# Posts by Steve

Total # Posts: 52,055

**Precal**

There are 5C3+5C4+5C5 ways to choose the coins. Now, are there any sums which can be made in more than one way? If so, then they'd be counted twice.

**Math**

-100 + 56 = -44 Mark it off on the number line

**Math**

42 - (-20) = 62

**Calculus**

Since all the answers use f'(x^2), I assume that you meant k(x) = f(x^2) k'(x) = f'(x^2)*2x k"(x) = f"(x^2)*2x*2x + 2f'(x^2) So, (d) check an example, such as k(x) = sin(x^2)

**Math**

that's "difference" not "deference" 58-(-88) = 146

**physics**

if it takes t seconds after the cop gets going, then 20(t+1) = (1/2)(2)t^2

**Geometry**

coin outcomes are 1/2 each If the spinner has n areas, then each has a 1/n chance of being hit So, P(H,C) = 1/2 * 1/n

**Math**

so, plug in your numbers ...

**Math**

surely you can think of two factors of 28. Do they differ by 3?

**Math**

surely you have the formula for the future value of such a cash stream. If not, here's a site with the formula and a calculator. http://keisan.casio.com/exec/system/1234231998

**6th grade Math**

(0*1+1*2+2*5+3*3+4*6+5*7+6*9)/(1+2+5+3+6+7+9) = 4.06 ? 4.1

**math**

and that's pi, not pie!

**Math (Calculus II) (Volume Revolution Setup)**

Yes, if you rotated about x = -6, r=x+6. As you say, using shells of thickness dx, v = ?[1,4] 2?(5-x)(?x-1) dx = 103?/15 You can check your answer by using discs (washers) of thickness dy. In this case, v = ?[1,2] ?(R^2-r^2) dy where R=(5-x)=(5-y^2) and r=1 v = ?[1,2] ?((5-y^2...

**math**

(a) yes, since 40+40+100 = 180 (b) no, since c^2 = a^2+b^2 (c) yes. Use 9 for the base. The other two sides sum to 10, so to make them fit, you have to lift up the vertex to form a triangle.

**Maths-Complex numbers**

we know that |z+2| <= |z|+|2| = 1+2 = 3 use similar logic on the other side.

**maths**

481, naturally but somehow I don't think that's what you really wanted to ask ...

**Math**

(x-48)/x < -8 You cannot just multiply by x as shown above, because x might be negative, which would then reverse the direction of the inequality. If x > 0, then x-48 < -8x 9x < 48 x < 48/9 so, 0 < x < 48/9 If x < 0, then x-48 > -8x 9x > 48 x >...

**MATHS**

you could start here: http://www.jiskha.com/display.cgi?id=1495379575

**math-precalculus**

well, you have sec 2? = 3 cos 2? = 1/3 2? = 1.231 or 2?-1.231 so, divide those by 2, and then add ? to get all 4 solutions.

**Mathematics - Limits**

Hmmm. Using L'hospitals rule, we have the value as (a - 1/2 (1+x)^(-1/2))/2x (1/4 (1+x)^(-3/2)/2 -> 1/8 So, I don't see how we can assign a value to a, because the limit is 1/8 for any value of a. Am I missing something?

**Algebra**

A = 30 + 0.10m B = 50 + 0.05m see what you can do with those equations.

**Physics**

recall that v^2 = 2as = 2E/m a = F/m so, s=2

**Math**

Type in your function here, and it will give you the roots and other info. http://www.wolframalpha.com/input/?i=-0.4x%5E2%2B8.3x%2B4.3 On this one, x^2+3x-10 = 0 you should have factored it: (x+5)(x-2) = 0 Then you would have seen that the roots are -5 and 2, rather than what ...

**Math**

no square matrix can have 8 elements!

**math**

not knowing the prices is a bit of a hindrance. However, if the prices were x and y, and there were s seniors, then sx + y(205-s) = 1930 So, plug in your values for the ticket prices, and solve for s.

**Math**

70 * (1-.25) * 1.06 = 55.65

**Math**

google Heron's formula

**math**

see your earlier post

**math**

solve the equations in pairs: 4x+y=1 2x-y=5 intersect at (1,-3) Now do the same for the other pairs: 4x+y=1 3y+3x=12 2x-y=5 3y+3x=12

**inequalities mathh 7a**

the monthly fee m is figured by subtracting the entry fee ($100) and dividing the rest by the number of months: m = (700-100)/12 = 600/12 = 50

**inequalities mathh 7a**

Your keyboard has all the symbols you need to type math problems.

**inequalities mathh 7a**

ever hear of real math symbols? r/2 - 6 = 14 r/2 = 20 r = 40

**math**

the two equations are y = 12x y = 120+2x

**Math**

if you just want to simplify it, add like terms: (14-12)x^2 + (3+11)x + (4-3) 2x^2 + 14x + 1

**Algebra**

2x^2-3x+1 = (2x-1)(x-1) x^2-6x+1 is a little harder, as it has no rational roots. So, use the quadratic formula to get x = 3±2?2

**Math**

There are 8 socks, so 3/8 * 2/8

**Math**

Assuming any number can follow any other, there are 50^3 combinations. If there are some restrictions, then that number will be reduced.

**Math**

write it as 5*3^n and you have 5*3^0 = 5*1 = 5 5*3^1 = 5*3 = 15 5*3^2 = 5*9 = 45 5*3^3 = 5*27 = 135 starts with 5, not 0

**geometry**

If she walked on the track, it's an arc of a circle. she walked 2/3 of the circumference. As you recall, the circumference of a circle of radius r is C = 2?r. An arc of a circle of radius r that covers an angle ? is s = r? So, the friend walked through an angle ? such that...

**Math**

If there are x students, then 3/8 x + 3/5 x + 20 = x now just solve for x

**Math**

just crank it out... sin(?-30°)+cos(?+30°) = sin? cos30 - cos? sin30 + cos? cos30 - sin? sin30 = ?3/2 sin? - 1/2 cos? + ?3/2 cos? - 1/2 sin? = (?3/2 - 1/2)sin? + (?3/2 - 1/2) cos? = (?3-1)/2 (sin?+cos?) = (?3/-1)/?2 (sin?/?2 + cos?/?2) = (?3-1)/?2 sin(?+45)

**Math**

if there were a adults, then the rest (8-a) were students. So, adding up the prices, 10a + 8(8-a) = 78

**Algebra**

correct

**Algebra**

The terms are powers of 3, so it's a geometric sequence with common ratio 3. So, S10 = 1(3^10-1)/(3-1) = 29524

**Algebra**

bring it on!

**Algebra**

(6-x)/(x^2+2x-3) -------------------- (x^2-4x-12)/(x^2+4x+3) (6-x)/(x+3)(x-1) ----------------------------- (x-6)(x+2) / (x+3)(x+1) (6-x)/((x+3)(x-1)) * ((x+3)(x+1))/((x-6)(x+2)) (6-x)(x+3)(x+1) ------------------------- (x-6)(x+2)(x+3)(x-1) -(x+1)/((x+2)(x-1)) -(x+1)/(x^2+x-2...

**Calculus**

Are you looking for a piecewise function? If so, a simple step function will do the job. f(x) = 0.1 for 0 <= x < 1 0.2 for 1 <= x < 2 0.4 for 2 <= x <= 2.5 You can do similar things with sloping lines, but you will get trapezoids or triangles to work with. If...

**math**

nah - that's what I got.

**math**

the distance traveled is the same: 126 km. So, since distance = speed * time, 6(b-c) = 126 4.5(b+c) = 126

**Algebra Help ASAP**

all correct. good work

**One more question Calculus**

the acceleration graph is apparently a(t) = 10-t so, that gives us v(t) = 10t - t^2/2 + C v(0)=0, so C=0 so, now we have s(t) = 5t^2 - t^3/6 + C s(0) = 10, so C = 10 s(t) = 10 + 5t^2 - t^3/6 s(30) = 0 So, the final position is 10 ft behind the starting line. However, the total...

**Calculus Please Check my answer**

You have found the net rate of change of the water (since E and F are measured in pints/minute) at t=5, not the amount of water. dA/dt = F(t)-E(t) = (t+7)/(2+t) - (ln(t+4))/(t+2) Unfortunately, ln(t+4)/(t+2) does not integrate using elementary functions. Also, without knowing ...

**Calculus really quick check**

I agree.

**Math**

If the bus's speed is x, then since time = distance/speed, 180/x = 240/(x+15)

**math**

I hope your answer to #5 is (b)

**Calculus Please Check my answers**

?[1,x] 2t+2 dt = t^2+2t [1,x] = (x^2+2x)-(1^2+2*1) = x^2+2x-3 ?[0,x] ?(4-t^2) dt = 1/2 ?(4-t^2) + 2 arcsin(t/2) [0,x] = 1/2 ?(4-x^2) + 2arcsin(x/2) - (1/2 ?(4-0) + 2 arcsin(0)) = 1/2 ?(4-x^2) + 2arcsin(x/2) - 1 The domain is clearly [-2,2] The range is [-?-1,?-1] http://www....

**math**

6*5*5 = 150 ft^3 Now just convert ft^3 to liters.

**Math**

well, you have the formula. So, v = s^3 = (3/4 in)^3 = 27/64 in^3

**Math**

If x passed both, then 45+52-x = 64

**geometry**

42

**algerbra**

all correct You can write >= for "greater than or equal" same for <= as intervals, the domain and range are x?[0,?) and y?[2,?)

**Calculus Please Check my answers**

Come to think of it, I changed my mind. g'(0.2) = f'(0.4)*2 = -8

**Calculus Please Check my answers**

#1 not D, since f' is always increasing, f" cannot be zero anywhere. So, no inflection points. I like C, since f'(0) = 0. f has a minimum, since f" > 0 and thus f is concave up. g(x) = f(2x), so g'(x) = f'(2x)*2 g'(0.2) = f'(0.2)*2 = 2*2 = ...

**caculus**

the length z of cord is z^2 = x^2 + 60^2 z dz/dt = x dx/dt when z=100, x=80, so 100 dz/dt = 80 * 5 dz/dt = 4

**Math**

a:p = 2:1 = 10:5

**Math**

2 negatives negative to odd power is negative

**Math**

log16 - 2logx

**Math**

what is the mean proportional for any two numbers?

**math**

consider a hexagon. The triangle's sides join alternate vertices of the hexagon.

**maths**

you want to divide the length into 9 equal parts. Place z 4/9 of the way from x to y.

**Calculus Please Check my answers**

#1 ok #2 4^x grows faster than any polynomial #3 I think more likely (-4,0), as in y = 2sin(pi/4 x) http://www.wolframalpha.com/input/?i=2sin(pi%2F4+x)

**English**

A little-known fact is that "shall" is used for first person: I shall arise and go now. "will" is used for 2nd and 3rd person. They will have a great time. These are reversed in the imperative mood. That is why the Ten Commandments read Thou shalt not steal...

**Math**

For the code, something like count=0 for i=1..2000 with student[i] do { read name,sex,score pct = score*5/2 sub45 = false if pct < 45 then &...

**math**

surely you meant If the diagonal of a square is 14 m, what is the area? well, if the diagonal is d, the area is d^2/2 because if the square's side is s, the area is s^2. The diagonal is s?2, so d^2 = 2s^2

**English language**

if the predicate contains two nouns, usually one will be a direct object, and other the indirect object. The indirect one answers the question: to whom? or for whom? so, what are your choices? Not all of the sentences contain both objects.

**English language**

usually, the first noun in the sentence is the subject. After that, the rest is predicate. What are your choices?

**maths**

YZ = 2(6 sin50°)

**maths**

how do the two figures relate? One diagonal of a rhombus does not tell you anything about its sides or angles.

**maths**

as with all triangles, the smallest angle is opposite the smallest side. So, it is opposite the side of 5cm, and tan? = 5/8

**math**

just plug it into the formula: 15000(1+.095/4)^(4*5)

**Math. HELP!**

#1. remember the slope-intercept form? y = mx+b #2. The slope is (1+3)/(3-1) = 2 So, the point-slope form is y-1 = 2(x-3) Now rearrange that into slope-intercept form. #3 -4y = 8x so, y = -2x for the equation y = kx, k is the constant of variation #4. y = kx, so y/x = k, a ...

**math**

S = a/(1-r) a = 2r, so 2r/(1-r) = 4 Now just solve for r, and you can then get A3 = ar^2

**math**

-48/6 = r^3 Now you can figure a, and then S11 = a(r^11-1)/(r-1)

**math**

a+39d = -54 S15 = 15/2 (2a+14d) S30 = 30/2 (2a+29d) So, 15/2 (2a+14d) + 30/2 (2a+29d) = 0 or, a+12d = 0 Now you have two equations, and you can find a and d, and then the sum S40.

**math**

no way for me to see. However, check your graph against this one: http://www.wolframalpha.com/input/?i=y%5E3+%3D+x-2

**mathematics**

but there are many online articles that show how to do it.

**Mathematics Algebra**

But (x^2 + 2x +4)/(x^2 - 2x - 4) does not always fit into the interval [1/3,3]. If x is very close to 1±?5 then the value is arbitrarily large f(3.2) = -129 plus, you have a typo. I think you mean [9.(3)^2x + 6.(3)^x + 4]/[ 9.(3)^2x - 6.(3)^x +4] using u = 3.3^x That ...

**maths**

huh?

**Mathematics**

"real"ize the denominator by multiplying top and bottom by 1-i sinx. Now you have [tan 2x - i(sinx+cosx)][1 - i sinx]/[(1-i sinx)(1+i sinx)] [tan2x - (sinx+cosx)(sinx) + i(junk)]/(1+sin^2 x) to get zero real part, we need tan2x = sin^2x + sinx cosx any multiple of ? ...

**Math**

For the code, something like count=0 for i=1..2000 with student[i] do { read name,sex,score pct = score*5/2 sub45 = false if pct < 45 then &...

**Math**

If there are x cars and y yellow cars, clearly the new car is yellow, since 1/9 > 1/10 y = x/10 y+1 = (x+1)/9 There are now 81 cars, with 9 yellow.

**trigonometry**

draw the diagram, and it is clear that the distance x is x/100 = tan 50°11'10"

**Math**

c+d = 500 .75c + .50d = 275

**math**

so, you used 304 ft of fence.

**Math, factoring**

Divide by (x^2-x-6) and you get quotient: x^2+(a+1)x+(2a+7) remainder: (8a+24)x + b+6(2a+7) For the remainder to be zero, we need a = -3 b = -6 So, x^4-3x^3-3x^2+11x-6 = (x+2)(x-3)(x^2-2x+1) The other roots are 1,1

**Algebra**

multiply the 2nd equation by 12 and you have 12y = -3x+20 or 3x+12y = 20 so, the two equations are the same.

**Math (Double checking trig identity)**

Nope. my bad. Still you were also wrong (a^2sec?)^2 - (a^2)^2 = a^4 (sec^2?-1) = a^4 tan^2?

**Math (Double checking trig identity)**

not at all. (a^2sec?)^2 - (a^2)^2 = a^4 (sec^4? - 1) = a^4 (sec^2?-1)(sec^2?+1) = a^4 tan^2? (sec^2?+1)

**Cornhole**

if you want names, puntarelli pamela kathryn huckaby

**Math**

No, ?35 is a number like any other, so it has to be included Just as 20+10x is not 30x, 20 + 10?35 is not 30?35 20?35 + 10?35 = 30?35 20+10?35 = 10(2+?35)