Saturday

December 10, 2016
Total # Posts: 47,042

**math**

it dropped 6K in 2 years. That's 3K per year. So, it started at 3K more than 20K. y = 23000 - 3000x

*October 22, 2016*

**Algebra**

5646000 * 1.047^n = 13000000000

*October 22, 2016*

**Algebra**

$15/hr

*October 22, 2016*

**Algebra**

do you not see the relation between hours and money?

*October 22, 2016*

**Math, Functions**

the domain of all polynomials is (-∞,∞) There is no value of x which is not allowed. Since the graph is a sloping straight line, it extends forever in both directions, so the range is also (-∞,∞)

*October 22, 2016*

**Math - correction**

nope. only 13 are boys. 22:13

*October 22, 2016*

**Math**

counting boxes, with x sold on Monday, x + x-3 = 15 12x is the number of pencils.

*October 22, 2016*

**math**

the lot will be 37.5 by 25 where the extra piece is parallel to the short sides.

*October 22, 2016*

**math**

the monthly charge for A is clearly 14.45 (0 minutes) So, for B, $(22.10-14.45)/45min = $0.17/min

*October 21, 2016*

**Math**

just multiply by 45 512.

*October 21, 2016*

**Math**

56000+9x = 16x

*October 21, 2016*

**calculus**

M'(x) = .0009x^2 - .132x + 4.64 Now just evaluate M' for your years.

*October 21, 2016*

**Algebra**

A is correct B is correct if the line goes through (-4,0)

*October 21, 2016*

**Math**

fix typo revenue = p*demand then take derivative.

*October 21, 2016*

**Calculus**

for f"=0 you need sin2x - cos4x = 0 sin2x - (1 - 2sin^2(2x)) = 0 2sin^2(2x) - sin(2x) + 1 = 9 (2sin2x+1)(sin2x-1) = 0 sin2x = -1/2 sin2x = 1 Count the solutions in the interval. Looks like ten to me.

*October 21, 2016*

**physics**

use the Pythagorean Theorem, since the directions are perpendicular.

*October 21, 2016*

**Maths**

45-(30+20-8) = 3

*October 21, 2016*

**math**

well, a*10^5 = 2000

*October 21, 2016*

**math**

(-28.9-(-79.9))/(19.125-12.75-(79.9/18.8)) = 24.0 ft/min

*October 21, 2016*

**Calvary School**

9/4 yards, 3 inches

*October 20, 2016*

**Physics**

8*10 + 2(-5) = (8+2)v

*October 20, 2016*

**algebra/AP physcis**

Can't be done. If you move the 1.2v^2/.6 to the left, you wind up with -.2v^2/.6 = 9.8 The left side is negative and the right side is positive. I suspect a typo somewhere, or an error in devising the equation.

*October 20, 2016*

**Calculus**

y = (x^2 +2x)^3(x+2) = x^3(x+2)^4 y' = 3x^2 (x+2)^4 + x^3 * 4(x+2)^3 * 1 = x^2(x+2)^3 (3(x+2) + 4x) = x^2(x+2)^3(7x+6)

*October 20, 2016*

**Math**

if p>0, then q<0 and -3p+4q is negative. So, p<0 and q>0 -3p > 0 4q > 0 so -3p+4q > 0

*October 20, 2016*

**Math**

huh?

*October 20, 2016*

**GCF**

940 = 20*47 420 = 20*21 940-420 = 20(47-21) = 20*26 = 520

*October 20, 2016*

**math**

f(g) = 4g-5 = 4(3x)-5 = 12x-5

*October 20, 2016*

**Math**

sides have ratio 36/22 areas have ratio (36/22)^2

*October 20, 2016*

**MATH**

x(t) = -t + t^3/6 v(t) = dx/dt = -1 + t^2/2 I think you can now answer the questions if you think a bit.

*October 20, 2016*

**Math**

well, the triangle's 3 sides can each intersect with two sides of the rectangle, making six intersections. Now see what you can do with the circle.

*October 20, 2016*

**Algebra**

3.29, naturally. cost = 3.29 * gals

*October 20, 2016*

**pre-calc**

Do you mean e^-t cost + e^-t sint = 0? If so, factor out the e^-t and you have e^-t (cost - sint) = 0 e^-t is never 0, so that leaves sint = cost or tant = 1 I think you can work with that, eh? Remember all four quadrants.

*October 20, 2016*

**algebra 1**

looks something like y = 2^x but you will have to adjust it a bit.

*October 20, 2016*

**Math**

nope. You lost a minus sign.

*October 20, 2016*

**Math**

1/3 Divided by 3 = 1/9 or 0.11111...

*October 20, 2016*

**Math**

1/3 Divided by 3

*October 20, 2016*

**physics need asap help**

add up the momentum vectors. Assuming the particles are moving in the +x and +y directions before the collision, with speeds u and v, <mu,0> + <0,2mv> = <3m*2sin37°,3m*2cos37°> now just solve for u and v.

*October 20, 2016*

**Math**

common ratios, common ratios... Find one ratio and use it to get the other dimension.

*October 20, 2016*

**Math**

the ratio of heights is the same as the ratio of ground distances, namely 25/10

*October 20, 2016*

**Math**

2/4 divided by 6

*October 20, 2016*

**A.P.**

Good question. If so, then the common difference must be zero, so x-2 = 5 x = 7 and the terms are 5,5,5 However, if the last terms is supposed to be x+2, then the terms are clearly 3,5,7

*October 20, 2016*

**math**

tell you what -- I'll show you one. Then you can work on the second. 6 dimes = 6*10 = 60 Now just replace some of the dimes with nickels.

*October 20, 2016*

**Check my answers**

Yes Ms. Sue

*October 20, 2016*

**Check my answers**

100% Correct for the Practice

*October 20, 2016*

**Check my answers**

1. B 2. B 3. C 4. D 5. B 6. A 7. B 8. A, C 9. A 10. C 11. C 12. A 13. D 14. A 15. C

*October 20, 2016*

**Check my answers**

I will check you answers give me a minute.

*October 20, 2016*

**maths sequence**

Recall that the sum of n terms is Sn = n/2 (first+last)

*October 20, 2016*

**Social Studies**

I think B. What do you think?

*October 20, 2016*

**Algebra**

Not quite sure what you means by "3/x 1/9" If I use conventional online notation, I could write something similar as 1/3 + 3/x - 9/x^2 = (x^2+3x-9)/x^2 Those do not simplify easily, so I expect I have gotten something wrong. Maybe you could clean it up a bit and ...

*October 20, 2016*

**math**

Or, since the two numeric terms have a ratio of r^2, so r^2 = 75/12 = 25/4

*October 20, 2016*

**algebra 2**

(c) obviously the vertex form (d) standard form, so you have the needed a,b,c coefficients. What do you think for (a) and (b)? Try doing them and see what happens.

*October 20, 2016*

**Math (Function differentiable)**

for the function to be continuous, the limit on both sides must be the same. They are both 5 (not 4!). So, your function is continuous. But that is not enough. Think of f(x) = |x|. It is continuous, but not differentiable at x=0. For it to be differentiable, the derivative on ...

*October 20, 2016*

**physics**

conserve momentum: Is the ball stationary, moving directly away/toward him? Pick your answer, and then just write the momentum equations before and after collision. Solve for v.

*October 20, 2016*

**Algebra 2**

no graph

*October 20, 2016*

**Algebra**

-4 <= 4(6y-12)-2y -4 <= 24y - 48 - 2y -4 <= 22y - 48 . . .

*October 20, 2016*

**Calculus**

revenue = price * demand If there are x price decreases, then price: p(x) = 5.00 - .20x demand: d(x) = 200 + 5x so, x = (d-200)/5 r(x) = p(x)*d(x) = (5 - x/5)(200+5x) = (5 - (d-200)/25)d = 13d - d^2/25 marginal revenue for demand is thus dr/dd = 13 - 2d/25 r'(120) = 13 - ...

*October 20, 2016*

**Math**

1/t + 1/(3t) = 1/9

*October 20, 2016*

**math**

so far, it's a tie!

*October 20, 2016*

**Algebra**

well, 1/5 + 2/5 = 3/5 4/5 - 1/5 = 3/5 so, (3/5)(x+y) hah!

*October 20, 2016*

**coordinate geometry**

Is that y = (√3)x+4 ? The other circle is clearly (x+2)^2 + (y+2)^2 = 4 and the way the question is worded, it appears that the first circle is also tangent to the x-axis at (1,0), making it (x-1)^2 + (y-k)^2 = k^2 Now it should not be too hard to find k so that the ...

*October 20, 2016*

**Math**

this is (almost) correct, since sec^2/tan^2 = 1/cos^2 * cos^2/sin^2 = csc^2 Actually, the derivative is -csc^2x cotx = cosx/sinx d/dx(cotx) = (-sinx*sinx - cosx*cosx)/sin^2x = -(sin^2x-cos^2x)/sin^2x = -1/sin^2x = -csc^2x Next time, show your work, so we can see what went wrong.

*October 20, 2016*

**Maths**

what's the trouble? Just write all that as math: (3/4)x + 3 1/2 = 6 1/2 - (2/3)x Now just solve for x.

*October 20, 2016*

**Trigonometry**

First, review your basic trig functions. Then see the related posts below.

*October 20, 2016*

**Math**

ummm, I'd say (3/5)*1200

*October 20, 2016*

**Math**

well, 30mi/10mph = 3 hr I assume you can figure what time that makes it...

*October 20, 2016*

**CALC**

If you found the mass, you're halfway there. M = ∫ ρ(x,y) dA The x-coordinate of the center of mass is just Mx = ∫xρ(x,y) dA / M and similarly for My.

*October 20, 2016*

**math**

If the polynomial has real coefficients, then the complex roots occur in conjugate pairs, so +3i is also a root. So, your description is a bit muddy, but I'd say y = (x-4)(x^2+9)(x-1) That is degree 4. If you want x=1 to be of multiplicity 2, then that would be y = (x-4)(x...

*October 19, 2016*

**Maths**

all those words and improper operators. You apparently want (x-3) to be a factor of x^4-3x^3+2x^2-2x+5 recall that the Remainder Theorem says that this will be the case if f(3) = 0. So, evaluate that.

*October 19, 2016*

**math**

1 by 81

*October 19, 2016*

**Algebra**

It lost 1/2 its value in 5 years. Using a linear model, that is 1/10 each year. So, in 7 years, it will have lost 7/10 of its initial value.

*October 19, 2016*

**math**

(a) first step: use scientific notation. (b) Assuming no typo, it is the same (1 times as great)

*October 19, 2016*

**math**

I'd double the 1st line and get -4x-2y = 2 -4x-2y = -1 Now what do you think?

*October 19, 2016*

**Maths**

fact #1: the two numbers are x and x+24 fact #2: ... then solve for x.

*October 19, 2016*

**Calculus**

profit = revenue - cost = demand * price - cost = x*p(x)-c(x) = x(.78+.003x) - (480-.32x+.0005x^2) = 0.0025x^2 + 1.1x - 480 marginal profit = dp/dx = 0.005x^2 + 1.1 p'(700) = 0.005*700^2 + 1.1 = 2451.1

*October 19, 2016*

**Gph**

If you know you are wrong, would it not behoove you to study a bit more? How about starting here: http://www.atmo.arizona.edu/students/courselinks/fall12/atmo336/lectures/sec1/humidity.html

*October 19, 2016*

**math**

(3x)(4x)(5x) = 20580 60x^3 = 20580 ...

*October 19, 2016*

**Algebra 2**

what? Can't plug in a value for x and evaluate? For example, f(23) = 4*23-2 = 90 direct variation means that y = kx So, plug in the numbers and find k.

*October 19, 2016*

**math**

well, what is 19+28 ? add that to 21

*October 19, 2016*

**Calculus**

1/9, since it's 3x^2 don't you read your own equations?

*October 19, 2016*

**Calculus**

This is just the Mean Value Theorem. Find c such that f'(c) = (f(4)-f(0)/(4-0) = 16/4 = 4 Since f'(x) = 2x, where does 2x=4? y-4 = 4(x-2) http://www.wolframalpha.com/input/?i=plot+y%3Dx%5E2,+y%3D4(x-2)%2B4,+y%3D4x

*October 19, 2016*

**Calculus**

f' = -1/x^2 + 1 = (x^2-1)/x^2 f'(1) = 0 so, using the point-slope form, y-2 = 0(x-1) or, y=2 http://www.wolframalpha.com/input/?i=plot+y%3D1%2Fx+%2B+x,+y%3D2,+0+%3C%3D+x+%3C%3D+2

*October 19, 2016*

**math**

in a town of 9000 people, 7000 died. What portion lived?

*October 19, 2016*

**geometry**

15 cos19° and 15 sin19° Looks like it's time to review the basic trig functions.

*October 19, 2016*

**math**

tanθ = 191/(4*5280) note that for small angles (in radians) tanθ ≈ θ

*October 19, 2016*

**MATH**

What are you measuring? assign variables for the quantities of interest. What are the constraints?

*October 19, 2016*

**calculus**

sure looks like -4 to me. You had to ask?

*October 19, 2016*

**Trig**

Apparently this is a bizarre attempt to write a fraction. Try using parentheses and virgules, as in (cscθ+1)/(cscθ-1) or something

*October 19, 2016*

**geometry**

draw two adjacent angles. I think the answer will be clear. If not, which statements do you know are true?

*October 19, 2016*

**physics**

The three blocks are accelerated in the amount of a = F/m = 14.3/(1.1+2.72+4.05) = 1.817 m/s^2 They all have that same acceleration. So, since the 1.1kg block is only pushing 2.72+4.05 kg, its force is F=ma Similarly, the last block is pushed with a force of 4.05a N

*October 19, 2016*

**Calc**

looks like ∫[-3,3] ∫[2,19-y] ∫[-√(9-x^2),√(9-x^2)] dy dz dx

*October 19, 2016*

**calculus**

you do need the () around 1/x, since powers are done first, and you don't want (e^1)/x 3+e^(1/x) is never zero, so there are no discontinuities because of a zero denominator. However, since 1/x is undefined at x=0, there is a jump there. lim as x->0- = 5/3 lim as x->...

*October 19, 2016*

**G.p.**

see the related questions below

*October 19, 2016*

**Geometric progression**

Well, you just didn't catch my typo. c*e = (1/3)r^3 * (1/3)r^5 = (1/3)* (1/3)r^8 = (1/3)(432) = 144 So, my correction was wrong. Clearly you did not examine my work to see where the mistake was made. You just threw up your hands and complained I was wrong. Bad form, I must...

*October 19, 2016*

**Geometric progression**

oops. (1/9)(432)

*October 19, 2016*

**Geometric progression**

(a)(ar^8) = a^2 r^8 = 432 c*e = ar^3 * ar^5 = ...

*October 19, 2016*

**A.p.**

If there are n means, then [x,3x] is divided into n+1 intervals. Thus, you have an arithmetic sequence where a = x d = (3x-x)/(n+1) = 2x/(n+1) x, x + 2x/(n+1), x + 4x/(n+1), ... x + 2nx/(n+1), x + 2(n+1)x/(n+1) = x, (n+3)/(n+1) x, (n+5)/(n+1) x, ... 3(n+1)/(n+1) x

*October 19, 2016*

**A.p.**

n/2 (2a + (n-1)d) = 3n^2+2n 2an + dn^2 - dn = 6n^2+4n dn^2 + (2a-d)n = 6n^2+4n d = 6 2a-d = 4, so a = 5 Tp = 5+(p-1)(6) = 6p-1 check: the AP is 5,11,17,23,29,... S1 = 5 = 3*1^2+2*1 S2 = 16 = 3*2^2+2*2 S3 = 33 = 3*3^2+2*3 S4 = 56 = 3*4^2+2*4 ...

*October 19, 2016*

**Algebra**

Just do a long division. Add what you need to make the remainder zero.

*October 19, 2016*

**Math**

3/15 are blue (in math)

*October 19, 2016*