Saturday

December 20, 2014

December 20, 2014

Total # Posts: 27,650

**algebra**

2(w + 3w-1) = 30 solve for w, and then figure the length
*December 2, 2014*

**Algebra**

-3/4 + 1/4 = -1/2 -9/10 + 2/5 = -1/2
*December 2, 2014*

**algebra**

116 - 6*8 = ?
*December 2, 2014*

**language arts**

Mine, too. Normally, the subjunctive is clearly indicated, as in So if you had any sense at all, you would avoid Storm Fever. (But, you clearly do not.) As phrased, it leaves open the possibility that you do have some sense, so based on that condition, you will avoid the movie.
*December 1, 2014*

**algebra**

well, just use what you're told. If the medium side is x, then x-7 + x + x+13 = 57 Now just find x and you're almost done.
*December 1, 2014*

**MATH**

7*5=35 2*4*35 = 280 (new area) 2(7*4+5*2) = 76 (new perimeter) 2(7*2+5*4) = 68 (new perimeter) You don't say which side is 2x and which is 4x larger No idea how to work this without double-digit numbers
*December 1, 2014*

**Math**

(97.65-45)/45 = ?
*December 1, 2014*

**Newtons Law of Cooling**

well, you know that dT/dt = k(68-T) dT/(68-T) = k dt ln(68-T) = kt + c1 68-T = e^(kt+c1) 68-T = ce^(kt) T = 68 - ce^(kt) We know that T(0) = 43, so c=25, and T = 68-45e^(kt) We know that T(12) = 55, so 55 = 68-45e^(12k) 45e^12k = 13 e^12k = (13/45) 12k = ln(13/45) k = ln(13/45...
*December 1, 2014*

**calculus**

You know that the average velocity is [y(3+h)-y(3)]/h So, just plug in the numbers and grind out the answers.
*December 1, 2014*

**Algebra 1**

use the vertical line test to determine if the relation(-6,-2), (-2,6),(0,3),(3,5) is a function
*December 1, 2014*

**Algebra 1**

What is the graph of the function rule? y=I5xI -2
*December 1, 2014*

**Algebra 1**

The ordered pairs(1,4), (2,16), (3,64), (4,256),(5,1024) represent a function. What is a rule that represents,this function? My answer is y=x to the 4th power
*December 1, 2014*

**Algebra 1**

The table shows the relationship between the number of players on a team and the minutes each player gets to play. Players minutes 7 35 8 30 9 25 10 20 Is the relationship a function that ism increasing or decreasing? Is the relationship a function that is linear or nonlinear...
*December 1, 2014*

**Algebra 1**

Write an equation that relates the number of parallelograms in the figure (n), to the perimeter of the figure (P). /_/ /_/_/ /_/_/_/ The top and bottom = 5 each and the sides = 4 each in each figure.
*December 1, 2014*

**Rounding up and down**

#1 0.34 #2 C 0.351 -> 0.4 0.562 -> 0.6 0.445 -> 0,4 #3 0.545 -> 0.55 Most people round up from 5. Some round to the closest even value, which would be 0.54 in this case. Follow the rules you were given.
*December 1, 2014*

**math**

treated in the related questions below.
*December 1, 2014*

**Calculus**

technically, since f(1) is not defined, you need to take the limits as x->1, but you will wind up with the same values.
*December 1, 2014*

**Calculus**

∫[-2,3] f(x) dx = ∫[-2,1] x dx + ∫[1,3] 1/x dx Now you can take it from there. Note that f(1) is not defined.
*December 1, 2014*

**Calculus**

well, which function is not constant?
*December 1, 2014*

**Calculus**

I assume you want the rectangles to have equal widths. Tat would be (5-1)/4 = 1. So, the left sides of the rectangles are at x = 1,2,3,4 the heights are f(1),f(2),f(3),f(4). So, evaluate f there and add up the areas.
*December 1, 2014*

**Math**

2(x+y) = 100 xy = 525 Carry on from there.
*December 1, 2014*

**Trig**

just recall your Pythagorean Theorem. It's also a distance formula.
*December 1, 2014*

**Maths**

This problem is discussed at http://www.beatthegmat.com/how-many-non-negative-integral-solutions-are-possible-t34635.html as you could have discovered with about 5 minutes' work.
*December 1, 2014*

**Pleaaaaaase help with differential graph question**

Sorry, no graphs here. But you know that since f(P) is dP/dt, when f is positive, P is increasing. As for the other items, no idea, since no picture is available. Sounds like P is some kind of exponential or logistic model. Better review those topics.
*December 1, 2014*

**physics**

8.0 rad / (2.56 rad/s) = 8.0/2.56 = 3.125 s Now convert that to minutes...
*December 1, 2014*

**Maths**

well, the first n terms of 2^(n-1) is 2^n - 1 The first n terms of 5n is 5 * n(n+1)/2 So, the sum you want is 2^n - 1 + 5n(n+1)/2
*December 1, 2014*

**Maths**

12 + 12+d + 12+2d = 6(1+r+r^2) But, d=r, so 36+3d = 6(1+d+d^2) 6d^2+3d-30 = 0 2d^2 + d - 10 = 0 (2d+5)(d-2) = 0 So, d=2 or -5/2 The two sequences are thus 12,14,16 6,12,24 with S3=42 or, 12,9.5,7 6,-15,75/2 with S3=28.5
*December 1, 2014*

**Math**

This is just an A.P. with a=4 d=4 The section in question has (64-4)/4 + 1 = 16 terms, so S16 = 16/2 (4+64) = 544
*December 1, 2014*

**College Geometry**

it's just a pyramid. And, of course Euler's formula holds.
*December 1, 2014*

**movie**

Hmmm. Try watching the movie! Failing that, if you are not familiar with America, just be aware that Tiffany & Co. is one of the most famous stores in the country. In particular, they sell fine jewelry and gifts.
*December 1, 2014*

**Precalc/Trig**

Suppose the nearer observer is at a distance x, and the tree has height h. Then, it is clear that h/x = tan30° h/(x+25) = tan20° Now, eliminate x and we see that h/tan30° = h/tan20° - 25 h(cot20°-cot30°) = 25 h = 25/(cot20°-cot30°) h = 24.62 ft
*December 1, 2014*

**Physics**

You can find the appropriate formula here: http://en.wikipedia.org/wiki/List_of_moments_of_inertia
*December 1, 2014*

**math**

The x- and y- changes from the first point to the last are 9 and 15 So, add 1/3 of those values to the first point.
*December 1, 2014*

**Math**

#1: 100000 * 1.1^5 #2: Sum of the first n integers is n(n+1)/2 #3 12 + 2*4/(1 - 1/3)
*December 1, 2014*

**calculus**

Oops. You are correct. So, make the fix and rerun the math.
*November 30, 2014*

**calculus**

The average rate of change between years a and b is (R(b)-R(a))/(b-a) Here, both intervals are 50 years, so things are convenient. You want to show that R(150)-R(100) ≈ 2(R(100)-R(50)) So, just plug in your numbers and see what you get.
*November 30, 2014*

**Differential equations in Calculus...plsssss help?**

so. you have dT/dt = -T/15 + 5 = -1/15 (T-75) dT/(T-75) = -1/15 dt ln(T-75) = -t/15 + k T-75 = e^(-t/15 + k) T-75 = ce^(-t/15) T = 75 + ce^(-t/15) Now, you are told that T(0) = 105, so 105 = 75 + ce^0 c = 30 T(t) = 75 + 30e^(-t/15) as desired in part (g) So, the coffee starts ...
*November 30, 2014*

**Differential equations in Calculus...plsssss help?**

Hard to see just what you wrote, but I'm guessing dT/dt = -1/(15T+5) (15T+5) dT = -dt No, that can't be right, if you want an exponential function. How about dT/dt = 15T+5 then we have dT/(3T+1) = 5dt ln(3T+1) = 5t+k 3T+1 = ce^5t T = (ce^5t - 1)/3 Hmm. Not that either...
*November 30, 2014*

**Math**

#1 You have an A.P. with a = 100 d = 50 S31 = 31/2(2*100 + 30*50) #2 You have a G.P. with a = 16 r = 1/4 So, find T7 and 2S6-16+T7 the strange sum is because you have just a drop on T1 and a bounce on T7, and not 7 complete round trips
*November 30, 2014*

**Math**

If there are x correct and y incorrect, then we have x+y=10 5x-2y = 29 Now just solve for x.
*November 30, 2014*

**Calculus**

That is correct. So, the limit is +∞, since the denominator near x=1 is always positive.
*November 30, 2014*

**Math please help!**

A glass of skim milk supplies 0.1 mg of iron, 8.5 g of protein, and 1 g of carbohydrates. A quarter pound of lean red meat provides 3.4 mg of iron, 22 g of protein, and 20 g of carbohydrates. Two slices of whole-grain bread supply 2.2 mg of iron, 10 g of protein, and 12 g of ...
*November 30, 2014*

**AP calc**

well we need the limit from both sides to be the same. f(2) = 2^2+3*2-1 = 9 So, we need -3bx+3 = 9 when x=2 -3b*2+3 = 9 -6b = 6 b = -1
*November 30, 2014*

**physics**

When a 1.50 kg block is 3.96 m above the base of the hill, the mass has a speed of 1.71 m/s. When the block reaches a point 0.750 m above the base of the hill, what is its kinetic energy?
*November 30, 2014*

**AB Calculus**

(a) is correct. Not sure how the sand maintains a constant plane cross-section, but we'll posit that. The volume of the conical pile is v = 1/3 π r^2 h If we assume that the aspect ratio of the cone does not change, then r = 8/23 h, so dr/dt = 8/23 dh/dt dr/dt = 3/4 ...
*November 30, 2014*

**ap calc**

and besides, everyone knows that for large values of 2, 2+2=5, for small values of 5.
*November 30, 2014*

**ap calc**

depends on the operator "+" and the base of the number system.
*November 30, 2014*

**geometry**

Just use what yo know about the interior and exterior angles. If there are n sides, you want 180(n-2)/n = 360/n + 100 180n-360 = 360 + 100n 80n = 720 n = 9 check: 180*7/9 = 40+100
*November 30, 2014*

**calc**

∫∫S f(x,y,z) dS we have f(x,y,z) = y z = 2/3 (x^3/2 + y^3/2) ∂z/∂x = x^1/2 ∂z/∂y = y^1/2 so the integral becomes ∫[0,5]∫[0,2] y √(x+y+1) dy dx Looks pretty straightforward now, right?
*November 30, 2014*

**Calculus**

Hmmm. -1/3 - 1 = -4/3 I get (B)
*November 30, 2014*

**AP Calculus**

Just use the quotient rule: h' = (f'g - fg')/g^2 = (14*1 - 0*1/7)/1^2 = 14
*November 30, 2014*

**Calculus**

the slope is 3/2, so y-3 = 3/2 (x-1) 2y-6 = 3x-3 3x-2y+3 = 0 Looks like (A) to me
*November 30, 2014*

**math**

9P5 = 9*8*7*6*5
*November 30, 2014*

**Calculus (Cross Section) again**

consider the stick as a stack of very thin slices of radius √(3x) and thickness dx. Apparently the whole stick is to be considered. v = 2∫[0,10] πr^2 dx where r^2 = 3x. So, v = 2∫[0,10] π(3x) dx v = 6π∫[0,10] x dx I think you can probably...
*November 30, 2014*

**Calculus (cross section)**

since the area of a triangle is 1/2 base * height, if the base is 2y, then we need only double the volume of half the solid, due to symmetry. v = 2∫[0,6] (1/2)(2√(36-x^2))(4) dx = 8∫[0,6] √(36-x^2) dx Now just plug and chug.
*November 30, 2014*

**Math**

Robyn: M=2d Shruti: M=3+d Since Shruti adds a constant 3 miles to her multiple of d, her relationship is not strictly multiplicative.
*November 30, 2014*

**percentages**

1.12*1.14 = 1.2768, or 27.68% change Doesn't matter how tall it was to start with, since we are given the changes as percentages.
*November 30, 2014*

**pre-cal**

h(t) = 5 + (90 sin60°)t - 16t^2 Now just plug in t=3.
*November 30, 2014*

**Gowng**

KE = 1/2 mv^2 You have m and v. So plug it in. First, you need to convert everything to cgs or MKS units.
*November 30, 2014*

**trig**

the radius of the circle is 12. So, the angle θ subtended by the chord is 2π/3. The area of the smaller segment is 1/2 r^2 (θ-sinθ) That should get you started.
*November 30, 2014*

**trigo**

If the nearer observation point is at distance x, and the peak has height h, then we have h/x = tan 31°30' h/(x+1000) = tan 27°45' Now eliminate x and solve for h.
*November 30, 2014*

**Calculus**

the slope at (1,4) is 2, so y-4 = 2(x-1)
*November 30, 2014*

**Calculus**

f = u^3, so f' = 3u^2 u' = 3sin^2(4x) cos(4x) (4) = 12 sin^2(4x) cos(4x)
*November 30, 2014*

**Calculus**

at x = -3, y=0, but y' does not exist, so (A).
*November 30, 2014*

**Calculus**

Looks like (D) to me. You know that y = (x+3)(x-1)^-1 each differentiation subtracts 1 from the exponent of (x-1), so the 2nd derivative must have (x-1)^-3
*November 30, 2014*

**Trigononometry and plaine sperical**

the reference angle is 60 degrees. It's in QIV, so check to see which functions are positive or negative.
*November 30, 2014*

**Math**

(f◦g)(x) = f(g(x)) Since f(x) = 2x^2+4, f(g) = 2g^2+4 = 2(x-3)^2+4 So, we want f(x) = (f◦g)(x) 2x^2+4 = 2(x-3)^2 + 4 x^2 = (x-3)^2 x^2 = x^2-6x+9 0 = -6x+9 x = 3/2
*November 30, 2014*

**Calculus**

y = sin(x+y) y' = cos(x+y)(1+y') y' = cos(x+y) + y'*cos(x+y) y' (1-cos(x+y)) = cos(x+y) so, (B)
*November 30, 2014*

**Calculus**

y = x sinx y' = sinx + x cosx y(0) = 0 y'(0) = 0 So the tangent line is y=0
*November 30, 2014*

**AP calculus**

x^2-2y^2+9x+8y-276=0 x^2+9x - 2y^2+8y = 276 x^2+9x+81/4 -2(y^2-4y+4) + 2(4) = 276 (x + 9/2)^2 - 2(y-2)^2 = 1153/4 This is just an hyperbola with vertices at (-9/2±√1153/2,2).
*November 30, 2014*

**Calculus**

there is a hole at x=1, but the limit on both sides is 3-1 = 2. It just happens that f(1) is not equal to the limit.
*November 30, 2014*

**trigonometry**

I guess I should not have phrased it in that way. The distance traveled is rθ, so divide rθ by the time it takes to travel through θ. That is, the linear speed v is v = rw Note the units v (cm/s) = r(cm) * w (1/s)
*November 30, 2014*

**trigonometry**

the distance s is s = rθ so, the speeds are ds/dt = r dθ/dt Your w is just dθ/dt, so plug in the numbers to get ds/dt
*November 30, 2014*

**Math**

The b's are 1 -2 4 -8 ...
*November 30, 2014*

**Math**

(330*2*33)/5280 = ?
*November 29, 2014*

**trigonometry**

the diameter of the wheel makes no difference. The angular velocity is 2π rad / 45sec = 2π/45 rad/s And that's pi, not pie !
*November 29, 2014*

**math**

well, f = 1/(x+2) f' = -1(x+2)^2 f" = 2/(x+2)^3 f(3) = -6/(x+3)^4 So, the Taylor series at x=0 is 1/2 - x/4 + x^2/8 - x^3/16 + ...
*November 29, 2014*

**math**

6x^2-3x-2y = -5 4x^2+2x-2y = -8 now subtract to get 2x^2-5x = 3 2x^2-5x-3 = 0 (2x+1)(x-3) = 0 ...
*November 29, 2014*

**trigonometry**

Since s = rθ, 5 = 40θ
*November 29, 2014*

**algebra 2**

How about using some parentheses, so it's clear just what you mean? √y+√(y+7)=7 √y = 7-√(y+7) y = 49 - 14√(y+7) + y+7 14√(y+7) = 56 √(y+7) = 4 y+7 = 16 y = 9 check: √9 + √16 = 7 √(x+11) = √x + 1 x+11 = x + 2...
*November 29, 2014*

**algebra**

If the distance is d, then since time = distance/speed, d/20 + d/30 = 12 d = 144 No idea which speed is coming or going.
*November 29, 2014*

**math calculus**

vertical asymptotes where (x-1) = 0 : x=1 horizontal asymptote at y = -x^2/x^2 = -1 extrema where f'=0 : at x=5 inflection where f"=0 : at x=7
*November 29, 2014*

**AP CALC. AB**

Recall the law of cosines. If the third side is x, then x^2 = 13^2 + 17^2 - 2*13*17 cosθ x^2 = 458-442 cosθ so, 2x dx/dt = 442 sinθ dθ/dt So, find x when θ=60°, and then just plug in the given numbers to solve for dx/dt.
*November 29, 2014*

**two step equation-fraction**

4 + m/3 = 5/6 multiply all by 6: 24 + 2m = 5 now subtract 24: 2m = -19 now divide by 2: m = -19/2
*November 29, 2014*

**Math**

nowhere sinx is defined everywhere x is defined everywhere so, sinx - x is also defined everywhere. The only time you can run into trouble is when dividing functions. Then if you try to divide by zero, things are undefined.
*November 29, 2014*

**Algebra B**

18x^3 / 6x^2 = 3x The rest don't matter - we just want the x term after the division.
*November 29, 2014*

**math**

(85-72)/72 = 13/72 = .18055 = 18.1%
*November 29, 2014*

**math**

5gal 1qt = 4gal 6qt = 4gal 5qt 2pt Now you can subtract 2qt 1pt and you have 4gal 3qt 1pt
*November 29, 2014*

**math**

2c+2c + 5oz+2oz = 4c 7 fl oz Just add the cups and oz separately; be sure to remember that 8 fl oz = 1 cup Extra credit: what is the difference between an oz and a fl oz?
*November 29, 2014*

**Algebra 2**

we see that y^2 = 17-4x, so 2x^2 + 5(17-4x) = 53 2x^2 - 29x + 32 = 0 2(x-2)(x-8) = 0 ...
*November 29, 2014*

**Algebra 2**

if you substitute y = 9-x^2, you have 16x^2 - 9(9-x^2)^2 = 144 9x^4 - 178 x^2 + 873 = 0 (x^2-9)(9x^2-97) = 0 Bot those quadratics are easy to solve, no? Each is the difference of two squares.
*November 29, 2014*

**TRIGONOMETRY**

If the width of the street is w and the height of the other house is h, then if you draw a diagram, it should be clear that 9/w = tan 60° (h-9)/w = tan 30° Now you can easily find w, and then h.
*November 29, 2014*

**Geometry**

Label the areas of the triangles with O as a vertex, in the order named above. If we call their areas a,b,c,d then we have a+b=5 b+c=9 c+d=10 d+a=6 I find no unique solution to that system of equations. I'll have to think on it some more, I guess. Some other geometric ...
*November 29, 2014*

**Additional maths**

(x-1)(2x+1)(4x+1) = 0 so, tanθ = 1 or -1/2 or -1/4 0 < θ < 18? That's a weird domain. Anyway, you now have the starting values, so plug and chug.
*November 29, 2014*

**math**

assuming the eye is on the top of the observer's head, it is clear that (22-1.5)/20.5 = tanθ so, what angle θ has tanθ = 1?
*November 29, 2014*

**math**

but, since you don't have a base-2 log button, you'll have to use the change of base formula: log23 = log3/log2 and you can use whatever base you want for that.
*November 29, 2014*

**math**

heck no -- use your own calculator, cantcha?
*November 29, 2014*

**math**

Assuming base 2 throughout, note that log4 = 2 log8 = 3 So, we have 1/4 log4 + 1/8 log8 + 3/8 log(8/3) + 1/4 log 4 1/4 (2) + 1/8 (3) + 3/8 (log8-log3) + 1/4 (2) 1/4 (2) + 1/8 (3) + 3/8 (3-log3) + 1/4 (2) 1/4 (2) + 1/8 (3) + 3/8 (3)- 3/8 log3 + 1/4 (2) 1/2 + 3/8 + 9/8 + 1/2 - 3...
*November 29, 2014*

**Maths**

well, first you had to get rid of the 2, so you could then concentrate on evaluating what was inside the brackets. You could also have processed the brackets by multiplying by 2 on the left first: 2(3x/5 - 1) = 10 6x/5 - 2 = 10 6x/5 = 12 6x = 60 x = 10
*November 29, 2014*

**Maths**

2(3x/5 - 1) = 10 3x/5 - 1 = 5 3x/5 = 6 3x = 30 x = 10
*November 29, 2014*

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