Saturday

July 2, 2016
Total # Posts: 41,550

**Math**

if you want the height, drop the altitude to the base. It divides the base into two parts, x and 12-x. Now, using the Pythagorean Theorem, you can find the height h using x^2 + h^2 = 8^2 (12-x)^2 + h^2 = 7^2 subtract and you get rid of h, leaving x^2 - (12-x)^2 = 64-49 24x - ...
*June 9, 2016*

**Physics- Help!!**

120 * (1200/100) = ?
*June 9, 2016*

**math**

do you see what a useless response that is? Restrict how? Anyway, I gave you the information you need. How about showing some effort of your own here?
*June 9, 2016*

**math**

infinitely many. Or did you want to restrict the available area? In the first quadrant, (4,4) is on the curve. There will be none to the left of (1,16) or to the right of (16,1). And don't count the lattice points on the curve...
*June 9, 2016*

**Pre-Cal: Domain**

the domains include all real numbers except where the denominator is zero. So, figure the resultant function, and then exclude any places where the denominator is zero. AND, exclude 0 and -7 because at those values f or g is undefined, so even though some mixture of them might...
*June 9, 2016*

**Calculus**

you are correct. 1/(x-1) -> ∞ I tried to pound a round peg into a square hole.
*June 9, 2016*

**Calculus**

note that if u = 1/(x-1) then what you have is sin(u)/u You have probably seen that this limit is 1, so follow the same argument. Or, try google. A good discussion is at math.ucsb.edu/~jcs/SqueezeTheorem.pdf
*June 9, 2016*

**Math/Dosage Calculations**

just add up the amounts (in fl oz): (1/3)(8) + (1/2)(16) + 3 = 41/3 fl oz. Now just convert that to mL. 1 oz = 29.57 mL Hmmm. I get 404 mL
*June 9, 2016*

**Math**

Just take each sentence and place the required digit in for one of the x's. Start with xxxx My hundreds digit is the last odd number. x9xx and move right along.
*June 9, 2016*

**Math**

w*2w = 1250 Now get w, then the length is 2w
*June 9, 2016*

**Math**

that would be the area of the inner cylinder, plus the area of the outer cylinder, plus the area of the two rings at the end. So, if R = outer radius r = inner radius h = length of pipe the total area is 2πrh + 2πRh + π(R^2-r^2) = 2πh(r+R) + π(R+r)(R-r...
*June 9, 2016*

**Math - Year 10 - Simult. Equations**

just plug in y=3x and you get x^2 + (3x)^2 = 10 Now x is easy, and then y=3x.
*June 9, 2016*

**geography**

how many time zones in between? That's the number of hours. Or, if you know the longitudes, since there are 24 hours in a 360° circle, each hour covers about 15°
*June 9, 2016*

**geometry**

since j is in the middle, mn = mj+jn = 6 3/4 + 6 3/4 = ?
*June 9, 2016*

**trigonometry**

If that fails, try cot (x-y)/2 and use your half-angle formula, and more parentheses next time.
*June 8, 2016*

**math**

sure, if so designated
*June 8, 2016*

**Math**

even without calculus, you can do these. Just think of what you know about lines and parabolas. Then you can verify your Algebra I with calculus. What do you think?
*June 8, 2016*

**math(bearing)**

There seems to be a lot of noise here If the bearing of Y from Z is 280, then the bearing of Z from Y is 280-180 = 100
*June 8, 2016*

**Pre-Cal (Help Plz)**

come on. x^2-4 is negative on the interval (-2,2) So, since |x^2-4| is always positive, that little arc below the x-axis is flipped up above it. Everywhere else, the two functions are identical. g(x) = -f(x) if |x| < 2 g(x) = f(x) if |x| >= 2
*June 8, 2016*

**Pre-Cal (Help Plz)**

take a look at the graphs and see what you can say. And why not use || for absolute value? It's not like you don't have the characters... http://www.wolframalpha.com/input/?i=plot+y%3Dx^2-4,+y%3D|x^2-4|
*June 8, 2016*

**math**

ok
*June 8, 2016*

**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*