Friday

July 1, 2016
Total # Posts: 41,529

**Trigonometry**

off course as measured how? Actual distance between target and position, or distance perpendicular to the intended course? In the first case, use the law of cosines. Both sides of the included angle are 152*2
*June 8, 2016*

**Trigonometry**

draw the diagram. It should be clear that if the height is h, h cot32° - h cot40° = 12
*June 8, 2016*

**Trigonometry**

draw the diagram. It should be clear that north: 83 cos37° west: 83 sin37°
*June 8, 2016*

**math**

looks good to me.
*June 8, 2016*

**Math - eh?**

NO IDEA! Where the heck is X?
*June 8, 2016*

**PreAlgebra**

x^2 = 22^2 - 7^2 now just crank it out
*June 8, 2016*

**Math**

your formula is correct, but your answer is not. Oh, well.
*June 8, 2016*

**Math**

a = 2πr(r+h) = 2π(9)(9 + 19) = 504π So, D How did you get 523?
*June 8, 2016*

**Math**

yes, it is correct.
*June 8, 2016*

**Science**

PV/T is constant, so you want T such that 1.82*18.7/353 = 1.41*25.1/T T ≈ 367
*June 8, 2016*

**aptitude**

clearing fractions and decimals, we have 125x + 5y = 21 3x + y = 3 now you can substitute y = 3-3x to get 125x + 5(3-3x) = 21 Now it's easy to solve for x, and then you can get y.
*June 8, 2016*

**Algebra**

Seems we want cube roots: ∛540r^3 s^2 t^9 = ∛27*20 r^3 s^2 (t^3)^3 = 3rt^3 ∛20s^2 so, yes, you are correct
*June 8, 2016*

**maths**

just plug in values for t: x(2) = 16*2 - 2*2^2 = 32-8 = 24 likewise for t=6
*June 8, 2016*

**math**

starting with 1, after chocolates, 3/5 remains after gift to sister, she has 8/9 * 3/5 = 8/15 after books, she has 1/4 * 8/15 = 2/15
*June 8, 2016*

**Math**

so, which ones did you get? Show some work, eh? This is a massive homework dump. And sorry, you will have to describe the graphics is possible.
*June 8, 2016*

**Trigonometry**

as usual, with fractions, cross-multiply, and you will see one of your standard identities.
*June 8, 2016*

**Trigonometry**

I assume you mean 4sin^2(3θ) - 3 = 0 sin^2(3θ) = 3/4 sin(3θ) = √3/2 3θ = 60° or 120° θ = 20° or 40° Since sin(3θ) has period 360/3 = 120°, add multiples of 120° to those values. You will find six solutions in the ...
*June 7, 2016*

**Trigonometry**

you have tanx = -2 so, the solutions will be in QII and QIV.
*June 7, 2016*

**Trigonometry**

since sin^2x = 1 - cos^2x, make that substitution, and then solve the quadratic for cos(x)
*June 7, 2016*

**Trigonometry**

No idea what a double identity is, but draw the triangle. The missing side is √24. In QII, that means x is negative, so cosθ = x/r = -√24/5 cotθ = x/y = -√24/1
*June 7, 2016*

**math**

clearly the major axis has length 13, and the minor axis has length 12. So a = 13/2 b = 6 c^2 = a^2-b^2 = 25/4, so c = 5/2 The center is at (13/2,0), so h = 13/2 and k=0 The foci are at (h±c,k) Since the major axis is horizontal, that means the equation is (x-h)^2/a^2...
*June 7, 2016*

**Math**

so, no, A is not correct.
*June 7, 2016*

**math**

a = 0.9 o so, o = a/0.9 = 1.111a so, the orange is 11% heavier than the apple. The grapefruit is just noise.
*June 7, 2016*

**math**

do you even look at what you type? That's "sides" not "size"! and "menion" just escapes me totally! If the two diagonals are x and y, then using the law of cosines, and the fact that consecutive angles are supplementary, x^2 = 6^2+9^2 - 2*6*9 ...
*June 7, 2016*

**Algebra**

It sounds like you want a recursive sequence, where T(n+1) = 4Tn - Tn^2 If so, then if Tn = 4*2.4 - 2.4^2 = 3.84 T(n+1) = 4*3.84 - 3.84^2 = 0.61
*June 7, 2016*

**math**

5000*1.01^8 + 10000*1.01^16
*June 7, 2016*

**Math Help Please**

Note which side is opposite to A and adjacent to A. Now review your basic trig functions. You can see that sinA = BC/AB cosA = AC/AB so you are correct. Extra credit: How did I figure it out, with no triangle to refer to?
*June 7, 2016*

**value**

you don't move on a bearing. You move on a heading. You observe on a bearing. Starting from R at (0,0), P moves to (-20/√2,-20/√2) Q moves to (15√3/2,15/2) now just find the distance between those points. Or, use the law of cosines. The angle between ...
*June 7, 2016*

**algebra**

12 mg / kg * 1kg/2.2lb * 120lb = 655mg
*June 7, 2016*

**maths**

What do you mean when you say when it is perpendicular to it ?? when what is perpendicular to what?
*June 7, 2016*

**Maths**

1+1 = 2 You have some other questions?
*June 7, 2016*

**Maths**

DBA = BDC = 28° DBC = 90° - DBA = 62°
*June 7, 2016*

**Math (money)**

(x - 200)(.80) - 131 = .20x x = 485 check: start: 485 dress: 200 leaves 285 bag: 57 leaves 228 shoes: 131 leaves 97 = 20% of 485
*June 7, 2016*

**MATH - ??**

Ouch! I got confused on the period/comma usage disparity. Or, I could just say I was figuring in km instead of meters!
*June 7, 2016*

**MATH**

x(1 - 1/3 - 2/5) = 6/5 4/15 x = 6/5 x = 9/2 check: 1/3 of 9/2 = 3/2 = 1.5 2/5 of 9/2 = 9/5 = 1.8 1.5 + 1.8 + 1.2 = 4.5 = 9/2
*June 7, 2016*

**maths**

if cos3Q = √3/2, sin3Q = ±1/2 so, tan3Q = ±1/√3 so, using the +1/2 value, tan6Q = 2tan3Q/(1-tan^2 3Q) = 2(1/√3)/(1 - 1/3) = (2/√3)/(2/3) = 3/√3 = √3 so, -√3 if sin3Q = -1/2
*June 7, 2016*

**Math (Trig)**

right on
*June 7, 2016*

**Maths**

200 trees means 199 intervals between. So, each interval is 396/199 meters. That means that the nth tree is at a distance from the first of Tn = (n-1)(396/199) Now just plug in 15 for n and you have your distance.
*June 6, 2016*

**math**

well, you could start off by writing that with math symbols: 2x/5 = 12 Now just solve for x.
*June 6, 2016*

**Math**

that equation describes a line. Was there something in particular you wanted to know about it? Getting rid of the fractions, it could also be written as 7x + 90y + 72 = 0
*June 6, 2016*

**Math**

not quite. It was simple interest, not compound. 4300(1+.018*15) = 5461.00
*June 6, 2016*

**Math**

well 40° is 1/9 of 360°, or the whole pizza.
*June 6, 2016*

**Math**

Draw a N-S line through A. From that line, the angle back to the start is 55°, and the angle to B is 64° 55+64 = 119 That means the interior angle must be 61°, since the entire N-S line is a straight angle.
*June 6, 2016*

**Math (*-*)**

a larger denominator means a smaller value. Think about it. If you cut a pizza into 4 pieces, 7 pieces, 10 pieces, which pieces are smaller?
*June 6, 2016*

**Math**

see related questions below
*June 6, 2016*

**trade science**

do you know the thermal expansion coefficient?
*June 6, 2016*

**math @Reiny**

I think I like your thinking. Sorry, danbaba.
*June 6, 2016*

**math**

(ii) 2/3 * 1/3 (i) 1 minus that
*June 6, 2016*

**Maths**

well, that would be 9*20000 cm Just change that to km or, to make things easier, 20000cm = 200m = 0.2km so, 9cm = 9*0.2 km
*June 6, 2016*

**7th Grade Math**

sorry that should be (a+b)/2 * h
*June 6, 2016*

**7th Grade Math**

There is missing information. You cannot get the area of the triangle without knowing its height. Then its area is 1/2 * base * height The area of a rectangle is width * height A trapezoid has two bases and a width (or height) -- you only name one. Then, a trapezoid with bases...
*June 6, 2016*

**7th Grade Math**

So, how did you get the area of 46 without finding the areas of the three shapes? You do know how to do those, I guess? Just show your work so the teacher knows you can do the areas. Why not show your work here, so we can see what's going on... Also, area is square inches...
*June 6, 2016*

**math**

draw the two triangles in standard position. A is a 5-12-13 triangle, and B is 8-15-17. sinA = 5/13 cosA = 12/13 tanA = 5/12 sinB = 8/17 cosB = -15/17 tanB = -8/15 Now just plug those values into your addition formulas for the trig functions.
*June 6, 2016*

**MATHS**

what, you can't plug in the digits and add? EGG+EGG = 899+899 and so on
*June 6, 2016*

**Maths**

well, what is 3/5 of 20? No idea? what is 1/5 of 20? ...
*June 6, 2016*

**abc**

since distance = speed * time, if Ashok's speed is a, and thw wind's is w, then we have (3/2)(a+w) = 60 (2)(a-w) = 60
*June 6, 2016*

**math**

Did you try actually plotting the line segment? Just plot (1,2) and then go 5 left and 5 right. A=(-4,2) B=(6,2)
*June 6, 2016*

**math**

hint: gradient 0 means y never changes.
*June 6, 2016*

**uncc**

x: 15 cos30 y: 15 sin30
*June 6, 2016*

**Math**

s = 1/2 at^2 = 1/2 t^(5/2)
*June 6, 2016*

**Pre-Calc**

sorry; the heading is 34.02° (It's 90-A), not 90+A
*June 5, 2016*

**Pre-Calc**

630 @ 𝑁 30° 𝐸 = <315.00,545.58> wind from the SE is in the direction NW, so 45 @ NW = <-31.82,31.82> Now, if the plane's ground speed is <x,y> we have <x,y> + <-31.82,31.82> = <315.00,545.58> x = 346.82 y = 513.76 ...
*June 5, 2016*

**Pre-Calc**

just convert the vectors to x-y components, add them, and you have the resultant. Can you get that far? If not, where do you get stuck?
*June 5, 2016*

**11**

Check your sum-to-product formulas. I expect it will fall right out.
*June 5, 2016*

**maths,science, physics**

just preserve momentum: (420)(-30) + (200)(30) = (420+200)(v) v = -10.65
*June 5, 2016*

**maths-calculus help me**

d/dx arccosh(u) = 1/√(u^2-1) so d/dx arccosh(x^2+1) = 1/√((x^2+1)^2-1) * 2x if y = arcsinh(x^2), dy/dx = 1/√(x^4+1) * 2x Not quite sure just what it is you are trying to do.
*June 5, 2016*

**Math2**

Draw a diagram. You have a pyramid with base ABC and summit at S. I am assuming that S is at the top of a vertical face containing BC. You have base angles A = 80 B = 61 so, C=39 side c = 400 So, now you can figure sides a and b using the law of sines. Hmm. Still can't pin...
*June 5, 2016*

**Math**

you have an isosceles triangle with both base angles 89.9° The base of the triangle is the diameter of earth's orbit. Knowing that, you can find the altitude of the triangle (the distance to the object).
*June 5, 2016*

**physics**

tanA = y/x = 29.0/-96.0 Note that it is in QII
*June 5, 2016*

**Vectors**

Your angle of 21.5° is right, but if you draw the diagram, you will see that the bird is roughly NW, so it must fly roughly SE. So, it is really -21.5° (relative to the x-axis), or a heading of 111.5° from due North.
*June 5, 2016*

**Vectors**

the distance d is found by d^2 = (4.8-1.5)^2 + 1.3^2
*June 5, 2016*

**physics**

acceleration = 20m/s ÷ 10s = 2 m/s^2 deceleration takes twice as long, so it is half as much.
*June 5, 2016*

**physics**

surely you have a formula which gives this value. If not, see the excellent article in wikipedia on "trajectory"
*June 5, 2016*

**Math**

MP*.84 = CP*1.05 MP/CP = 1.05/0.84 = 1.25 so, the original markup was 25%
*June 5, 2016*

**MATH**

Draw a diagram. It is clear that the distance of chord from center: d^2 + (6/2)^2 = 5^2 the chords are 2d apart.
*June 5, 2016*

**maths**

18*2 + x = 27*3
*June 5, 2016*

**itc c++**

what, you don't have a c++ system?
*June 5, 2016*

**math**

The beam length b is b^2 = 192^2 + (70+10)^2
*June 5, 2016*

**Math - surface area**

well, the area is πr^2 = 4π yd^2 a first approximation is thus 4π*13
*June 5, 2016*

**maths**

24^2 + 48^2 + h^2 = 56^2
*June 5, 2016*

**math**

3b+c = 14t b+6t = 1c so, find c in terms of t: c = 14t-3b = 14t-3(c-6t) c = 14t-3c+18t 4c = 32t c = 8t
*June 5, 2016*

**math**

If the 4 groups have w,x,y,z respectively, then w = x+4 x = y-1 z = 2x w+x+y+z = 20 Now just crank it out.
*June 5, 2016*

**Math(Please Help Assignment Due Today)**

man. of all the possible sets, AAA does not guarantee congruence -- only similarity.
*June 4, 2016*

**Math(Please Help Assignment Due Today)**

well, you've posted this a few times. Did you think of googling your question? Many hits. Maybe you could start here: http://www.regentsprep.org/regents/math/geometry/gp4/ltriangles.htm
*June 4, 2016*

**math**

not quite. You figured the straight-line distance, but that does not answer the question. Read it again carefully.
*June 4, 2016*

**physics**

The ratio of KE after the bounce to before the bounce is 1/2 m(.8v)^2 ----------------- 1/2 mv^2 = .8^2 = .64 so, the ball will bounce .64 as high as the original height
*June 4, 2016*

**geometry**

so, add 5 to x and subtract 4 from y. Not so hard, eh?
*June 4, 2016*

**maths**

f(x) = 2x Looks like y= ax, with a=2 or y = 2f(x) if f(x) = x not quite sure just what it is you're asking. enter your expression at wolframalpha.com and see the graph.
*June 4, 2016*

**science**

the difference is .11 hours .11/3.15 = 0.0349 ≈ 3.5%
*June 4, 2016*

**mathematics**

As always, draw a diagram. AB/18 = tan35° CD^2 = 18^2 + 24^2 from C, tanθ = AB/24 from D to C, tanθ=24/18, so the bearing is 360-θ
*June 4, 2016*

**mathematics**

well, the common denominator is 2(y-1), so that gives you 1(y-1)+3(2) ----------------- 2(y-1) ...
*June 4, 2016*

**Math**

but, you know how to count money, right? 7 quarters = $1.75 need 11 more coins, so start using smaller ones. 6 quarters and 5 nickels = $1.75 need 7 more coins 5 quarters and 10 nickels = $1.75 need 2 more coins. 4 quarters and 15 nickels - $1.75 need 2 coins less 4 quarters, ...
*June 4, 2016*

**Math-Calc**

so, the critical numbers are the values where f'(r) = 0, right? A fraction is zero if its numerator is zero. So, what is r if 7 - 10r^2 = 0?
*June 4, 2016*

**Statistics**

Never mind -- I'm bored already.
*June 4, 2016*

**maths again plz**

since 1 hour is 60 minutes, 35'25" = 0.59 of a complete circle. the circumference is 2pi*r, so the distance traveled is 2pi*8.8*0.59 = 32.64 cm
*June 3, 2016*

**maths-need help asap**

yes, with no radius, it's difficult to find the volume. The problem has been garbled, apparently. Both the cylinder and the cone are incompletely specified. And that doesn't even get into the bad grammar...
*June 3, 2016*

**maths-need help asap**

just find the volume in cm^3 Then recall that 1L = 1000cm^3
*June 3, 2016*

**math**

(5/3)^2 * (3/4)^2 = (5/4)^2 = 1.5625
*June 3, 2016*

**Math**

well, let's see... cos(x) starts at the top, and has amplitude 1. You want to start at the bottom, with amplitude 30. cos(kx) has period 2π/k. You want period 2. So, what is k?
*June 3, 2016*

**algebra**

xy=20
*June 3, 2016*