Wednesday

September 2, 2015
Total # Posts: 30,698

**Precalc**

we know sinØ = y/r and cosØ = x/r and tanØ = y/x then for the given: sinØ = x = y/r rx = y ---> r = y/x 3cosØ = y = 3x/r ry = 3x (y/x)y = 3x y^2 = 3x^2 y = ±√3 x
*June 10, 2015*

**math help trigonometry**

yup, that's what you typed the first time. And I pointed out to you that that is not an identity.
*June 10, 2015*

**math help trigonometry**

tan x cos x/sin x=1/sin x cos x or tan x cos x/sin x=1/(sin x cos x) ? For either case, LS = tanx(cosx/sinx) = tanx cotx = 1 For any given x, RS ≠ 1 Your equation is NOT an identity, so it cannot be proven to be true.
*June 10, 2015*

**Algebra**

2nd: 10x^2 + 70x + 100 = 10(x^2 + 7x + 10) = 10(x+5)(x+2) 3rd: 2x^2 + 9x + 4 = 0 (2x + 1)(x + 4) = 0 carry on 4th: real easy ... (x-5)^2 - 81 = 0 (x-5)^2 = 81 take √ of both sides x-5= ±9 x = 5 ± 9 = 14 or -4
*June 10, 2015*

**Geometry**

If I read this correctly, the radius of his model is 6 inches Volume of sphere = (4/3)π r^3 = (4/3)π(216) = 288π cubic inches or appr 904.8 cubic inches SA = 4πr^2 = 4π(36) = 144π inches^2 or appr 452.4 inches^2 to double the radius results in ...
*June 9, 2015*

**Geometry**

As the question suggests, the line joining your two points is a diameter. So the centre is the midpoint, which is ((-5+11)/2 , (-11+19)/2 ) = (3, 4) radius = √(3+5)^2 + (4+11)^2 = √(64+225) = √289 = 17 equation of circle: (x-3)^2 + (y-4)^2 = 289 if x = 24 21^...
*June 9, 2015*

**Geometry**

Where is E ? Where is P ? Are your given points A,B,and C on the circle?
*June 9, 2015*

**Geometry**

Have you learned about b^2 - 4ac, called the discriminant, of the quadratic equation? If the discriminant is zero, there is only one solution, so evaluate b^2 - 4ac. If the discriminant is positive, there are 2 real and different solutions, If the discriminant is negative, ...
*June 8, 2015*

**Math**

5/36 is correct
*June 8, 2015*

**Geometry**

from the linear equation: x = 7y - 75 intersect the two relations, x^2 + y^2 = 225 (7y-75)^2 + y^2 - 225 = 0 49y^2 - 1050y + 5625 + y^2 - 225 = 0 50y^2 - 1050y + 5400 = 0 divide by 50 y^2 - 21y + 108 = 0 (y - 12)(y - 9) = 0 y = 12 or y = 9 if y = 12, x = 9 if y = 9 , x = -12 ...
*June 8, 2015*

**calculus1**

Did you make a sketch? Here is Wolfram's take on the situation. http://www.wolframalpha.com/input/?i=plot+y%3D3x%2Cy%3Dx%2Cy%3D4-x You need the intersection of the lines I assume you want the area of the enclosed region. the intersection points are (0,0), (1,3), and (2,2) ...
*June 8, 2015*

**math**

perimeter = 1 length + 2 width + (1/2)circumference of circle = 16 + 2(7) + (1/2)(2π)(8) = 30 + 8π = appr 55.13 using a better value than 3.14
*June 8, 2015*

**Algebra**

Joanie's equation is (x+16)(x-2) = 0 x^2 + 14x - 32 = 0 , in hers the -32 is wrong, the 14x is correct Kelvin's equation is (x+36)(x-2) = 0 x^2 + 34x - 64 = 0 for his, the 34x is wrong , but the -64 is right the teacher's equation is x^2 + 14x - 64 = 0
*June 3, 2015*

**math**

take two identical rulers and lay them across each other at 30° to get a mental image of the problem. Is the overlapping area not a rhombus with acute angles of 30° and obtuse angles of 150° ? So make that sketch, all 4 sides are equal, label them x draw a vertical...
*June 3, 2015*

**math**

I don't know what your diagram looks like, but try this I will assume that this is a question dealing with vectors. let O be the centre of you NS, EW grid draw a line OB with a heading of N65°W , and mark it x km/h this will be your resultant vector draw in your wind ...
*June 3, 2015*

**Math**

arggh, go with Damon on the first question, I read it as 1/3 of the time, not 1/3 hour less
*June 3, 2015*

**Math**

rate at first part --- x mph rate for return --- x + 5 mph time on first part = 120/x time on 2nd part = 120/(x+5) (these should be entries in your chart) now it says: time on return = (1/3) time of first part 120/(x+5) = (1/3)(120/x) divide by 120 1/(x+5) = 1/(3x) 3x = x+5 2x...
*June 3, 2015*

**Algebra**

One 12 oz can = 12(29.6) ml or 426.4 ml so a six pack = 6(426.4) ml or 5116.8 ml cost of 2 L bottle = $2.19/2000 ml = .001095 $/ml cost of six pack = 2.99/5116.8 = .00058 $/ml draw your conclusion
*June 3, 2015*

**Algebra**

1 litre costs 1.31 1 gallon = 3.7854 L 12 gallons = 12(3.7854) L 12 gallons cost CND $1.31(12)(3.7854) = CN $59.51 1 CND = US 1.02 C 59.51= US (1.02)(59.51) = $60.70 cost per gallon in US$ = 60.70/12 = $5.06 actual cost TODAY: 12 gallons cost 12(3.7854)(1.06) = $48.15 CND $1...
*June 3, 2015*

**algebra**

I bet you meant: x^2/(x+4) ÷ ( (x-2)/(x^2+x-12) ) = x^2/(x+4) * (x+4)(x-3)/(x-2) = x^2(x-3)/(x-2) , x ≠ -4,2
*June 3, 2015*

**Geometry.**

Steve is right, pardon the senior moment, and my diagram even has D corresponding with A
*June 3, 2015*

**Geometry.**

angle A = 180-121-35 = 24° since angle FDE corresponds with angle CAB angle D = 35°
*June 3, 2015*

**Calculus**

At a stop, the velocity is zero, so t^3 - 3t^2 + 3t = 0 t(t^2 - 3t + 3 ) =0 t = 0 or t^2 -3t + 3 = 0 t^2 - 3t + 3 = 0 has no real solution to slow down a(t) would have to be negative a(t) = 3t^2 - 6t + 3 let's see where it is zero t^2 - 2t + 1 = 0 , after dividing both ...
*June 3, 2015*

**probability**

9 cars, 3 of which are red so prob(red,red) = (3/9)(2/8) = 1/12 you do the decimal stuff
*June 2, 2015*

**Math**

cos^4 Ø - sin^4 Ø + sin^2 Ø = (cos^2 Ø + sin^2 Ø)(cos^2 Ø - sin^2 Ø) + sin^2 Ø = (1)(cos^2 Ø - sin^2 Ø) + sin^2 Ø = cos^2 Ø
*June 2, 2015*

**Math**

Took me a while, but I was sure that Steve had answered this for you http://www.jiskha.com/display.cgi?id=1432887186 and again here http://www.jiskha.com/display.cgi?id=1432801604
*June 2, 2015*

**Algebra**

I have a feeling you meant: (x+2)/8 + 1 = 4x/2 multiply each term by 8 x+2 + 8 = 16x 10 = 15x x = 10/15 = 2/3
*June 2, 2015*

**Math**

the first chair can be filled in 8 ways, for each of those the 2nd chair can be filled in 7 ways, (one is seated) and for each of those, the 3rd chair can be filled in 6 ways, so number of ways = 8*7*6 = 336 ways
*June 2, 2015*

**Algebra**

9y/7 = 12y - 24 times 7 9y = 108y - 168 99y = 168 y = 168/99 y = 56/33
*June 2, 2015*

**Math**

nope, what you wrote is the opposite of what you had That is like saying can 5 be written as -5 ?
*June 2, 2015*

**Math**

one problem, you should have brackets around n-1 so a(n) = 5(-2)^(n-1) otherwise, good job
*June 2, 2015*

**Algebra**

divide both sides by -9 (4-k)/12 = -(k+1)/8 cross-multiply 8(4-k) = -12(k+1) 32 - 8k = -12k - 12 4k = -44 k = -11
*June 2, 2015*

**Alegebra**

multiply each side by 12 7(w+2) = 4(w-7) 7w + 14 = 4w - 28 3w = -42 w =-14
*June 2, 2015*

**Algebra 1**

You first started asking this question under the name of Gabs, now you are Amy The above question is clearly based on some previous problem which we cannot see. e.g. how can we possible know how many men and how many women are in the cast?
*June 2, 2015*

**math**

I agree
*June 2, 2015*

**Math**

let's start with dimes ---- x then quarters ---- x+8 nickels ------- 5(x+8) now to the VALUE of these 10x + 25(x+8) + 5(5)(x+8) = 700 10x + 25x + 200 + 25x + 200 = 700 60x = 300 x = 5 so 5 dimes 13 quarters 65 nickels check: 5(10) + 13(25) + 65(5) = 700 my answer is correct
*June 2, 2015*

**math**

You started off ok, but then your logic exploded. You only have to consider the unit digit listing only the unit digit: 2^1 -- 2 2^2 -- 4 2^3 -- 8 2^4 -- 6 2^5 -- 2 2^6 -- 4 2^7 -- 8 2^8 -- 6 .... notice they cycle 2,4,8,6 If the exponent is evenly divisible by 4, the last ...
*June 2, 2015*

**mathh**

number of 5's --- x number of 7's ---- y 5x + 7y = 130 5x = 130-7y x = (130-7y)/5 x must be a whole number, so 130-7y must be divisible by 5, only if 7y is a multiple of 5 to get the least number of stamps we should use as many of the 7's (the y) as we can if y = ...
*June 2, 2015*

**Math**

in effect, each day the turtle gains 2 ft after 1st day/night -- 9 after 2nd day/night -- 7 after 3rd day/night ---5 after 4th day ---- since he crawls 5 ft during the day, he is at the top at the end of the 4th day.
*June 2, 2015*

**Algebra 1**

ok, I misread your question total number of castings = 35 number of casting with the one brother and sister = 1 number of castings with the one brother and two sisters = 1 x 2 = 2 prob (bother and sister in casting) = 3/35
*June 2, 2015*

**Algebra 1**

missing information. In what way are the sibling choices affecting the casting? e.g. can the 2 siblings not be cast together?
*June 2, 2015*

**Algebra 1**

no, any of the 7 men could have the leading role, for each of those, there would be 5 choices for the female role, so number of ways = 7x5 = 35
*June 2, 2015*

**Algebra 2**

Nothing wrong in your steps, except at the end when you have x = 3 or x = -2, we can't allow x = 3, since we then would have divided the original by zero, which we can't do so the only answer is x = -2 Had you factored like in the previous .... (x+5)(x-3)/(x-3) = 3 x+5...
*June 2, 2015*

**Algebra 2**

(x^2+bx-18)/(-2x+4) = 4 = 4/1 same as before, cross-multiply x^2 + bx - 18 = -8x + 16 now let's sub in the x = -17 (-17)^2 - 17b - 18 = 136 + 16 -17b = 152 + 18 - 289 -17b = -119 b = 7
*June 2, 2015*

**Algebra 2**

So I read that as: (x^2-7x+12)/(x^2+x-12) = 3 (x-3)(x-4)/((x+4)(x-3)) = 3 (x-4)/(x+4) = 3/1 cross-multiply 3x+12 = x-4 2x = -16 x = -8
*June 2, 2015*

**Algebra**

give it a try, this one is really easy
*June 1, 2015*

**alegbra2**

50 = 40(1.08)^n 50/40 = 1.08^n n log 1.08 = log(5/4) n = log(5/4) / log 1.08 = 2.899 or appr 2.9 years check: 40(1.08)^2.9 = 50.002 Your expression is incorrect
*June 1, 2015*

**Algebra**

times 6 -66(5-b) = 3b -330 + 66b = 3b 63b = 330 b = 330/63 = 110/21
*June 1, 2015*

**Math**

The LCM of 7,11, and 30 = 7*11*30 = 2310 So 2311 would not divide by any of those, in each case there would be a remainder of 1
*June 1, 2015*

**Math**

multiply each term by 11 10v + 110 = 88(v-7) 10v + 110 = 88v - 616 -78v = -726 v= 121/13
*June 1, 2015*

**Maths**

I assume you want to find the surface area a) two circle + a rectangle = 2π(5^2) + 2π(5)(13) = 50π + 130π = 180π = appr 565.49 cm^2 (Nobody would use π as 3.1 in 2015) b) if it is open at one end, the use only one circle and repeat my steps
*June 1, 2015*

**math**

amount of 45% stuff --- x ml amount of 15% stuff ---- 780-x mls .45x + .15(780-x) = .35(780) times 100 to get rid of decimals 45x + 15(780-x) = 35(780) take over from here
*June 1, 2015*

**Math**

your first response tells me you are having difficulties evaluating a function for a given x x = -1 , y = 3(-1)^2 - 5(-1) - 5 = 3+5-5 = 3 so the point is (-1,3) x = 0 , y = -5 , so the point is (0,-5) x = 1 , y = 3(1)^2 -5(1) - 5 = -7 so the point is (1,-7) x = 2 , y = -3 , ...
*June 1, 2015*

**Math**

make a sketch of the graph using x = -1,0,1,2,3
*June 1, 2015*

**Math (check answers)**

correct , see http://www.wolframalpha.com/input/?i=plot+y+%3D+16x+%2B+150%2C+y+%3D+-x%5E2+%2B60x+-+10
*June 1, 2015*

**math**

trust yourself!
*June 1, 2015*

**Linear quadratics and inequalities**

I meant to say: ".. they intersect at a negative x and a value of x which would result in a negative height."
*June 1, 2015*

**Linear quadratics and inequalities**

I graphed your functions using Wolfram and they don't intersect at a negative x and a value of showing which would show a negative height. check your equations, or else the question is bogus http://www.wolframalpha.com/input/?i=plot+y%3D-0.1x%5E2+%2B2.9x%2C+y+%3D+-x%2B10
*June 1, 2015*

**math**

816 is correct, but I am interested to know why you decided on A.
*June 1, 2015*

**math**

amount of dextr... ----- x amount of water ------ 980-x .55x + 0(980-x) = .275(980) .55x = 269.5 x = 490 mls so you need 490 mls of the dextrose and 490 mls of water (notice that 27.5% = (1/2) of 55% )
*June 1, 2015*

**maths**

cos^2 30° cos^2 45° + 4sec^2 60° + (1/2)cos^2 90° - 2tan^2 60° these are all special angles and you should memorize the trig ratios of the 30-60-90 and the 45-45-90 triangles = (√3/2)^2 (1/√2)^2 + 4(2)^2 + (1/2)(0) - 2(√3)^2 = (3/4)(1/2...
*June 1, 2015*

**mathematics**

What's the question? How many kids are in the pool? What is the water temperature? ??
*June 1, 2015*

**Trigonometry**

Did you make your sketch?? If so, then you will see that tan41° = 50/d d = 50/tan41 = ....
*June 1, 2015*

**Math Algebra 1111**

Whatever the domains of F(x) and G(x) are they would be the same for (F+G)(x) but for (F/G)(x) you would have to restrict any values for which G(x) = 0 In effect you stated that with an example
*June 1, 2015*

**Math**

mine: 3x^2 - 15x + 6 = 0 x^2 - 5x + .... = -2 + ..... x^2 - 5x + 25/4 = -2 + 25/4 (x-5/2)^2 = 17/4 (x - 5/2)^2 - 17/4 = 0 suppose I multiply by 3 3(x-5/2)^2 - 51/4 = 0 looks like they were both correct
*June 1, 2015*

**math**

let the original number be 10a + b where a is the tens digit and b is the units digit. New number: 10(a+4) + b-4, where 4 < b < 9 10(a+4) + b-4 = 2(10a + b) 10a + 40 +b-4 = 20a + 2b 10a + b = 36 a = (36-b)/10 the only value of b which would make 36-b a multiple of 10 is ...
*June 1, 2015*

**Math**

sin^-1 x is the angle Ø so that sinØ = x or x/1 so you have a right-angled triangle with opposite side x and hypotenuse 1 then y^2 + x^2 = 1 y = √(1 - x^2) and cos(sin^-1 x) = cosØ = √(1 - x^2) test for some value of x let's say x = .1234 ...
*June 1, 2015*

**Math**

wecome, please stick to one name, switching names makes it hard to go back in replies
*June 1, 2015*

**Math**

RS = 1 + sin^2 x/cos^2 x = (cos^2 x + sin^2 x)/cos^2 x , using common denominator = 1/cos^2 x = sec^2 x = LS
*June 1, 2015*

**Math**

As you can see from Steve's graph, your "critical" values are x = -1,2,3 so you have the following line segments 1. for x ≤ -1 : let x = -2, y = 10 let x = -1 , y = 7 , slope = -3 y-7 = -3(x+1) -----> y = -3x + 4 2. for -1 ≤ x ≤ 2 let x = -1...
*June 1, 2015*

**lotto math equation**

I have no clue what that is supposed to mean.
*June 1, 2015*

**Algebra II**

no, you have to do the division first, even though the devision is not exact. 4² + [15 ÷ 9 - 3(2)] = 16 + [ 5/3 - 6] = 16 + [-13/3] = 35/3 or 11 2/3 http://www.wolframalpha.com/input/?i=4%C2%B2+%2B+%5B15+%C3%B7+9+-+3%282%29%5D+
*May 31, 2015*

**alegra2**

wolves(t) = 1350 (1.1)^t , were t is number of yrs since 2005 wolves(10) = 1350(1.1)^10 = ....
*May 31, 2015*

**alegbra**

value = 34750 (.88)^7 = ...
*May 31, 2015*

**algebra**

length --- x width ---- x height ---- x+5 x(x)(x+5) = 72 x^3 + 5x^2 - 72 = 0 I tried ±2, and ±3 x = 3 works so by synthetic long division x^3 + 5x^2 - 72 = 0 (x-3)(x^2+ 4x + 24) = 0 the quadratic has no real solution, so x = 3 the prism is 3 by 3 by 8
*May 31, 2015*

**Math**

But why not lay the cone in sideways so that the base of the cone rests against the end of the box So the base of the cone fits the 3 cm by 3 cm end and its height is 6 cm radius = 1.5 cm Volume = (1/3)π(1.5)^2 (6) = 4.5π cm^3 (cosine's cone is only 2.25π cm^3)
*May 31, 2015*

**Math**

Your major mistake is in line 4 You are basically saying log(A-B) = logA/logB there is no such relation! From line 3: y=4log2(8x-56)-12 = 4log2(8(x-7)) - 12 = 4log28 + 4log2(x-7) - 12 = 4(3) + 4log2(x-7) - 12 = 4log2(x-7) , where x>7
*May 31, 2015*

**Algebra 1**

Where does the 126 come from? the original question had √136
*May 31, 2015*

**Algebra 1**

short leg --- x longer leg -- x+2 x^2 + (x+2)^2 = (√136)^2 x^2 + x^2 + 4x + 4 = 136 2x^2 + 4x - 132 = 0 x^2 + 2x - 66 = 0 I will complete the square: x^2 + 2x + .... = 66 + ... x^2 + 2x + 1 = 66 + 1 (x+1)^2 = 67 x+1 = ±√67 x = -1 + √67 , since x > 0...
*May 31, 2015*

**precal**

x/r = cosØ ---> x = rcosØ y/r = sinØ ---> y = rsinØ rcosØ = rsinØ sinØ/cosØ = 1 tanØ = 1 Ø = π/2 which makes sense to me, x = y is a straight line with (0,0) as the intercept and a 45° angle &...
*May 31, 2015*

**Maths**

If 4 buns and 4 cakes cost $4 ,then 1 buns and 1 cake costs $1 --> 4b + 4c = 4 b + c = 1 you want 5 buns + 5 cakes = 5b + 5c = 5(b+c) = 5(1) = $5 The fact that 3 buns cost as much as 2 cakes is not needed and is simply a distraction. It would matter only if you needed the ...
*May 31, 2015*

**trignometry(maths)**

done, see previous post
*May 30, 2015*

**maths**

I assume you mean: (4cosØ - sinØ)/(2cosØ+sinØ) = 4/5 the way you typed it, the statement would be false. given: tanØ = 4/3 make a sketch of the right-angled triangle. You should recognize the 3-4-5 right-angled triangle, if not r^2 = 3^2 + 4^...
*May 30, 2015*

**Algebra**

amount invested at 9% --- x amount invested at 5% --- 1900-x .09x + .05(1900-x) = 143 multiply by 100 to get rid of decimals 9x + 5(1900-x) = 14300 take over from here
*May 30, 2015*

**Math help please thank you**

amount needed of the 25% stuff --- x L amount needed of the 50% stuff ---- 500-x L .25x + .50(500-x) = .35(500) times 100 25x + 50(500-x) = 35(500) 25x + 25000 - 50x = 17500 -25x = -7500 x = 300 L state your conclusion
*May 30, 2015*

**Math**

The slopes of perpendicular lines must be negative reciprocals of each other, so... what do you think ?
*May 30, 2015*

**Math**

No idea what you mean by "x's cannot repeat" The definition of a function is found in every textbook that deals with that topic
*May 30, 2015*

**Hanahan**

I would subtract them to get 10y = 80 y = 8 back into the second: 4x + 24 = -40 4x = -64 x = -16
*May 30, 2015*

**linear systems**

To have an infinite number of solutions, both equations are really the same equation if we multiply the first by 3, 2x + 3y = 48 comparing with the 2nd: kx + 3y = 48 there would be an infinite number of solutions if k = 2 One solution, k ≠ 2 to have no solution, the two ...
*May 30, 2015*

**Math**

I assume you mean $10.22 So the rate is 10.22/145.99 = .07 or 7% y = .07x
*May 30, 2015*

**Math word problem**

number of $1 coins --- x number of quarters ---- y x + y = 21 100x + 25y = 1275 --> 4x + y = 51 subtract them: 3x = 30 x = 10 then y = 11 He has 10 dollar coins and 11 quarters check value: 100(10) + 25(11) = 1275 cents = $12.75
*May 30, 2015*

**maths**

I don't understand your notation
*May 30, 2015*

**Math please help!!**

did you make a sketch? let the shorter diagonal be x by the cosine law: x^2 = 78^2 + 90^2 - 2(78)(90)cos35° ... x = 51.8 to the nearest tenth let the longer diagonal by y y^2 = 78^2 + 60^2 - 2(78)(90)cos145° etc
*May 30, 2015*

**maths**

tan70 = tan(90-20) = cot 20 but cot 20° = y then 1 + tan^2 70 = 1 + cot^2 20° = 1 + y^2
*May 30, 2015*

**math**

so he has 63/7000 m^3 and 1 m^3 is 100^3 cm^3 so he has 63/7000 * 1000000 cm^3 or 9000 cm^3 available let the length be h area of cross-section = 12(15) - (10)(12) = 60 cm^2 60h = 9000 h = 150 cm long or 1.5 m
*May 30, 2015*

**math**

first of all: (x - 1/x)^3 = x^3 - 3x + 3/x - 1/x^3 so we will need x^3 if x = 1 - √2, then x^3 = (1-√2)^3 = 7-5√2 , I will leave it up to you to expand it So we have : (7 - 5√2) - 3(1 - √2) + 3/(1 - √2) - 1/(7 - 5√2) rationalize the ...
*May 30, 2015*

**math**

1/(1+√2) + 1/(√2 + √3) + 1/(√3 + √4) rationalize each term: 1/(1+√2) = 1/(1+√2) * (1-√2)/(1-√2) = (1-√2)/-1 = √2 - 1 1/(√2 + √3) = 1/(√2 + √3) *(√2 - √3)/(√2 - √3...
*May 30, 2015*

**math**

x = 1 - √2 , then x^2 = 1 - 2√2 + 2 = 3 - 2√2 then (x - 1/x)^2 = x^2 - 2 + 1/x^2 = 3-2√2 - 2 + 1/(3-2√2) = .... rationalizing 1/(3-2√2) 1/(3-2√2) =1/(3-2√2 * (3+2√2)/(3+2√2) = (3+2√2)/1 so back in ... = 3 - 2&#...
*May 30, 2015*

**Algebra/Math**

"above" ---> greater than temp > -15
*May 29, 2015*

**math**

shorter piece --- x longer piece ---- 3x-2 solve for x: x + 3x-2 = 10 take it from here
*May 29, 2015*