# Posts by Steve

Total # Posts: 50,437

**Math**

A net is an unfolding of a line drawing of the pyramid. Each line represents an edge of the original figure. So, they will give a way to determine the dimensions and thus the area and volume of the pyramid. If you google "square pyramid net" you will see lots of ...

**Math**

well, what do the lines on the net represent? The surface area is the area of the square base, plus 4 isosceles triangles.

**Trigonometry**

see your previous post, in related questions below. Also, recall that tanA = sinA/cosA

**math please help me out fam**

#1,#2 use any online grapher, such as wolframalpha.com #3 well, what do you have to add to 4 to get to -4? (4,-4) + (-8,?) = (-4,7) #4. Put the words into algebra: T: (x,y) -> (x-4,y+1) ... #5 no figure available #6 to reflect in the x-axis, change the signs of the y-...

**physics help**

Knowing the radius of the circle is not enough. Is it a full semi-circle, or a smaller section? A speed bump 1cm high will not do much.

**calculus**

if the NS sides have length x and the EW sides have length y, then if he spends the entire $500, 2*4x + 2*8y = 500 or, 2x+4y=125 A = xy = x(125-2x)/4 This is just a parabola, with its vertex (maximum area) at (125/4, 15625/32) For #b, y=1878/x, so the cost c is 8x+16y = 8x+16(...

**Trigonometry**

cosA = -3/5 so A is ins QIII sinA = -4/5 now the rest is easy...

**math help? burnet help.**

1/(4x) takes 1/4 as long to get to the same y, so the graph is compressed by a factor of 4 horizontally. the rest is easy ...

**Algebra 1**

how is 3^5^2 a radical? Looks like exponents to me. 3^5^2 = 3^(5^2) = 3^25 no idea where a enters the picture ...

**Algebra 1**

If x sold early, and y sold on the day, then x+y <= 800 6x+9y >= 5000

**Physics**

review your text. I'm sure it says that 1/R = 1/15 + 1/10

**math**

There are 10 fruits. P(banana) = 1/10 since there is only one banana. After the 1st pick, there are only 9 fruits left. Now it should be clear what P(pear) is. P(both) = P(banana) * P(pear)

**Math**

let u = 3^x. Then you have u - 1/u - 1 = 0 u^2-u-1 = 0 solve for u, and then x = log3u

**Math**

measuring from the tower, if the distance away is x, and the rocket had risen to a height h when first observed, h/x = tan25° (h+4)/x = tan66° Now you can solve for x.

**math**

note that x^2+25 = (x-5i)(x+5i) The exponents on the factors are the multiplicities.

**Precalculus**

(x+1)(x-(4-5i))(x-(4+5i)) now just expand that

**Math**

(a) 500(1/2)^t (b.1) 500(1/2)^(25000/5730) = 500(1/2)^4.363 = 24.3 see what you can do from here.

**Precalculus**

y = (x-4)(x-7i)(x+7i) now just expand that.

**math 1**

rubbish. If you followed the steps, what were they?

**math 1**

so, what were your calculations? If we are to analyze why your answer may be wrong, show us how you got it.

**AP physics (multiple choice)**

v is changing direction, and a is also changing direction. so, what do you think?

**Just Curious**

take a look at some of the questions and the responses. You decide.

**Precalculus**

Assuming you mean real coefficients, then D

**Precalculus**

complex zeroes occurs in pairs, so ...

**math**

X/Y = A/B X/Y + 1 = A/B + 1 (X+Y)/Y = (A+B)/B (X+Y)/(A+B) = Y/B

**Math**

area scales as the square of the ratio. So, the square of the ratio is (125/20) = 25/4. So, the ratio is 5/2 Volume scales as the cube of the ratio, or 125/8

**math**

yes no - positive powers only. the leading term is usually the highest power. The constant term is just a number, with no variables.

**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