Saturday

November 22, 2014

November 22, 2014

Total # Posts: 26,888

**math**

assuming you want integer dimensions, just find the factors of 2460: 2x2x3x5x41 Now, you can mix and match, as long as you use them all once each. For example, 2*3 x 2*5*41 = 6x410
*September 27, 2014*

**Math**

better explain what it is you want to do. "9x" is an incomplete expression.
*September 27, 2014*

**ap physics**

depends on your speed, doesn't it? I guess you weren't paying attention, either in class or in your posting.
*September 27, 2014*

**math**

dunno - what is a family of graphs? enter your functions in the box at wolframalpha.com and it will show the graphs. Then you can play around with different parameters.
*September 27, 2014*

**Calculus**

I think you both misread the question. We want a line tangent to ln(x) which passes through (0,0). Since the slope of the tangent is 1/x, we want a line through (0,0) and (h,ln(h)) with slope 1/h. Clearly, at x=e, y=1 and the slope is 1/e, so y = x/e is the tangent at (e,1) ...
*September 27, 2014*

**Math problems.**

multiplication and division first, then addition and subtraction, all done left to right: 15 – 6 • 9 ÷ 2 + 10 15 – 54 ÷ 2 + 10 15 - 27 + 10 -12 + 10 -2 Evaluate parenthesized expressions first of all: 4 • (6 + 3) + 5 4 • (9) + 5 36 + 5 41
*September 27, 2014*

**physics**

See an almost identical problem in the first related question below.
*September 27, 2014*

**math**

0 is 4 to the left of 4 So, the other end's x value is 4 to the right of 4, or 8. Do y the same way.
*September 27, 2014*

**math**

No, since area = side^2 It is not a constant multiple of the side length.
*September 27, 2014*

**Physics**

At time t > ∆t, position of A: v(t-∆t) + 1/2 (v/∆t)t^2 position of B: 1/2 a(t-∆t)^2 So, now you need to find when the positions are the same: v(t-∆t) + 1/2 (v/∆t)t^2 = 1/2 a(t-∆t)^2 2vt - 2v∆t + (v/∆t)t^2 = at^2 - 2a&#...
*September 27, 2014*

**math**

since 3 < pi < 4, √9 < pi √16 In fact, since pi^2 = 9.86, √9 < pi < √10
*September 27, 2014*

**Trig**

clearly, x^2/a^2 + y^2/27^2 = 1 at x=24, y=9, so 24^2/a^2 + 9^2/27^2 = 1 24^2/a^2 + 1/9 = 1 24^2/a^2 = 8/9 a^2 = 24^2 * 9/8 = 648 x^2/648 + y^2/729 = 1
*September 27, 2014*

**algebra**

No idea what "if a sheet is 0.003 8" means. Also, say 1 cm^3 for volume.
*September 27, 2014*

**Math**

it's because x^2+6 has no factors in common with x^2+3
*September 27, 2014*

**algebra 2**

#3 since one equation tells us that y = -3x+6, use it in the other equation: 2y = 10x-36 2(-3x+6) = 10x-36 -6x+12 = 10x-36 -16x = -48 x = 3 Do #4 the same way
*September 27, 2014*

**Algebra**

If the short piece is x, then x + 4x = 20 Now find x, and then 4x.
*September 27, 2014*

**Algebra 2**

all I did was clear the fractions, just as you would if you were doing, say, x/2 = 3/7 7x = 3*2 After that, I just expanded terms and simplified. You never said what you wanted to do with the fractions. Naturally, if y=4 and x=1, (y-1)/(x+3) = 3/4
*September 27, 2014*

**Algebra 2**

(y-1)/(x+3) = 3/4 4(y-1) = 3(x+3) 4y-4 = 3x+9 3x-4y+13 = 0 Not sure where you want to go with this.
*September 27, 2014*

**Physics**

The range of the projectile is R(x) = 15.5^2 * sin(84°)/9.81 = 24.36 m The height is y(t) = 15.5 sin 42° t - 4.9t^2, so it takes 2.12 seconds to fall back to the initial height. So, the runner has to run toward the thrower at (30-24.36)/2.12 = 2.66 m/s
*September 26, 2014*

**Math Homework**

(fg)(x) = f(x)*g(x) = (x^2)(4x-5) = 4x^3 - 5x^2 (g/f)(x) = g(x)/f(x) = (1/x^2) / (1/x) = 1/x^2 * x/1 = x/x^2 = 1/x
*September 26, 2014*

**Pre-Calc**

g(f) = 5-f = 5-(3x+5) = -3x f(g) = g+2 = (4-x^2)+2 = 6-x^2 f(g) = g^2 = (x-1)^2 = x^2-2x+1 Hmmm.
*September 26, 2014*

**Math help, anyone?**

#6,7 are positive, the rest are negative
*September 26, 2014*

**Calculus**

#1 using l'Hospital's Rule, ( t^(1/3) - 1)/(t^(1/2) - 1) -> (1/3 t^2/3)/(1/2 t^1/2) = 2/3 t^-1/6 = 2/3 at t=1 If y=2^x, y' = ln2 2^x at x=0, that's just ln2 So, the line is y-1 = ln2(x) so, y = ln2 x + 1 So, the slope is ln2 = 0.693 See http://www....
*September 26, 2014*

**Pre-Calc**

I would draw it just like x^2, but with the vertex at (4,-2) instead of (0,0).
*September 26, 2014*

**Pre-Calculus**

(f◦g)(x) = f(g(x)) Since f(x) = 8x, f(g) = 8g = 8(2x+1) = 16x+8
*September 26, 2014*

**Trigonometry**

I'd use law of cosines to find the third side, then use Heron's formula for the area.
*September 26, 2014*

**algebra 2**

really? If x = 3, y=3*3-4 = 5
*September 26, 2014*

**algebra 2**

for #1, it's easy to substitute. You have two values for y, so set them equal. 3x-4 = -2x+11 5x = 15 x = 3 so, y=3x-4 = 5 For #2, 4x-y=1, so y=4x-1 Plug that in, and we have 5x+2y = 5x+2(4x-1) = 24 5x+8x-2 = 24 13x = 26 x = 2 so, y = 4x-1 = 7 There are lots of graphing ...
*September 26, 2014*

**math - typos**

x = (7y-8)/(3y-1) x(3y-1) = 7y-8 3xy-x = 7y-8 y(3x-7) = x-8 y = (x-8)/(3x-7)
*September 26, 2014*

**Math Problem (please help)**

wherever there is an x, plug in a 3 and then simplify: f(3) = (2*3^2 + 8*3)/(5*3^3 + 40) = (18+24)/(135+40) = 42/175 = 6/25
*September 26, 2014*

**Trigonometry**

if the third side is 2x, then x = 14 sin 55º, and the height is h = 14 cos 55º, so the area is hx = (14 cos55º)(14 sin55º) = 7 sin 110º
*September 26, 2014*

**Math Problem (please help)**

what's the trouble? Just plug in the values. For example, (fg)(x) = f(x)*g(x) = (7x-9)(8-x) = -7x^2 + 65x - 72 Do the others in like wise.
*September 26, 2014*

**Pre-Calc - Oops**

Bzzzt. E
*September 26, 2014*

**math**

well, did you try any of them? for (a) 1 divides 1 2 divides 18 3 divides 183 4 divides 1836 5 divides 18365 6 divides 183654 7 does not divide 1836549 so, try the others
*September 26, 2014*

**MATHS**

You will probably like the answer better if you enter .5^(1/1590) ...
*September 26, 2014*

**Pre-Calc**

I assume you mean √(x+2) + 5 That would be √x shifted left 2 and up 5 So, (b) x^2 -x^2 is flipped upside-down -x^2-3 is then shifted down 3 So, (b) again
*September 26, 2014*

**Math Problem (please help)**

(x+3)(252/x - 2) = 252 x = 18 Check: 18*14 = 21*12 = 252
*September 26, 2014*

**AP Physics**

a^2 = 15^2 + 6.5^2 a = 16.3 I expect you can determine the direction.
*September 26, 2014*

**sat**

3/12 and that's "three pitchers are" not "three pitcher's is"
*September 26, 2014*

**Algebra**

20+8x
*September 26, 2014*

**Calculus**

If the man's shadow has length s when his distance from the pole of height h is x, then using similar triangles, (x+s)/h = s/6 6x+6s = hs s(h-6) = 6x so, (h-6) ds/dt = 6 dx/dt ds/dt = 6*30/(h-6) = 180/(h-6) Now, we were given 15/h = 6/6, so h=15 Thus, ds/dt = 180/9 = 20 ft...
*September 25, 2014*

**math**

well, I guess you know which is the correct choice, eh? consider it this way. car A is 15 miles ahead when car B starts. B is only going 10 mi/hr faster than A, so it will take 15/10 hours to make up the difference. As Reiny showed, that means it takes 1.5 hours.
*September 25, 2014*

**math**

the mistake was in not showing the arrangement.
*September 25, 2014*

**Urgent geometry help**

Since there is no conditional specified, there can be no converse. However, given your interpretation, your converse is correct. The "will" isn't really necessary at all, if you consider coming and leaving as sort of chronic events, one conditional on the other.
*September 25, 2014*

**Algebra 1**

|n-34| <= 8
*September 25, 2014*

**Algebra**

|p-53| <= 5
*September 25, 2014*

**algebra**

clearly the minimum cost is at the vertex of the parabola, which is at x = -b/2a = 18.1/1.8
*September 25, 2014*

**AP Physics**

28 @ S45W = (-19.8,-19.8) 56 @ E = (50,0) 84 @ E53N = (50.5,67.1) Add them up to get (80.7,47.3) So, we need (-80.7,-47.3) to get back to (0,0), a displacement of 93.5 m
*September 25, 2014*

**Physics**

Speedboat A negotiates a curve whose radius is 100 m. Speedboat B negotiates a curve whose radius is 221 m. Each boat experiences the same centripetal acceleration. What is the ratio vA/vB of the speeds of the boats?
*September 25, 2014*

**Math-please check**

#1 of course not: |-6|=|6| but -6≠6 #2 same reasoning. y could be -8. You can say, therefore that if |a|=|b|, then a^2 = b^2.
*September 25, 2014*

**math...**

I think you meant b-9 > 20 Now add 9 to both sides to get b by itself: b > 29
*September 25, 2014*

**Math Help.**

clearly, x <= 1, so 3x <= 3 3x+1 <= 4 Looks like D to me. C would have an open circle on the 1, rather than a filled dot.
*September 25, 2014*

**geometry**

oldv = 1/3 pi r^2 h Now replace r with 4r and h with whatever fraction of h, say, 1/n, it has become. new v = 1/3 pi (4r)^2(1/n * h) = 1/3 pi * 16r^2 * h/n = oldv * 16/n
*September 25, 2014*

**Please math help**

A n n-10 10-n 7-n n-7 n-7 = -3, so n=4 B yes, it always works, as shown above
*September 25, 2014*

**Algebra**

7p^2+16=2151 7p^2 = 2135 p^2 = 305 p = √305
*September 25, 2014*

**Math**

Things would be a lot clearer if you wrote CN^2 = 13^2 + 18^2 - 2*13*18 cos 45 = 169 + 324 - 468*√2/2 = 162.07 CN = 12.73
*September 25, 2014*

**trignometry**

y^2*sin(x)*sin(z)/2*sin(y) area = z/2 * y sinX but, z/sinZ = y/sinY, so z = ysinZ/sinY area = ysinZ/2sinY * y sinX = y^2 sinZ sinX / 2sinY
*September 25, 2014*

**algebra**

(5.1*10^3)*(3.2*10^3) (5.1)(3.2)*(10^3)(10^3) 16.32 * 10^6 But that's not scientific notation, so you need to shift the decimal point to 1.632 * 10^7
*September 25, 2014*

**Algebra 2**

a = 2pi r h + 2 pi r^2 = 2pi((3x-2)(x+3) + (3x-2)^2) = 2pi(3x^2+7x-6 + 9x^2-12x+4) = 2pi(12x^2-5x-2)
*September 25, 2014*

**MATH**

#1 ok #2 First, recall that ALL |x| are positive. So, make sure all the plain old negative values come first, then the positives. For example, D is obviously not it, because the numeric values are 7,4,3,0,-3,-7 In fact, it's exactly the reverse of what you want. #3 ok #4 |...
*September 25, 2014*

**Math**

4x + 2x = 15 x = 2.5 So, 5 cups of nuts, 10 cups of chips.
*September 25, 2014*

**maths**

26*(48)+(-48)*(-36) 26*(48)-(48)*(-36) Now factor out the 48 (48)(26-(-36)) (48)(62)
*September 25, 2014*

**ALGEBRA 2**

1-x >= -3 x <= 4 5x-1 >= 19 5x >= 20 x >= 4 Looks like all real numbers.
*September 25, 2014*

**ALGEBRA 2 - Oops**

Oops - mangled that one, eh? -1 <= 3+2x < 11 -4 <= 2x < 8 -2 <= x < 4
*September 25, 2014*

**ALGEBRA 2**

-1 <= 3+2x < 11 -4 <= 2x <= 14 -2 <= x <= 7
*September 25, 2014*

**TRIANGLE**

hard to solve without knowing which angle is 59°. If A or C, 8/sin59° = 12/sin(C or A), so then you can figure B and use law of sines again to get AC. If B, AC^2 = 8^2 + 12^2 - 2*8*12 cos 59°
*September 25, 2014*

**Maths**

x+y = cscA+cosA+cscA-cosA = 2cscA x-y = cscA+cosA-(cscA-cosA) = 2cosA 2/2cscA = sinA 2cosA/2 = cosA All clear now?
*September 25, 2014*

**math**

Assuming he started at the pole, after 5 seconds he is 7.5 m away. So, if the shadow has length s, use similar triangles to get (s+7.5)/3.8 = s/0.95 Now just solve for s.
*September 25, 2014*

**Math**

correct. Just as 47 is in between 36 and 49, the root is in between 6 and 7.
*September 24, 2014*

**Algebra, Please Help!**

no way to tell. There are lots of circles with a given center. Just think of a target with a bulls-eye. You will need some other piece of data, such as a point on the circle.
*September 24, 2014*

**Math**

cube has volume 4^3 = 64 sphere has volume π/6 * 4^3 = 64π/6 Now just subtract the volumes: 64(1 - π/6) Actually, the wording is a bit vague. Since the sphere has replaced some of the cube, the space is still occupied, so the whole cube constitutes the souvenir...
*September 24, 2014*

**calc 3**

(a) study your text, or (b) google for examples, such as http://www.math.ucla.edu/~ronmiech/Calculus_Problems/32A/chap12/section6/812d55/812_55.html and follow the same steps.
*September 24, 2014*

**Algebra II Honors**

Any line in the yz plane becomes a plane perpendicular to the yz plane, in xyz space. Think of the point y=3. When propagated parallel to the x-axis (perpendicular to the y-axis), it becomes a line. Same for a line propagated through space. x+y=0 is the same thing, parallel to...
*September 24, 2014*

**Math-calculus**

what don't you get? You have the equation for d(t). Just do the math. For the graph, see http://www.wolframalpha.com/input/?i=40%2B60cos%28%28pi%2F2%29%28t-2%29%29 For the rest, just solve for t under the given conditions: (b) 40+60cos(PI/2(t-2)) = 76 (c) is d(t) on its ...
*September 24, 2014*

**math-pre calc**

To average 48 mi/hr, you need to cover 106 miles in 2.208 hours. The first half took 53/36 = 1.472 hours So, how fast do you need to go to cover the other 53 miles in (2.208-1.472) hours?
*September 24, 2014*

**SAT Math**

how about (A)? 3 is not divisible by 2.
*September 24, 2014*

**Pre-Calc**

clearly, y = a(x+5)^2 - 2 Now just plug in the point and solve for a. But, you want y=ax^2+bx+c, so just expand and collect terms to wind up in that format.
*September 24, 2014*

**Algebra 2**

You are correct
*September 24, 2014*

**Calculus**

y = 3x^2 - x - 5 y' = 6x-1 So, at (1,-3), the slope of the normal is -1/5, and the line there is y+3 = -1/5 (x-1) y = -1/5 x - 14/5 So, where does the line intersect the parabola? -1/5 x - 14/5 = 3x^2-x-5 -x-14 = 15x^2-5x-25 15x^2-4x-11 = 0 (15x+11)(x-1) = 0 So, the line ...
*September 24, 2014*

**Geometry**

R+S = 90 U+V = 90 so, y-2 + 2x+3 = 90 2x-y + x-1 = 90 2x+y = 89 3x-y = 91 5x = 180 x = 36 and now you can take it from there.
*September 24, 2014*

**Pre-Calculus**

the slope is 5/2, so y+3 = 5/2 (x-2) Now rearrange the terms as needed
*September 24, 2014*

**Math 112**

equal has many meanings, most of them to do with numbers. Maybe equal area, equal perimeter, etc. congruent is for geometric properties, not numeric values. For instance, we say two angles are congruent when they have equal measures.
*September 24, 2014*

**geometry**

v = (40)(20*8 + 20*7/2)
*September 24, 2014*

**Math 112**

since both your expressions are identical, I think there's a typo. If you meant for the operators to be "equal" and "congruent", then the answer is that AB and CD are geometric figures. As such, they can be congruent, not equal. They are congruent if ...
*September 24, 2014*

**algebra 2**

Just draw the line 3x+y = 5 then, shade the area below the line, where, naturally, 3x+y < 5 See http://www.wolframalpha.com/input/?i=3x%2By%3C5
*September 24, 2014*

**Math--Teach me?**

work your way from the inside, to get rid of parens and brackets: 2[4(9 – 7) + 1] ÷ 3 2[4(2) + 1] ÷ 3 2[8+1]÷3 Now do the multiply and divide, left to right: 2[9]÷3 18÷3 6
*September 24, 2014*

**Pre-Calculus**

correct
*September 24, 2014*

**Pre-Calculus**

correct
*September 24, 2014*

**Algebra**

just recall that orthogonal means that u•v = 0. So, (a) (1,2)•(k,k) = k + 2k = 0, so k=0 (b) (1,2,1)•(k,2k,4) = k+4k+4 = 0, so k = -4/5 Kind of boring examples, IMO.
*September 24, 2014*

**Pre-Calculus**

Ummm, y = mx+b is called the slope-intercept form because you can just read them off: slope = m y-intercept = b Now can you pick the correct choice?
*September 24, 2014*

**Algebra**

you just need to d a lot of these problems. Look for common factors. 9w-w^3 clearly w can be factored out, so w(9-w^2) Now you have to recall the difference of squares is x^2-y^2 = (x-y)(x+y), so w(3-w)(3+w) As for the rest, I have no idea what 3h2 the 2 power + 1+3a even ...
*September 24, 2014*

**calculus**

z = √(x^2+y^2)+xy ∂z/∂x = x/√(x^2+y^2) + y ∂^2z/∂x^2 = y^2/(x^2+y^2)^(3/2) ∂^2z/∂x∂y = 1 - xy/(x^2+y^2)^(3/2) Since z is symmetric in x and y, the y derivatives are the same, just using y instead of x. As usual, check my ...
*September 24, 2014*

**trig**

review your basic trig definitions, and draw a diagram. You will see that (h-1.6)/50 = tan 36° Now just solve for h.
*September 24, 2014*

**Calculus I**

Since T(t) is the temperature after t hours, N(T(t)) is the number of bacteria after t hours
*September 24, 2014*

**calculus**

f(x) is decreasing when f'(x) < 0, so just solve 6x^2+6x-12 < 0 x^2+x-2 < 0 (x+2)(x-1) < 0 The roots are at -2 and 1, and the parabola opens upward, so f'(x) < 0 between the roots. f(x) is decreasing on (-2,1) Note that it is not [-2,1] because at the ...
*September 24, 2014*

**math,algebra**

3^-1 = 1/3, so you have (x^3/3)^2(12x^5) = (x^6/9)(12x^5) = 4/3 x^11
*September 23, 2014*

**calculus**

(6x^2+4y^2)/(3x+7y)=3 we know that (u/v)' = (u'v-uv')/v^2, so ((12x+8yy')(3x+7y)-(6x^2+4y^2)(3+7y'))/(3x+7y)^2 = 0 36x^2 + 84xy + 24xyy' + 56y^2y'-18x^2-12y^2-42x^2y'-28y^2y' = 0 because the denominator is not zero y'(24x+56y^2-42x^2-28y...
*September 23, 2014*

**Trigonometry**

this is just (1-cos^2)^2 = (sin^2)^2 = sin^4
*September 23, 2014*

**MATH ASAP!!**

1/(1-√7+√5) multiply top and bottom by 1-√7-√5 (1-√7-√5)/((1-√7)^2 - 5) = (1-√7-√5)/(1-2√7+7-5) = (1-√7-√5)/(3-2√7) now multiply by 3+2√7 (1-√7-√5)(3+2√7)/(9-28) (1-√7...
*September 23, 2014*

**Math**

since sine = y/r, and r is always positive, sine is positive where y is positive.
*September 23, 2014*

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