Thursday

October 8, 2015
Total # Posts: 34,405

**Find x algebra help**

no easy algebraic way to do this. However, with a little thought, you should be able to come up with 3^2-1 = 2^3
*October 8, 2015*

**Pre-Calculus**

well, just plug in the values: f(x)-f(1) = (-3x+2)-(-3*1+2) = -3x+3 Now divide by x-1 and you have (-3x+3)/(x-1) = -3(x-1)/(x-1) = -3 For the other, the same thing: f(x+∆x)-f(x) = 2(x+∆x)-2x = 2x+2∆x-2x = 2∆x Now divide by ∆x and you have 2∆...
*October 8, 2015*

**math**

a < b x >= y etc. If you have some specific questions, it would be a bit more useful to see how they can be answered.
*October 8, 2015*

**Algebra**

Looks like 9!-1 to me. Now, if you want (9!-6!)/(9-3)! that's 503. What is that supposed to model?
*October 8, 2015*

**Math**

If the faster person takes x hours, then the slower person takes (x+1) hours. So, how much of the job gets done in an hour? 1/x + 1/(x+1) = 1/5 Now just find x.
*October 8, 2015*

**Calculus 1**

the light turns through angle θ where dθ/dt = 4π. If the light hits x km away from P, then x = 4tanθ So, knowing that tanθ = 1/4, plug in the numbers and find dx/dt: dx/dt = 4 sec^2(θ) dθ/dt
*October 8, 2015*

**calculus**

I'll let you do the discussion. You just need to recall that f' = x^3 - 3x f" = 3x^2 - 3 f is increasing where f' > 0 f is concave up where f" > 0 f has max/min where f'=0 f has inflection where f" = 0
*October 8, 2015*

**Geometry**

as with the supplementary angles problem, 4x = (90-x)+10
*October 8, 2015*

**Geometry**

two angles are supplementary if they add to 180. So, if one angle is x, the other is (180-x), and you have 2x = (180-x)-36 2x = 133-x 3x = 144 x = 48
*October 8, 2015*

**Calculus 1**

If the hands are separated by an angle θ, then using the law of cosines, the distance z between the tips is z^2 = 6^2 + 9^2 - 2(6)(9) cosθ The hour hand rotates at a speed of 2π/1440 radians/min, and the minute hand moves at 2π/60 radians/min. So, figure &#...
*October 8, 2015*

**Calculus 1**

at the moment in question, we have a 6-8-10 triangle. You know that tanθ = y/x = √(100-x^2)/x or, x tanθ = √(100-x^2) So, tanθ dx/dt + x sec^2θ dθ/dt = -x/y dx/dt Now just plug in your values to find dθ/dt
*October 8, 2015*

**Calculus 1**

You know that when the water has depth y, the surface of the water has radius (4/6)y So, at depth y, the volume of water is v = π/3 r^2 y = π/3 (4/9) y^3 dv/dt = 4π/9 y^2 dy/dt That's if no water is draining out. SO, adjust dv/dt when you get your answer above.
*October 8, 2015*

**math**

amt = ears/3
*October 8, 2015*

**HELP HELP CALCULUS**

u=5e^x u' = 5e^x y = e^u y' = e^u u' = e^(5e^x) * 5e^x or, y' = 5e^(x+5e^x) y" = 5(1+5e^x)e^(x+5e^x)
*October 7, 2015*

**Geometry**

I've solved this one ---thanks
*October 7, 2015*

**Geometry**

find the image, of o(-2,-1) after two reflections first across the line y=-5 and then across the line x=1
*October 7, 2015*

**Math**

well, geez -- get rid of the parentheses and collect terms! 4+2(x+5) / 14-2(x+5) = 4+2x+10 / 14-2x-10 = 14+2x / 4-2x = (7+x)/(2-x)
*October 7, 2015*

**Math**

v = pi r^2 h so, find h when pi*18^2 h = 31000
*October 7, 2015*

**math**

rearrange to y = 3x-6 That is called the slope intercept form. I suspect you can now tell the slope, eh? For the y-intercept, just set x=0 and find y.
*October 7, 2015*

**algebra**

(9^2)^p = 9^(2p) so, 2p = 8
*October 7, 2015*

**algebra**

3^5 = 243 Just going that far, you can see that none of the choices will do. I suspect a typo in your presentation.
*October 7, 2015*

**math**

I'll assume you mean (x+160)/5 = 2/(3x-2) (x+160)(3x-2) = 10 3x^2+476x-330 = 0 Now you can use the quadratic formula. If that's not what you meant, how about some parentheses to clarify the grouping?
*October 7, 2015*

**algebra**

(2.8 * 10^7)/(1.4 * 10^3) = (2.8/1.4)*10^(7-3) = 2.0*10^4
*October 7, 2015*

**ALGEBRA**

I think you'd do well to review your scientific notation. Any explanation I give here will just be a duplication of what you have already (I hope) read.
*October 7, 2015*

**ALGEBRA**

14,200,000 10,700,000 ---------------- 24,900,000 = 2.49x10^7 or, 1.42x10^7 + 1.07x10^7 = 2.49x10^7
*October 7, 2015*

**ALGEBRA**

3^2 * 3^-5 = 3^(2-5) = 3^-3 = 1/3^3 Your notation is very sloppy.
*October 7, 2015*

**algebra**

x = ±√20
*October 7, 2015*

**Math**

direct: y=kx or x=ky inverse: xy = k where k is a constant. So, check 'em out.
*October 7, 2015*

**math**

102 = 96*1.0625 So, 102 is 6.25% larger than 96
*October 7, 2015*

**Algebra 2**

an expression has no solution. An equation has a solution. Plugging in x = -9/4 results in a value of 3/56. Not sure whether that's what you want. Leaving in x as-is, the expression evaluates to (-3x-6)/14
*October 7, 2015*

**Alg2.**

what, you can't check it? Apparently not. In fact, your results row should be 4 -2 -1 1/2 7/4 25/8
*October 7, 2015*

**Calculus**

I think you mean y = √(11x + √(11x+√(11x))) y^2 = 11x + √(11x+√(11x)) If u^2 = 11x, u' = 11/(2√(11x)) y^2 = u^2 + √(u^2+u) 2y y' = 2u u' + (2u+1)u'/(2√(u^2+u)) 2y y' = u' (4u+1)/(2√(u^2+u)) Now ...
*October 7, 2015*

**Algebra**

Rotating 90° about the origin takes P:(x,y) -> P':(-y,x) So, plug in your numbers.
*October 7, 2015*

**Algebra**

A segment with endpoint a(3,1) and b(-1,-1) is rotated 90 degrees about its origin what are the new of a and b prime?
*October 7, 2015*

**math**

1000 * 1.05^18
*October 7, 2015*

**Algebra**

just solve for x when h=0. Clearly, it is only when x=0. If you mean h(x)=-32x^2/(140^2+x) If you meant it as written, h(x)=-32x^2/140^2 + x = x(1 - 64x/1225) h=0 when x = 1225/64
*October 7, 2015*

**Please help with MATH**

been there, done that. http://www.jiskha.com/display.cgi?id=1444251037
*October 7, 2015*

**Geometry**

Each quadrant contains 90°, starting from 0°. QI is 0-90° QII is 90-180° Or, going the other way, QIV is 0 to -90° QIII is -90 to -180° So, now what do you think?
*October 7, 2015*

**Geometry**

You are correct. An angle has its own measure, regardless of whether it is anywhere near another angle.
*October 7, 2015*

**Math**

assuming she answered all the questions, and got x wrong, then there were (50-x) correct. 2(50-x) - x = 85 Now just find x.
*October 7, 2015*

**Algebra 2**

you know the possible roots are ±8,±4,±2,±1 So, look for the easy ones first, using synthetic division. It's easy to find that there are no rational roots.
*October 7, 2015*

**math**

well, how many hours in a day? -42.4 * 24
*October 7, 2015*

**Math**

there is a 1/3 chance of rolling a 3 or a 5. So, 1/3 of the time, you are likely to get that. What is 1/3 of 260?
*October 7, 2015*

**Algebra**

p(x) = 25x-500
*October 7, 2015*

**Math**

No idea. I can't quite even figure just what the question is. It appears you are working on multiplying fractions. 1st model: 7/20 2nd model: (7/20)(6/7) = (7*6)/(7*20) = 3/10 3rd model: (7/20)(27/35) = (7*27)/(20*7*5) = 27/100 4th model: (7/20)(4/5) = (7*4)/(20*5) = 7/25
*October 7, 2015*

**Math**

If you're taking square roots, the number must be a perfect square to give a rational root. 64 = 8^2, so √64 = 8 is rational Similarly, since 64 = 4^3, ∛64 = 4 is rational. For small numbers, it's easy to check for possible roots, as there aren't that ...
*October 7, 2015*

**math**

40 men @ 8 hrs/day = 320 man-hrs/day 60 men @ 6 hrs/day = 360 man-hrs/day So, it will take 320/360 = 8/9 as long, or 80/3 days
*October 7, 2015*

**Calculus 1**

for a sphere, a = πd^2 da/dt = 2π dd/dt -6 = 2π*11 dd/dt So, dd/dt = -3/(11π) cm/min
*October 7, 2015*

**math**

3+4+5 = 12 360 = 12*30 that should help
*October 7, 2015*

**math**

first off, what is standard form? Those look pretty standard to me already.
*October 7, 2015*

**physics**

s = 3.2t + 1/2 at^2 3.2(3.3) - a/2 (3.3^2) = 16 a = 1.0 so, a = 1/9.8 g So, sin(theta) = 1/9.8 theta = 5.86 degrees
*October 7, 2015*

**math**

well, the sets of three factors of 48 are 1,1,48 1,2,24 1,3,16 1,4,12 1,6,8 2,2,12 2,3,8 2,4,6 3,4,4 Looks like the 10th student will have to duplicate one of those solids.
*October 7, 2015*

**Calculus**

Okay. If the man is x meters from the pole, and his shadow's length is s meters, then using similar triangles, we have (x+s)/5 = s/1.7 So, that means that x = 5s/1.7 - s or, s = 0.51x So, ds/dt = 0.51 dx/dt = 0.51*1.6 = 0.82 m/s
*October 7, 2015*

**Simultaneous equation**

well, using the values 1,3,5 how can you make them add up to 1 if not all are positive?
*October 7, 2015*

**Simultaneous equation**

well, if we want integer solutions, it is clear that not all the values can be positive. However, squares are positive, and we know that 1^2+3^2+5^2 = 35 You will need to make one or two of the variables negative, so you sum to 1, but that should work for the cubes as well.
*October 7, 2015*

**MATH HELP PLEASE**

k(x) = (x−1)^2 − 6 Clearly the vertex is at (1,-6), The x-intercepts are not integers, The y-intercept is at (0,-5) So, pick any integer value except 0 or 1 for x, and plug it in! k(-1) = -2 k(10) = 75 k(-100) = 10195 You can use decimal or irrational values as ...
*October 7, 2015*

**6th Grade Math**

If there were x boxes at $2, the rest (35-x) were $3 boxes. So, add up the sales for everything: 2x+3(35-x) = 92
*October 7, 2015*

**math**

since the relationship is proportional, the distance d and the time t are related by d = kt where k is a constant. So, if you double the time, you have k(2t) = 2(kt) = 2d so, the distance also doubles. Or, you can think of it like this. d/t = k, a constant If you now have 2t, ...
*October 7, 2015*

**Algebra**

Well, you did some calculations (some of them incorrect), but at the end, you didn't even come close to answering the question that was asked. k(x) = (x−1)^2 − 6 Clearly the vertex is at (1,-6), The x-intercepts are not integers, The y-intercept is at (0,-5) So...
*October 7, 2015*

**Math**

if the differences are constant, the sequence is arithmetic. They are 1,2,3,4 So, the sequence is not arithmetic. If the ratios are constant, the sequence is geometric. They are 3/2, 5/3, 8/5, 3/2 So, the sequence is not geometric either.
*October 7, 2015*

**math**

0x3 + 18x1 1x3 + 15x1 2x3 + 12x1 ... 6x3 + 0x1
*October 7, 2015*

**maths**

(6/2)(2a+5d) = 6 (a+d)(a+4d) = -80 or, 2a+5d = 2 a^2+5ad+4d^2 = -80 Now just solve for a and d. One sequence is -14, -8, -2, 4, 10, 16 sum = 6 (-8)(10) = -80 There is another value for a as well
*October 7, 2015*

**Angle of depression need help also**

if the angle of depression is decreasing, then the plane is flying away from the platform. 30s * 600km/hr * 1hr/3600s = 5km So, if you draw a diagram, you can see that if the plane's height is h km, h cot45° - h cot60° = 5
*October 7, 2015*

**What must be added maths need help**

if the needed amount is z, then x/y + z = y/x z = y/x - x/y You can massage that into other forms if you like. Not sure why this caused you trouble. If I ask what needs to be added to 6 to make 10, you naturally say 10-6 = 4 right?
*October 7, 2015*

**Maths**

cubics are not simple to solve, in general. Descartes' Rule of Signs says that there are no negative roots. I suggest a graph or Newton's method for finding the roots. What methods have you studied? f(0) = -2.4 f(1) = 3.49 f(3) = 3.0 f(4) = -0.9 f(10) = -1.08 f(11) = ...
*October 7, 2015*

**Calculus 1**

after t hours, car One has traveled 40t miles car Two has traveled 30t miles Using the Pythagorean Theorem, the distance between the cars is 50t miles. Looks like the distance is growing at a constant 50 mi/hr. No calculus even needed.
*October 7, 2015*

**algebra**

the values to be excluded from the domain are those which make the denominator zero, since division by zero is undefined. so, where is 8x(x-1) = 0?
*October 7, 2015*

**Math**

so, how does your chart look? I assume you have bands of various widths, in various colors.
*October 7, 2015*

**Holy Batman!**

40000*1.16^3
*October 6, 2015*

**Math**

where is 8x^2-8x = 0?
*October 6, 2015*

**algebra 1**

If the son takes x hours, 1/7 + 1/x = 1/4
*October 6, 2015*

**Word problems**

since the height y is given by y = 5.0 + 10.5t - 4.9t^2 just solve for t in 5.0 + 10.5t - 4.9t^2 = 0
*October 6, 2015*

**Algebra**

Ah, but divide by what? If there are r rows, each row has r+9 seats. So, r(r+9) = 1960 r^2+9r-1960 = 0 factors of 1960 are many, but two which differ by 9 are 40 and 49. So, (r+49)(r-40) = 0 There are 40 rows of 49 seats.
*October 6, 2015*

**Pre-algebra**

because the slope of a vertical line is undefined.
*October 6, 2015*

**Math**

Sort of, but murkily. Since the players can be divided into groups of more than one player each, the size of the groups is a divisor of the number of players. So, not prime.
*October 6, 2015*

**adv functions**

-6-√2 <x< -6+√2 -√2 < x+6 < √2 (x+6)^2 < 2 x^2 + 12x + 36 < 2 x^2 + 12x + 34 < 0
*October 6, 2015*

**math**

just take the reciprocal. hr/gal = 1 / (gal/hr)
*October 6, 2015*

**Music**

Bzzzt. Try C
*October 6, 2015*

**Music**

I think I prefer tempo, which describes how fast the music is. Notation just describes how it is written; nothing about the actual music. The others are musical terms, but are not parameters of the music itself.
*October 6, 2015*

**Geometry**

(-2,-1) -> (-2,-9) -> (4,-9)
*October 6, 2015*

**math- 165 calculas**

Of course x=9 is incorrect. If you cut off a 9" corner from the 18" square, the volume is zero! Probably not the maximum value. dv/dx=0 at another value of x. Use that one.
*October 6, 2015*

**math- 165 calculas**

v = x(18-2x)^2 = 4x^3-72x^2+324x Now just find where dv/dx=0 and that will be the maximum (or minimum!) volume. I'm sure you will be able to tell which one.
*October 6, 2015*

**Math (Algebra)**

The scores were the same, so 18m+19 = 15m+31 Now solve for m, the # points for each multiple-choice question.
*October 6, 2015*

**math- 165 calculas**

here the fraction approaches 5x^2/7x^3 = 5/7x -> 0 If the denominator has higher degree than the numerator, the horizontal asymptote is always y=0.
*October 6, 2015*

**math- 165 calculas**

since the numerator is of higher degree than the denominator, there is no horizontal asymptote. As x gets huge, the lower powers become insignificant, and the fraction approaches x^3/x^2 = x So, there is a slant asymptote, but ho horizontal asymptote. The graph is at http://...
*October 6, 2015*

**Algebra**

clearly, the scale is (10 in):(150 ft) = (10 in):(150*12 in) = 1:180
*October 6, 2015*

**physics**

surely your text (or google!) gives an explanation of how to work with circuits and resistors. Do you have some particular problem, which the concepts can be used to solve? The basic principle is that for resistors in series, the current is the same for all of them. for ...
*October 6, 2015*

**Pre Calculas**

#7 since 0<1, f(0) = 0^2 = 0 #8 double the 1st and your system becomes 6x-2y=0 5x+2y=22 Now eliminate y by adding the two equations. #9 If you try elimination here, you get: 24x-6y=15 -24x+6y=14 Can you see that there is no solution to this system?
*October 6, 2015*

**trigonometry**

You have a nice equation. Is there something we're supposed to do with it? As I recall, w represents the width of a river, and you have the angles at A and B. What is there that you cannot do? Just plug in the values of A and B, and the evaluate the numbers to get w!
*October 6, 2015*

**MATH**

or, equivalently, since the numerator is 2(2x+4) r = (2x+4)/(5x+1)
*October 6, 2015*

**Physics**

speed: (130+65)m/(16.0+4.3)s = 9.61 m/s velocity: (130-65)m/(16.0+4.3)s = 3.20 m/s
*October 6, 2015*

**math**

well, Manny is now 18, so ...
*October 6, 2015*

**maths**

yadda yadda yadda. The equations say it all. The words are redundant. If runner 1 passes runner 2, their distances are equal: 0.2t(t-5.1)(t-9.1) = 2.4 + 0.75t t ≈ 10.1 Check out the graphs at http://www.wolframalpha.com/input/?i=0.2t%28t-5.1%29%28t-9.1%29+%3D+2.4+%2B+0.75t
*October 6, 2015*

**pre-cal**

since we have real coefficients, and 2i is a root, so is -2i. If 1/2 is a root, one factor of p(x) is (2x-1). So, p(x) = (2x-1)(x-2i)(x+2i) = (2x-1)(x^2+4)
*October 6, 2015*

**math**

recall the law of cosines, and you will see that θ can be found using 38^2 = 30^2 + 27^2 - 2(30)(27)cos(π-θ)
*October 6, 2015*

**Math**

well, you have 5*4 = 20 How many more to get 30?
*October 6, 2015*

**Algebra**

if they traveled x hours at 20 mi/hr, then they went (260-20x) miles at 70 mi/hr Since time = distance/speed, (260-20x)/70 + x = 9 x = 7.4 check: they traveled 1.6 hr * 70 mi/hr = 112 mi 7.4 hr * 20 mi/hr = 148 mi Total: 260 mi
*October 6, 2015*

**Calculus/Math**

you have (1-2x)/(√(1+x^2)-1) multiply top and bottom to get (1-2x)(√(1+x^2)+1) / (√(1+x^2)-1)(√(1+x^2)+1) = (1-2x)(√(1+x^2)+1)/x^2 = (1-2x+√(1+x^2)-2x√(1+x^2))/x^2 = 1/x^2 - 2/x + √(1/x^4+1/x^2) - 2√(1+1/x^2) as x→&#...
*October 6, 2015*

**MATH**

7(4z-3) > 59 28z - 21 = 59 28z > 80 z > 20/7
*October 6, 2015*