the integral? You want to integrate twice? I'll just go with ∫(7tan^2x+12) dx ∫(7sec^2x-7+12) dx ∫(7sec^2x+5) dx and that's elementary, right
*February 5, 2015*

**Calculus**

(a) treat x and y as constants and you have 5x^4m^4 + y^3 (b) v' = 6t+8 v" = 6
*February 5, 2015*

**Calculus**

just divide each term by t^4 and you have 7/t^2 + 6/t^3 - 15/t^4 If that doesn't look familiar, think of it as 7t^-2 + 6t^-3 - 15t^-4 and then just apply the power rule as normal.
*February 5, 2015*

**Calculus**

f" = x^2 f' = 1/3 x^3 + c f'(0)=3, so c=3 and f' = 1/3 x^3 + 3 Now do that again going from f' to f.
*February 5, 2015*

**College Algebra**

in each case, all the 1st coordinates are the domain, and the 2nd coordinates are the range. If no element of the domain appears more than once, it is a function.
*February 5, 2015*

**Algebra**

If x is at 3%, the remainder (9000-x) is at 2.6%. So, add up the interest: .03x +.026(9000-x) = 250
*February 5, 2015*

**Math**

correct. It is a polynomial: y = -x^3
*February 5, 2015*

**algebra**

Assuming a month is 4 weeks, after x weeks, Sparky weighs 4 + (5/2)(x/4) So, when do you have 4 + 5/8 x = 14 That is the number of weeks, not months. Clearly it is linear, not exponential. (2.5 lbs/month)
*February 5, 2015*

**Maths**

a+b+c+d = 116 a+5 = b-4 = 3c = d/2 22,31,9,54
*February 5, 2015*

**Algebra II**

√-18x^5y^4 = √9x^4y^4√-2x = 3x^2y^2√2 i
*February 5, 2015*

**Calculus**

You have ∫√((x-1)/x^5) dx = ∫√((x-1)/x) 1/x^2 dx If you let u = (x-1)/x du = 1/x^2 dx and you have ∫√u du = (2/3) u^(3/2) = (2/3) ((x-1)/x)^(3/2) + C
*February 5, 2015*

**Maths Problem**

a = b+x m = b+6 a+b+m = 9x (b+x)+b+(b+6) = 9x 3b+x+6 = 9x 3b = 8x-6 b = (8x-6)/3 I guess you can finish it off now.
*February 5, 2015*

**chemistry**

PV=kT Since P is constant, V/T = k/P is constant if V is decreased by a factor of 361/400, T must decrease by the same factor to have the quotient remain constant. 298.15 * 361/400 = 269.08K It seems odd that the temperature goes down as the gas is squeezed, but remember that ...
*February 5, 2015*

**algebra**

5/(1/2 * 4/7) = 17.50
*February 5, 2015*

**math**

you want the slope of the line through (-3,-4) and (-5,6) That is 10/-2 = -5 Not quite sure why the negative time values. I guess that means minutes before noon.
*February 5, 2015*

**algebra**

37 @ $1 = $37 so, there are $14 extra. That means there were 14 $2 and 23 $1 coins.
*February 5, 2015*

**english?**

yes, that's sort of like English.
*February 5, 2015*

**math**

2(w + 2w+6) = 102
*February 5, 2015*

**math**

pi * (100/2pi)^2 = 2500/pi ft^2
*February 5, 2015*

**math**

correct, except for the spelling of phrase.
*February 5, 2015*

**math**

6(-3) = -18
*February 5, 2015*

**math**

since the angle from 240-180=60 feet away is 45 degrees, the height is 60. SO, if the angle is x, we have tan(x) = 60/240 Now just find x.
*February 4, 2015*

**math**

v(t) = 30-32t so, when is 30-32t = 10 (going up)? 30-32t = -10 (coming down)?
*February 4, 2015*

**Survey of Calculus**

That is correct.
*February 4, 2015*

**math**

must be 42
*February 4, 2015*

**solid**

CS=5, so the distance is √(25+144) = 13
*February 4, 2015*

**math - eh?**

and the answer is . . . 42
*February 4, 2015*

**Math**

Depends on the dimensions of the patio, and whether partial stones (after cutting) may be used. Just the simple answer is 24*100^2/30^2 = 267.7 stones.
*February 4, 2015*

**Math**

after h hours, you want √((18h)^2+(24h)^2) = 150 30h = 150 h = 5
*February 4, 2015*

**math 4th grade**

3 1/2 + (2+ 4 1/2) 3 1/2 + 4 1/2 + 2 3 + 1/2 + 2 + 4 + 1/2 3+2+4 + 1/2 + 1/2 9 + 1 10 or, you can rearrange the numbers in other ways.
*February 4, 2015*

**math**

If each notebook cost x, then 3x + 2*3.95 + 1.23 - 0.75 = 15.85 Now just solve for x.
*February 4, 2015*

**Math**

just start working your way from the inside out. How far do you get? for example, [(4x^2)^-1]^-2 = (4x^2)^(-1 * -2) = (4x^2)^2 = (4)^2 (x^2)^2 = 16x^4 do each factor like that and then simplify.
*February 4, 2015*

**Calculus 2**

you are correct. My bad. Good catch.
*February 4, 2015*

**Calculus 2**

∫ e^x √(1+e^(2x)) dx Note that if u = e^x, you have ∫ u√(1+u^2) du Now, if v = 1+u^2, dv = 2u du, and you have 1/2 ∫ √v dv That I think you can handle, eh?
*February 4, 2015*

**Math**

#1 x^2 = 2x-1 x = 1 #2 2(x-1) = 3x+1 x = -3
*February 4, 2015*

**Math**

(2+2+2)/2 - 2 = 1 2+2+2 - 2/2 = 5 2+2+2 + 2/2 = 7 2*2*2 + 2/2 = 9
*February 4, 2015*

**Math**

1 false a^x is exponential, for any positive a. #2 false, since 1^x is constant.
*February 4, 2015*

**Calculus 2**

(x^3+5x^2+12)/((x^2)(x^2+4)) = (x+2)/(x^2+4) + 3/x^2 So, ∫(x^3+5x^2+12)/((x^2)(x^2+4)) = ∫ (x+2)/(x^2+4) + 3/x^2 dx = 1/2 log(x^2+4) + arctan(x/2) - 3/x + C
*February 4, 2015*

**Math**

Nope. try 4n-6 < 0.5(-n) "opposite" is a vague term. Usually it means negative. You have interpreted it as reciprocal. In either case, "the product of four and a number" is 4n six less than that is 4n-6, not 6-4n.
*February 4, 2015*

**physics**

both are 3 significant digits.
*February 4, 2015*

**Math**

why divide? Just add up all the times. Dividing by 4 will give the average time per activity, but no one asked for that. Dividing 4 by the sum is meaningless. It gives the number of activities that can be done in a second, but that's of no use to anyone.
*February 4, 2015*

**math**

2/5 + 1/6 + 1/3 = 12/30 + 5/30 + 10/30 = 27/30 So, what's not covered?
*February 4, 2015*

**Algebra**

23 nickels is right, but that wasn't what you were asked to find.
*February 4, 2015*

**From Theory to practice**

Q: What's the difference between theory and practice? A: In theory, there is no difference, but in practice, there is!
*February 4, 2015*

**algebra**

(4x+6)(4x-6) = 16x^2-36
*February 4, 2015*

**Math - Absolute Value Inequalities**

B is correct. |x-2| means the difference between x and 2. That difference must be less than 3, so 2-3 < x < 2+3 -3 < x-2 < 3
*February 4, 2015*

**math**

4^2 = 16 5^2 = 25 so, 4.5 is a good first guess. 4.5^2 = 20.25 too big 4.4^2 = 19.36 4.45^2 = 19.80 4.46^2 = 19.89 4.47^2 = 19.98 4.48^2 = 20.07 So, you want something between 4.47 and 4.48 keep going.
*February 4, 2015*

**College Alegbra**

w(2w-7) = 400
*February 4, 2015*

**trig**

who is r? If x is the angle between p and q, then 12^2 = 8^2+10^2-160cos(x) Now subtract x from q's heading (not bearing!)
*February 4, 2015*

**geometry**

since the linear scale has doubled, the volume has grown by a factor of 8.
*February 4, 2015*

**Algebra 1B**

surely your text has examples and explanations. And remember that google is your friend.
*February 4, 2015*

**algebra**

for F(x) = ax^2+bx+c, the vertex lies on the axis of symmetry, at x = -b/2a. so, for your function, that is at x = -2/-2 = 1 f(1) = 7 so, the vertex is at (1,7). See the graph at http://www.wolframalpha.com/input/?i=-+x^2+%2B2x+%2B+6 Recall that the quadratic formula says that...
*February 4, 2015*

**algebra**

there are oodles of online graphing utilities. wolframalpha.com desmos.com come to mind. Just type in your function.
*February 4, 2015*

**algebra**

to find the y-intercept, set x=0: f(0) = 1 To find the x-intercept(s) use the quadratic formula.
*February 4, 2015*

**Calculus II**

use integration by parts. Let u = x, du=dx v = e^-4x dx, v = -1/4 e^-4x ∫ udv = uv - ∫v du so, ∫(x)(e^-4x dx0 = -1/4 xe^-4x + 1/4 ∫e^-4x dx = -1/4 xe^-4x - 1/16 e^-4x = -1/16 e^-4x (4x+1) + C
*February 4, 2015*

**math**

f(g) = 3g+4 = 3(bx-4)+4 = 3bx-8 g(f) = bf-4 = b(3x+4)-4 = 3bx+4b-4 Clearly, b = -1
*February 3, 2015*

**Geometry**

I get 7/2
*February 3, 2015*

**Math - Operations with Algebraic Vectors**

|i+3j-k| = √(1+9+1) = √11 Better go back to where it shows how to figure the length of a vector.
*February 3, 2015*

**MATH**

If there are n terms, n/2 (5+61) = 957 n = 29
*February 3, 2015*

**Math**

12x-6y=21 3x+6y=9 now add them up and the y's are eliminated. Should be easy going from there.
*February 3, 2015*

**math (geometry)**

42
*February 3, 2015*

**calculus**

y' = x-3 P0 = (1,4) ∆y = y'(x0)*∆x = (-2)(0.1) = -0.2 P1 = (1.1,3.8) ∆y = y'(x1)*∆x = (-1.9)(0.1) = -0.19 P2 = (1.2,3.61) ∆y = y'(x2)*∆x = (-1.8)(0.1) = -0.18 P3 = (1.3,3.43) Read the fine article at wikipedia, which shows a ...
*February 3, 2015*

**calculus**

(√(x+4)-2)/x using l'Hospital's Rule, that gives (1/(2√(x+4)))/(1) = (1/(2*2))/2 = 1/4
*February 3, 2015*

**Math**

Note that the last digit of 3^n repeats: 3,9,7,1,... 3^997 = 3^996*3, so it ends in 3 so, ...
*February 3, 2015*

**Math Work**

4(1/2)(sin(pi/3+5pi/6)-sin(pi/3-5pi/6)) Now just add the fractions and plug in the values.
*February 3, 2015*

**geometry**

the area is a = πr^2 + 2πrh So, πr^2+100πr = 3000 r^2+100r = 955 r = 8.779 So, r is about 9
*February 3, 2015*

**Geometry**

CZ extended to AB is a median. The centroid is 2/3 of the way from the vertex to the opposite side. So, you have an isosceles triangle with base=6 and altitude=6
*February 3, 2015*

**Math**

I'd use (4-1)/(1+1) = (3+1)/(3+1) Don't know why all the choices have a -2 in them, since none of the points contains a -2.
*February 3, 2015*

**Math Help - DQ?**

I interpret the instruction to use the Half-Angle formula as sin(15°) = √((1-cos30°)/2) = √((1-√3/2)/2 = √(2-√3)/2 = (√6-√2)/2 Granted, the half- and double-angle formulas are two sides of the same coin, but ...
*February 3, 2015*

**math**

If you mean f(x) = (x^2-4)/(x-2) if x<2 then since x<2, (x^2-4)/(x-2) = x+2 So, f(x) = x+2 if x < 2 ax^2-bx+3 if 2 <= x < 3 4x-a+b if x >= 3 So, as x->2-, f(x)->4 That means we must have 4a-2b+3 = 4 at x=2 9a-3b+3 = 12-a+b at x=3 a = 7/2 b = 13/2 f(x...
*February 3, 2015*

**math**

(6p+4b+3c)(1.2)=(3p+8b+6c)=3p+2(4b+3c) So, if Mohan spent 2x on puffs and y on the others, (2x+y)(6/5)=x+2y 6(2x+y)=5(x+2y) 12x+6y=5x+10y 7x=4y y = 7x/4 Now, we want x/(2x+y) = x/(2x+7x/4) = x/(15x/4) = 4/15 Mohan spent 8/15=53.3% of his money on puffs. check: (8/15+7/15)(6/5...
*February 3, 2015*

**algebra**

(2/5)x + x = 21 (7/5)x = 21 x = 15 The two numbers are 6 and 15
*February 3, 2015*

**algebra**

just plug in -2 for all the x's: f(-2) = 2(-2)+1
*February 3, 2015*

**math**

all are functions. For #4, suppose that u and v produce the same y value. That is, mu+b = mv+b then that means that mu = mv u = v In other words, if you have the same y-value, it can only be produced by one x-value. That means that you have a function.
*February 3, 2015*

**math**

correct. but stop guessing.
*February 3, 2015*

**math**

(c)
*February 3, 2015*

**math**

y"=6x-24 = 6(x-4) you want y"=0
*February 3, 2015*

**math**

they are where y' = 3x^2-12x+9 = 0
*February 3, 2015*

**math**

.42x = 210
*February 3, 2015*

**Pre-Calculus**

x-axis: (x,y)->(x,-y) y-axis: (x,y)->(-x,y) but, since h(x) is an odd function h(-x) = -h(x) reflecting h(x) through either axis flips it to the same thing. (x,x^3)->(x,-x^3) (x,x^3)->(-x,x^3) = (-x,-(-x)^3)
*February 3, 2015*

**Pre-Calculus**

swap x and g and then solve for g: x=(g+7)^3+4 x-4 = (g+7)^3 ∛(x-4) = g+7 g = ∛(x-4) - 7 check. If h is g inverse, g(h(x)) = h(g(x)) = x g(h) = (h+7)^3+4 = ((∛(x-4) - 7)+7)^3+4 = (∛(x-4)-7+7)^3+4 = ∛(x-4)^3+4 = x-4+4 = x and similarly for h(g)
*February 3, 2015*

**math**

and you'd be correct. But why guess? A brief check on the definition of a function would make it certain.
*February 3, 2015*

**Pre-Calculus**

nothing. x^2 is an even function. Reflecting it about its axis changes nothing. (x,y)->(-x,y) since f(x) = f(-x), the graph is unchanged.
*February 3, 2015*

**Pre-Calculus**

you surely know that this is a parabola which opens up, so there is no local maximum. So, find the f values at the ends of the interval, and pick the larger for the global max. No interval? No maximum.
*February 3, 2015*

**Pre-Calculus**

x then y: (x,y)->(x,-y)->(-x,-y) y then x: (x,y)->(-x,y)->(-x,-y) f(x) -> -((-x)^3+2(-x)^2-(-x)+5)
*February 3, 2015*

**Pre-Calculus**

reflect: (x,y)->(-x,y) f(x) -> -x^3+2x^2+x+5 shift: (x,y)->(x+4,y) f(x) -> -(x+4)^3+2(x+4)^2+(x+4)+5 = -x^3-10x^2-31x-23 see the graphs here: http://www.wolframalpha.com/input/?i=plot+y%3Dx^3%2B2x^2-x%2B5%2C+y%3D-x^3-10x^2-31x-23 the shift isn't real clear, but...
*February 3, 2015*

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