# Posts by steve

Total # Posts: 50,410

**math**

in terms of Friday time, Sat at 2:26:33:07 = 26:26:33:07 So, just subtract 26:26:33:07 -7:44:35:16 to allow for borrowing, that's the same as 25:85:92:107 -7:44:35:16 --------------- 18:41:57:91

**Algebra**

note that each term has an x. So, factor it out: x(x+6)

**math**

2x^2-2x 2x^2-15

**math**

try adding: 2/7 + 1/3 = 13/21 so, 13/21 - 1/3 = 2/7

**maths**

1/4 = 16r^6 r = 1/2 16,8,4,2,1,1/2,1/4

**Geometry**

they are similar

**math**

a = 4?r^2 da/dt = 8?r dr/dt v = 4/3 ?r^3 = ar/3 dv/dt = 1/3 (r da/dt + a dr/dt) Now just plug in your numbers

**MATHS**

ever heard of math notation? (sin^4x-cos^4x)/(1-sin^2x) = (sin^2x+cos^2x)(sin^2x-cos^2x)/cos^2x = tan^2x-1

**Maths**

google is faster. For any pyramid, v = 1/3 Bh where B is the area of the base.

**Maths**

v = 1/3 Bh, so 100 = 1/3 B * 12 Now solve for B, the area of the square base, and then the side is easy

**calculus**

assuming you want to integrate, let u=ln(3+x) dv = x dx du = 1/(3+x) dx v = x^2/2 ?u dv = uv - ?v du ?x ln(3+x) dx = x^2/2 ln(3+x) - 1/2 ?x^2/(3+x) dx Now just divide, and you will have easy stuff left.

**calculus**

let u = (lnx)^2 dv = x^-3 dx du = 2lnx/x dx v = -1/2 x^-2 ?u dv = uv - ?v du ?(lnx)^2/x^3 dx = -(lnx)^2/(2x^2) - ?(-lnx)/(x^3) dx Now repeat with u=lnx and you will be left with just a power of x to integrate.

**Precalculus**

since complex roots occur in conjugate pairs, (x-3)^2 (x-(5+i)) (x-(5-i))

**Precalculus**

y = (x-2)^2 (x+4)^3 now just expand that.

**Fraction**

48

**Math**

Let's make things easy by ending up with 21L of the 1:20 solution. That will contain 1L of pure solution. (2/7)x + 0(21-x) = (1/21)(21) x=7/2 So, mix 3.5L of the 2:5 mixture with 17.5L of water

**physics homework**

depends on the spring constant. asking twice won't help ...

**Algebra**

1200 * e^(.05*6) = 1619.83 too bad you didn't include your work ...

**Math**

2^2(7+2^2)+0*1 71-(20+7) There are lots of ways

**Mathematics**

If (x-1) is supposed to be the altitude, then (x+6 + 3x-4)/2 * (x-1) = 119 x = 8

**Maths**

Huh? LCM(140,n) has to be at least 140...

**Maths**

you are correct. I misread the problem. Can you see where?

**Maths**

(2x+2x+2x)-(10+19+29)+x = 90 Hmmm. I suspect a typo.

**Chemistry**

see related questions below

**Science**

(30m/s)/(60s) = 0.5 m/s^2

**physics**

The resistance of the parallel component is 30R/(R+30) If R' is the equivalent resistance of the circuit, R' = 1/(1/30 + 1/R) + 6 = 36(R+5)/(R+30) The current through the 6? resistor is 120/R' = (10R+300)/(3R+15) The voltage drop across R is [30R/(R+30)]/[36(R+5)/(...

**Math**

T13/T6 = ar^12/ar^5 = r^7 (3/16)/24 = 1/128 = (1/2)^7 so, ...

**maths**

506 base 6 does not exist, since base 6 numbers consist only of the digits 0-5.

**Math**

use the law of sines to show that the base is divided into equal lengths. Then that will imply that the angles at the base of the altitude are also equal, so they are both 90 degrees.

**math**

50 * (1788/3725)

**algebra**

just solve the equation x^2 + (1/2 x + 11)^2 = (2x+1)^2 x = 8

**college algebra**

C(x) is just a parabola. Its minimum value is at the vertex, where x = -b/2a = 7

**math**

90 years is 3 half-lives. That means that 1/8 of the original amount remains. So, the 1 mg that has decayed represents 7/8 of the original amount. So, starting out at 8/7 g, 1/7 g = 14.27mg remains. ---------------------------------- after 4 half-lives, 1/16 of the original ...

**college algebra**

Sorry, that -3072 means its initial height (at t=0) is 3072 feet below ground.

**Math**

since |y| > y, y < 0, so |y| = -y x>|y|, so x+y>0 ==> |x+y| = x+y |x+y|-|x|+|y| = x+y-x-y = 0

**Math**

4^x = 6 2^x = ?4^x = ?6 8^x = (2*4)^x = 2^x * 4^x = 6?6 or, 8^x = (4^(3/2))^x = (4^x)^(3/2) = 6^(3/2) = 6?6

**Math**

(A+B)/2 = A + B/10 A/2 + B/2 = A + B/10 A/2 = 2/5 B A = 4/5 B A.B = 4.5

**Calculus**

2?r^2 + 2?rh = 20 so, h = 10/?r - r the volume is v = ?r^2h = ?r^2(10/?r - r) = 10r-?r^3 so, find r when dv/dr=0

**Math**

If the walk has width w, then since the parkland occupies 80% of the area, (30-2w)(5-2w) = 0.80*30*5 w = 0.4396m

**Calculus**

h+4s = 14 v = hs^2 = (14-4s)s^2 so find s when dv/ds = 0

**Physics**

since V is constant, and PV=kT, P/T = k/V is constant. So, you want P such that P/(273+25) = 7/(273+15)

**math**

because 1/c = c^(-1)

**Math**

check your prior post in the related questions below ...

**Calculus please help!**

consider f'(1) where f(x) = ?x where you should have written the limit as (?(1+h) - ?1)/h

**Pre calculus**

p(x) = (x-11i)(x+11i) = x^2 - (11i)^2 = x^2+121

**Physics**

distance = speed * time so, time = distance/speed

**Math(PLEASE HELP ME ASAP)**

22.50 * 1.15 = 25.875

**Math(PLEASE HELP ME ASAP)**

well, how did you figure it out?

**Math**

well, the base has area 132/11 = 12 so ...

**Calculus**

(a) c = 13(4x+y+4x+y) + 5(3y) = 112x+41y (c) xy=9, so y=9/x, and c(x) = 112x + 369/x now find x where dc/dx = 0

**Math**

huh? Just solve (75+x)(225-5x) = 16000 x = 5 so, ...

**Precalculus**

a) ok b) The average cost is the cost per unit, or c?(x) = c(x)/x So, c?(x) = 0.02x + 0.5 + 40/x The marginal average cost is thus c?'(x) = 0.02 - 40/x^2

**Algebra - rats**

I see other typos. The final answer is correct, but let me just do it right: Since the vertex and focus both lie on the line x = -2, the axis of symmetry is vertical, meaning the equation is (x-h)^2 = 4p(y-k) Since the focus above the vertex, p is positive. You know that the ...

**Algebra - typo**

As you noticed, the axis is vertical, on he line x = -2

**Algebra**

Since the vertex and focus both lie on the line x = =2, the axis of symmetry is horizontal, meaning the equation is (x-h)^2 = 4p(y-k) Since the focus above the vertex, p is positive. You know that the parabola x^2 = 4py has the vertex at a distance p from the focus. Here, that...

**Math, Algebra**

If you drop an altitude to the center of the base, then looking from the side you have a right triangle with legs 4 and 22. So, the hypotenuse of the triangle (the slant height of the pyramid, and the altitude of each triangular face) is h = ?(4^2+22^2) = ?500 = 10?5 That ...

**MAth help connexus**

c - review the section on functions, esp. the vertical line test.

**math - trigonometry**

one walks on a heading, not a bearing. a) Note that angle Q is 90 degrees, so you have a right triangle. Easy... b) This is the correct use of "bearing." If the bearing is x, then tan(x+25) = 18.7/11.4 A diagram will show you why.

**Math Measurment unit test**

1. A = bh 2. A = bh/2, so looks good.

**math**

2/s = 1/4 + 1/5

**math**

well, the volume of water flowing in every second is 1.2cm * 350cm/s = 420 cm^3/s Now divide the volume of the tank by the inflow rate to get the time (in seconds)

**calculus**

The 13 is just a scale factor, and has no bearing. You know that ? ? 1/10^n = 0.11111111 = 1/9 n=1

**Maths**

If the 5th is 4 times the 3rd, then r=2 a+ar = -4 a+2a = -4 a = -4/3 now take it from there.

**math**

Let the man run to point P, which is x meters upstream from the line AB. He swims across, being swept y meters downstream, and then runs to B. We know that he is in the water for d/(v/3) = 3d/v seconds, so y = 3d/v * v = 3d. His path is thus ?(d^2+x^2) meters on land, d meters...

**math**

a?

**math**

again c? aa

**math**

c¯

**math**

before: 00AF

**math**

before: AFc after: cAF

**math**

c-bar ̅

**Math**

well, the last digit is 5 start there.

**Physical science**

2Na + 2H2O = 2NaOH + H2 now just multiply everything by 3/2

**Math**

y = x-12

**Math**

write them as 4x-6y = -48 x+6y = 18 add them together and you get 5x = -30 ...

**chem**

assuming a constant temperature, PV = kT is constant. So, you want P such that P*2.94 = 0.988*3.35

**Math**

y = 1+2x 3x+(1+2x) = 11 ...

**chem**

assuming a constant temperature, PV = kT is constant. So, with 5/4 the pressure, you get 4/5 the volume

**Math**

square both sides write cos^2 = 1-sin^2 Then you have a quadratic in sinx

**physic**

how long does it take to fall 160m? The horizontal speed does not affect that. 4.9t^2 = 160 Now, the horizontal speed does not change, so the body travels 20t meters

**Math1**

You want t years, where 2100(0.95)^t = 600 0.95^t = 2/7 t = log(2/7)/log(0.95) = 24.42

**Pre-cac**

well, even without completing the square, you know the vertex is at x = -b/2a = 3/2a So, a(3/(2a))^2 - 3(3/(2a)) + 5 = 15 5 - 9/(4a) = 15 a = -9/40 But, in the spirit of completing the square, ax^2-3x+5 = a(x^2 - 3/a) + 5 = a(x^2 - 3/a + (3/2a)^2) + 5 - a(3/2a)^2 = a(x - 3/2a...

**Maths**

4000*1.015^2 = 4120.90

**Math**

a. correct b. The volume with the cuts made and the sides folded up is v = (12-2x)(18-2x)x = 4x^3-60x^2+216x If you have calculus, then dv/dx = 12(x^2-10x+18) dv/dx=0 when x=5-?7 ? 2.35 If no calculus, then a graphical or numeric method is needed.

**Math**

(x,y)-> (x-4,y+1) is correct. But you still botched it: (-3,1) -> (-3-4,1+1) = (-7,2) + means ADD: 1+1 = 2, not 0! same for subtracting: -3-4 = -7 -3+4 = 1 -1 is just plain bogus, man.

**Math**

left 4 means subtract. as you said, up 1 means add. But even if your translation were correct, you did not do the math right 1+1 ? -2 Try again. You're getting the +'s and -'s mixed up some.

**Math**

naturally, we can see no graphs here. What are your choices for the others?

**Math**

(a) trés simples, non? (b) 2x cosy - x^2 siny y' - cosy y' = 0 y' = (2x cosy)/(cosy + x^2 siny) So, at (0,?), y'=0 (c) so, the tangent is a horizontal line, and the normal is a vertical line through (0,?).

**Pre-calculus**

I will assume a typo, and work with x^2+10x+4y+9 = 0 That may not be what you had in mind, but you can follow the method using your corrected equation. x^2+10x+4y+9 = 0 4y = -x^2-10x-9 y = -1/4 x^2 - 5/2 x - 9/4

**Math 120**

6+2 + 2 (conversion+safety) 6+1 + 3 6+2+2 (two safeties) 3+3+2+2 2+2+2+2+2

**math**

the nice thing about Ø is that Ø-1 = 1/Ø

**math**

sinx = cos^2x sinx = 1-sin^2x sin^2x + sinx - 1 = 0 now solve for sinx, and pick the positive solution.

**Calculus - Newton's Method**

well, f'(x) = 1-sinx So plug that into Newton's method. Here is a nice web site: http://keisan.casio.com/exec/system/1244946907

**calculus**

you will need to use integration by parts, twice. 1st step: u = t^2 dv=sin(2t) dt du = 2t dt v = -1/2 cos(2t) That makes the first integration by parts: ?u dv = uv - ?v du ?t^2 sin(2t) dt = -1/2 t^2 cos(2t) + ?t cos(2t) dt Then repeat, letting u = t dv = cos(2t) dt You should ...

**Calculus**

I suspect a typo, since what you have written is just f(x) = x/(x+x^2) = 1/(1+x) f' = -1/(1+x)^2 f" = 2/(1+x)^3 Anyway, fix that, find f", and recall that f"=0 means an inflection point f" > 0 means concave up f" < 0 means concave down

**calculus**

use integration by parts. That is just the product rule in reverse. d(uv) = u dv + v du u dv = d(uv) - v du ?u dv = ?d(uv) - ?v du ?u dv = uv - ?v du So, here we just let u = (lnx)^2 dv = dx du = 2lnx * 1/x dx = (2lnx)/x v = x ?(lnx)^2 dx = x(lnx)^2 - ?2lnx dx Now repeat, this...

**math**

x >= 15000/12 = 1250 So, consider how long a ride might take, and how long the park is open each day.

**math**

5 divides evenly any multiple of 5. Not sure what you mean by "5 divide into equaly" [sic]

**Help please I am lost, algebra**

if g(14) = -6, then that means you have the point (14,-6) on the graph. Using the point-slope form, then, you get y+6 = -5/4 (x-14) Now massage that to the slope-intercept form: y+6 = -5/14 x + 35/2 y = -5/14 x + 23/2

**Math**

The first model is linear because the changes are constant. In fact, y = x/2 - 1 For the second, check the differences: 1st: 5.1, 20.4, 81.6, 326.4 2nd: 15.3, 61.2, 244.8 If t were quadratic, the 2nd differences would be constant. Since they are also growing rapidly, you ...

**Functions, please help**

2700x+21000 = 83400 just solve for x 2010 is 7 years after 2003, so just plug in x=7 for P(x)