Posts by STEVE
Total # Posts: 51,217
it would be e^(sin x^2)*2x - e^(cosx) * (-sinx) This is just the chain rule via the back door. If F(t) = ?f(t) dt then ?[u,v] f(t) dt = F(v) - F(u) Now take the derivative, remembering the chain rule.
#1. No consider f(x) = e^(-x^2) #2. No consider f(x) = -x^2 See what you can do with #3.
Algebra 2 check my last answer?!
where did you learn to write fractions? 2?/3 is the way to write it. However, your choice is not quite correct. You forgot the reflection part. So, the correct answer is f(x) = ?10cos(2?/3 x)+10 -10cos(2?/3 x) has max/min at 10,-10 and is a reflection of cos(2?/3 x) Now add 10...
S1: 2 | 1^2 + 7*1 2 | 1+7 2 | 8 true do likewise for n=2,3
I shall use x instead of whatever character you chose to copy/paste. The graph of y = a sin bx has amplitude = |a| period = 2?/b so, what do you get?
quite a few maybe the easiest to use is wolframalpha.com desmos.com fooplot.com https://rechneronline.de/function-graphs/ are all good sites with lots of versatility
for any general question, google is always a good place to start. https://www.google.com/search?q=estimated+probability&ie=utf-8&oe=utf-8
15/x = 3/4 If the areas are in the ratio 9:16 then the sides (which are the square root of the areas) are in the ratio ?9:?16 3:4 = 15:20
the sides are in the ratio 3:4, so ...
wolframalpha.com is very easy to use. Just type in your function, and it will show the graph. For example, http://www.wolframalpha.com/input/?i=4sin(3t)%2B2
Algebra 2 check
almost. If the max is 12 and the min is -6 then the midpoint is at (12-6)/2 = 3 So, f(x)=9sin(6?x)+3 your choice has max = -6+9 = 3 min = -6-9 = -15
you really expect someone to do this whole thing for you? How about you pick some piece of it that causes trouble, and show some ideas you have so far. I'd bet that there are code samples online for just about any part of it that you can break out.
kind of tricky, eh? Let x = e^u. Then u = lnx du = 1/x dx, so dx = x du = e^u du Now the integral is ? sin(5lnx) dx = ? sin(5u) e^u du = ? e^u sin(5u) du and you can apply your formula.
oops. My answer was the amount earned in the 39th week. To find how much he has earned during the whole 39 weeks, use the sum formula: S39 = 39/2 (2*135 + 38*5) = 8970
135 + 38*5
the slope is dy/dx = (dy/dt)/(dx/dt) = (sint + tcost)/(cost - t sint) y'(?) = (0-?)/(-1-0) = ? at t=?, x=-? and y=0 so, using the point slope form of the line, y = ?(x+?)
clearly, x^2+y^2 = 1 The question is, how much of the circle is graphed? ? x y -? -1 0 0 0 1 ? 1 0 we only get the top half of the circle.
5/3 as many days, so 5/3 as much pay.
well, you know the slope on one side is 1.4 and the slope on the other side is -1.4 You have the point (12,1.2) so the region has two lines for its boundary: y = 1.4(x-12)+1.2 = -15.6+1.4x y = -1.4(x-12)+1.2 = 18-1.4x Type in the functions at wolframalpha.com and you can see ...
Well, clearly you did not bother to evaluate the integral yourself. I made a typo, and the answer is 864?/35 instead. Just to check, we can apply the theorem of Pappus. The area of the triangular region is (1/2)(6/5)(12/7) = 36/35 The centroid is at x=12, so the distance ...
v = ?2?rh dx where r=x and h=y v = ?[78/7,90/7] 2?x(6/5 - 7/5 |x-12|) dx = 8634?/35
too bad you don't show your work. Just doing the decimal values, I get 8.97436 I agree with your evaluation of Reiny's final line, so let's see what went wrong. Did you bother to check his math? His only mistake was to omit the leading minus sign. Had you corrected...
Since LM and KN are parallel, alternate interior angles ?LKM and ?LMK are congruent. So, ?LKM is isosceles. That means that KL=LM. So, since LM:KN=8:9, we can say that LM=8x and KN=9x. That means that KL=MN=8x, and thus 8x+9x+8x+8x = 132 33x = 132 x = 4 The legs KL and MN are ...
(4+5)/100 = 9/100
small oops: 20/3 = 6.6
just divide, as with any fraction. 18/5 = 3.6 20/3 = 6.3 It just means that the value left of the decimal point will be greater than 0.
You have not made it clear what ?1 and ?2 are. Try using letters, such as ?1 = ?KNM or something Also, which sides are the parallel bases? If they are KL and MN, then the figure is a parallelogram. Is PKLMN the perimeter? Or is P some other point outside the trapezoid. You ...
Where does the Chlorine come in? CH3OH = 12 + 3*1 + 16 + 1 = 32
Math help? Someone help me?
Still no. try (A) If there is a min or max, the curve folds back on itself, so it is no longer invertible. It fails the horizontal line test.
no, y is on a bearing of 60° from x. The man travels on a heading of 60° Draw a diagram. In ?XYZ, you have angles X = 30° Y = 45° Z = 105° so, using the law of sines, y/sin30° = 200/sin105° = z/sin105°
#1: y = 3-2x so, 5x-(3-2x) = 11 7x = 14 x = 2 now you can find y using either equation.
see your earlier post
If you draw the diagram, since 25 sin35º = 14, then the angle ? south of W is tan? = (18-14)/20 = 4/20 ? = 11.3º
lowest height = 4 - 3 = 1 greatest height = 4 - 3/2 = 5/2 amplitude = (5/2 - 1)/2 = 3/4
Algebra 2 check my last answer?!
and we have a winner! B it is
looks good to me
algebra 2 help asap
just replace x with 3. Same as you have done for the last year or so in algebra... f(3) = 81 + 3e^(-0.7*3) = 81 + 3e^-.21 = 81 + .81058 = 81.81058 Better get out your calculator (physical or online) for these!
math please help ASAP
see your previous post. Why did you repost with the same error in typing? And are you sure this is algebra 2, and not calculus? Maxima and rates of change are usually not doable with just algebra.
algebra 2 please help :(
NO! x = 0.73
algebra 2 please help :(
Hmmm. f'(ln9/3) = 15 not 10 or 20
algebra 2 please help :(
some parens help make things clearer online: f(x)=20/(1+9e^(-3x)) f'(x) = 540e^(3x)/(9+e^(3x))^2 the greatest growth rate occurs when f' has a maximum. So, f"(x) = 1620e^3x * (9-e^3x)/(9+e^3x)^3 since the denominator is never zero, f" is zero when e^3x = 9 x...
x+y+z = 26 y = 3x+1 z = 5x-2 hat do you get?
If there were originally b boys and g girls, .60b + .90g = .75(b+g) .9g = .6b+12 solve for b It's a bit less cumbersome if you get rid of those pesky decimals 60b+90g = 75b+75g 3g = 2b+40 60b + 30(2b+40) = 75b+25(2b+40) b = 40
math help urgent please
I disagree. If f(x) is the cost for x tickets, then each ticket costs (.50x+10)/x = .50 + 10/x and ... ?
if it goes on without repeating, it is irrational. If it reaches a digit of group of digits that repeat forever, it is rational.
and is 16.5 closer to 15.8 than 16?
#1 ok #2 Any decimal which repeats is rational #3 nope. ?150 = 25?2 -- irrational #6 no diagram #7 How can you possibly say that the pole is 16 times as high as the length of the cable? You are clearly just guessing here. When you get an answer, take a step back and see ...
160*0.9/6 = 24
Algebra help?? Stuck!
I get 6220 That 6200 is unlikely, since the first two terms don't ad to zero...
I don't get this math!
The sequences are 1,2,4,8,16,32,64,128,256,512,1024 1,2,4,9,16,25,36,49,64,81,100 1024/100 = 256/25 = (16/5)^2
a) note that you just keep subtracting 2, so an = 1+2(n-1) = 2n-1 b) Hmmm. Differences are 8, 2, 3, 9, ... .-6, 1, 6, ... ....7, 5 ... Not a quadratic. The terms are 8, 8(1/4), 2(6/4), 3(12/4), ... The numerators are 1,6,12. Again, not much good there. The sequence is very ...
The AP numbers are: a, a+d, a+2d The GP numbers are: b, br, br^2 Add the terms, and you get a+b = 32 a+d+br = 26 a+2d+br^2 = 32 a+a+d+a+2d = 51 Now you have 4 equations for 4 variables. Use your favorite method to solve them (I suggest substitution), and you end up with AP: 5...
Looks like I had mismatched subscript tags. I meant #1. quadratic sequences have a constant second difference. In this case, 4. The sequence and differences are -1, 2, 9, 20, 35, ... ...3, 7, 11, 15, ... .....4, 4, 4, ... So, the 1st differences form an arithmetic progression...
math- Please help
#1. quadratic sequences have a constant second difference. In this case, 4. The sequence and differences are -1, 2, 9, 20, 35, ... ...3, 7, 11, 15, ... .....4, 4, 4, ... So, the 1st differences form an arithmetic progression: 3 + 4k So, a1 = -1 an+1 = an + 3+4(n-2) or, an+1 = ...
Draw a diagram. If the length is x, then 142.9/x = cos 32.1°
x = Rohit's age now y = Roshni's age now x = 2y x-5 = 3(y-5) Now just solve for x and y
10*1/13 + 2*1/13 + 1*1/13 - 2 = -1 better go back to class and stop gambling
x(2x+2) = 1200
You have $13.67 dollar balance. You would divide that total by the .25 per minute. $13.67/.25 = 54.58 minutes or 54 minutes.
solve the equation cos theta - tan theta * cos theta = 0 for 0< theta < 2 pi My answer is pi/4,5pi/4
tan .09 = .09 so, the solutions are .09 + n? I suspect a typo. Did you mean tan^-1(-.09) ? and also on the last post?
Find the values of the inverse function in radians. tan^-1(0.09) answer choices: a.-0.09+2pi n b. no such angle exists c.-1.48+ pi n d. -0.09+ pi n My answer is D
Nope. Those answers are in QI and QIV The solutions are in QI and QII: sin(.71) = .65 so the solutions are .71 + 2n? pi-.71 + 2n? = 2.4 + 2n?
Find the values of the inverse function in radians. 1. sin^-1(0.65)? answer choices: a. 0.71+2pi n and -0.71+2pi n b. 0.71+2pi n and -3.85+2pi n c. 0.86+2pi n and -0.86 +2pi n d. -0.61+2pi n and 2.54+2pi n My answer is B
P * 0.034/12 * 13 = 50 solve for P
that would just be ?[0,30] 12 e^(0.2t) dt = 24,146
I assume you mean x-intercepts. Since e^0 = 1, you just need to solve x^2-9x+20 = 0 Back to Algebra I, now ...
add 4y to both sides add 14 to both sides divide by 4
Algebra 2 help please.
you don't want any maxima or minima, because that means the graph bends back on itself, meaning it fails the horizontal-line test. (A) is what you want. the domain is [0,pi]
Let t=0 at the given instant. The slope of the line is tan? (with ? measured clockwise from the x-axis) tan? = (7-3t)/(5+2t) = 7/5 sec^2? d?/dt = -29/(5+2t)^2 74/25 d?/dt = -29/25 d?/dt = -29/74 ------------------------------ second method ------------------------------ The ...
Calculus Please Check my answer
looks good to me.
well, tan? = (24-1.5)/15 as seen from the man's eyes (assuming they are on the top of his head!)
Calculus Please Check my answer
looks good to me. Or, using shells, v = ?[0,1] 2?(y+1)(e^y-1) dy
These are all easy, since the line segments are all horizontal or vertical. Just figure the lengths and compare. For example, AB = 2-(-4) = 6 They will be very easy to do if you just plot the points.
Huh? There's nothing to do next. I gave you the first integral. The 2nd one was just for extra credit.
The curve is maybe a bit easier to visualize if you write it as x = 4sin(y). The area is ?[0,?/2] (4-x) dy = ?[0,?/2] (4-4sin(y)) dy = 2?-4 ?4-4sin(y) dy = 4y + 4cos(y) This is a bit easier than using vertical strips, where the area is ?[0,4] arcsin(x/4) dx since integrating ...
Sorry. There's still a typo. Now you have two x terms. Don't you read what you post? As typed, your two circles do intersect. http://www.wolframalpha.com/input/?i=plot+x%5E2+%2B+y%5E2+-+10x+-+8y+%2B+18+%3D+0+,+x%5E2+%2B+y%5E2+-+8x+-+4y+%2B+14x+%3D+0 Rather than just ...
well, f(x) is continuous, so the MVT applies. (f(1)-f(-1))/(1-(-1)) = (-8-8)/2 = -8 So, you want c where f'(c) = -8 3c^2-9 = -8 c = ±1/?3 Note that at f'(0) = -9, so it is too steep. You were just guessing, there, right? Just to show that our c values work, note...
I don't know, but I'm sure google does ...
I suspect still a typo, since there are two x terms. Sorry, friend.
5?700 = 5?(100*7) = 5(?100)(?7) = 5*10 ?7 = 50?7
are you ever going to fix the typo? Why keep wasting time with the same flawed posting? x^2 + y^2 - 8y - 4y + 14 = 0
Math (written answer)
T > -15 it doesn't get much simpler than this. Better review the section in your text... or google inequalities
?3/4 (x^5/5 - x^4/2 + x^3/3) [0,1] = ?3/4 (1/5 - 1/2 + 1/3) = ?3/4 * 1/30 = ?3/120
Not quite. You have figured the volume consisting of squares. For the triangle, the base is x-x^2, so the altitude is (x-x^2)?3/2. The volume is then ?[0,1] (1/2)(x-x^2)(x-x^2)?3/2 dx = ?[0,1] ?3/4 (x-x^2)^2 dx = ?3/120
You know the sides are in the ratio 1:?3:2 Just switch angles to figure sin60
MATH HELP PLZ
what question? In any case, I'm sure it will mainly involve counting how many data fit the criteria, and then dividing by the total.
2sinx + cosx = 0 2sinx = -cosx tanx = -1/2 Now you know there will be two solutions, in QII and QIV
Algebra check plz
Hmmm. What does they need to add a base shipping charge of $5 mean to you?
well, count up the ones with 4 C's and divide by the total number.
Or, try this way. The tangent line must be perpendicular to the radius which meets it. Also, the distance from the center to the line must be r. Since the center of the circle is at (a,b), if it touches the line at (h,k), we have m = (a-h)/(mh+c-b) That must have a single ...
If the line is tangent, the solution to the system of equations has a single solution. (x-a)^2 + (y-b)^2 = r^2 (x-a)^2 + (mx+c-b)^2 = r^2 x^2 - 2ax + a^2 + m^2x^2 + 2m(c-b)x + (c-b)^2 = r^2 (m^2+1)x^2 + (2mc-2mb-2a)x + a^2+(c-b)^2-r^2 = 0 To have a single solution, the ...
x^2 + y^2 - 10x - 8y + 18 = 0 x^2-10x + y^2-8y = -18 x^2-10x+25 + y^2-8y+16 = -18+25+16 (x-5)^2 + (y-4)^2 = 23 The second equation has a typo. Fix it and complete the squares to find its center. Then compare the distance between centers to the sum of the radii.
(a) 4/9 * 3/4 = 1/3 (b) 1 - 1/4 - 1/3 = 5/12
x = 2/10 x + 1/2 x + 12 6/20 x = 12 x = 40
since the scale ratio remains constant, x/135 = 1/20
(10/12 - 2/3) / (1/6) = 2 days
math series2(please helpassessment tomorrow)
If M = bL^3/12 replace b by 1.015b and L by 0.98L and you have (1.015b)(0.98L)^3/12 = (1.015*0.98^3)bL^3/12 = 0.955M so, M has decreased by 4.5% I see no reason to resort to binomial approximation. But, if you want to try it, and compare results, feel free.