How can I solve it? Please someone help?
When given such questions there is a thumb rule that you must follow. Count the number of variables that make up the equation --> 2 in this case X and Y.Count the minimum number of variables that you need to arrive at the final solution --> 2 in this case X and Y. Generally the number of equations you will be given will be the same as this number. Now that we know the above 2 points, the 3rd point is your freedom to do whatever you want to do within the rules of mathematics with the equation to arrive at a 3rd equation that will get you closer to your solution. So think how you can use basic operations like addition, subtraction, multiplication and division to reduce complex terms like xy (in this case). That way you'll leave yourself working with simpler terms like x and y. Once you get there you have 2 approaches. To find your way back from the final equation and try and get it to this 3rd equation OR do something to the 3rd equation to get to the solution equation and in the process get your result.I'm pretty sure this is exactly what your teacher taught you to do. But Mathematics is best understood and mastered only and only by practice! PS: It is pretty strange that so many people actually solved the problem and posted all the steps to get to the final answer instead of nudging the ask-er towards the solution or the approach to it! Especially coz this pic is clearly a homework question of an elementary school kid. (And I didn't know there were tags for homework questions) Damn!! If only I could learn to cycle by letting my dad ride it for me instead. Thanks Quorans for your effort to drum out another student in the assembly line of "copy-paste" workers! :-)
Can someone help me solve An=An-1+10, A1=29?
An is the nth term. An-1 is the (n-1)th term So An = An-1 + 10 means that each term is 10 more than it's predecessor. So the the first four terms are 29, 39, 49, 59.
Can someone help me solve (5/3x) + 1 = 7/6?
(5/3x)=7/6-1 5/3x=1/6 x=(1/6) / 5/3 x=1/6 x 3/5 x=3/30 x=1/10
Can someone help me solve this integral?
Note that [math]\int dx/(ax+b) = \log(ax+b)/a[/math] and is not rational, whereas for [math] n \geq 2, n\in \mathbb{N} [/math], [math] \int dx/(ax+b)^n [/math] is rational.When you try to split the integrand in your problem into partial fractions, you will have terms of the form [math] 1/x, 1/x^2, 1/(x+1),1/(x+1)^2,1/(x+1)^3 [/math] - out of these, [math]1/x, 1/(x+1) [/math] will give irrational integrands. Thus, these cannot appear in your partial fraction expansion. Therefore [math]\frac{f(x)}{x^2 (1+x)^3} = \frac{a}{x^2}+\frac{b}{(x+1)^2}+[/math] [math] \frac{c}{(x+1)^3} [/math][math] \implies \ f(x) = a(x+1)^3+b x^2(x+1)+c x^2 [/math]Imposing the condition that [math] f(x) [/math] is quadratic and its value at 0 is 1, we obtain[math] a+b = 0, a = 1, b =- 1 [/math]Now, you can easily evaluate [math]f'(x) = 3a(x+1)^2+b x^2+2b x(x+1)+2cx^2 [/math] and thus [math] f'(0) = 3a = 3 [/math].
Physics: Can someone help me solve this?
Thanks for A2A...Question:A convex lens of focal length 100 cms and a concave lens of focal length 10 cms are placed coaxially at a separation of 90 cms.If a parallel beam of light is incident on convex lens, then after passing through the two lenses, the beam will,(1) Converges(2) Diverges(3) remains parallel(4) DisappearsAnswer: (3) remains parallelExplanation:The parallel beam of light first hits the convex lens,Then, the light converges.The converging beam of light hits the concave lens,Then, the beam of light gets parallel.(Recommended: Draw a rough diagram in a paper to get a clear view)Additional tips i like to add here:-(a) Convex lens => converges(b) Concave lens => diverges(c) two convex lens => diverges(d) two concave lens => converges(e) bi-convex lens => converges(f) bi-concave lens => divergesSome help links:(i) The Anatomy of a Lens(ii) Molecular Expressions: Science, Optics, and You: Light and Color(iii) Lens Basics
Can someone help me Solve these questions?
The lesson here is order or operations Do opertions inside brackets first then: My Dear Aunt Sally= Multiple Add Subrtact then Divide 1+2[3+2(4)]= so, in [] there is a multiply so: 1+2[3+8]= now add in btrackets 1+2[11]= rest is all add and order is unimportant. ------------ [2+3(5-2)]= do the inner parens () [2+3(3)]= now My Dear Aunt Sally and multiply [2+9]= add to get eleven ----------------------------- [(5x2)-2)]= is this a typo ? [(10)-2)]= [8]= ========================== 24-2[5-(3x1)]= inner most parens ( ) 24-2[5-(3)] brackets next 24-2[2] My Dear Aunt Sally means Multiply next 24-4 = 20 final answer. 42/2[4+3(1)]= 42/2[4+3]= 42/2[7]= My Dear Aunt Sally means multiple before divide. 42/14=
Can someone help me solve this math question?
The question is poorly formulated. I think your teacher means that you only sum the elements of [math]A_n[/math] that are integers. There is indeed no nice closed form to express harmonic numbers, and there would be no reason for the sum of elements of [math]A_n[/math] to be integers. Don’t read further if you don’t want me to spoil the answer given this information.Using this we get [math]f(n) =\sum_{k|n+1} (-1+\frac{n+1}{k})[/math]. We have [math]5001=3\cdot 1667[/math], so these are the only divisors. Hence,[math]\begin{align} f(5000)= -1+3-1+1667 = 1668.\end{align}[/math]Edit:Of course we are interested in [math]f(500)[/math], not [math]f(5000)[/math], as pointed out in the comments. [math]501 = 3 \cdot 167[/math]. So the only divisors are [math]167, 3, 1[/math] (I forgot [math]1[/math] above). So we get[math]f(500) = -1+501-1+167-1+3= 668[/math]
Can someone help me solve this inequality? [math]3-5x\le-\frac45[/math]
It's been a while since I had to do this, but here's my attempt. I believe each step is self-explanatory.3 - 5x <= -4/53 <= 5x - 4/53 + 4/5 <= 5x3/5 + 4/25 <= x(15+4)/25 <= x19/25 <= xx >= 19/25Or in decimal:x >= 0.76
Can someone help me solve these binary to decimal problems?
Go to start>programs>accessories. Choose calculator. Then chose view and choose scientific. From there you can convert to hex, oct, decimal, and binary. If you need to do this without a calculator its not hard. Write out. 256 128 64 32 16 8 4 2 1 Each one you check is a number (add them) 256 128 64 32 16 8 4 2 1 0 0 1 0 0 0 0 0 1 =65 dec S0, 001000001 bin = 65 dec When dealing with IP address, each group is one group of these. So 192 would be 256 128 64 32 16 8 4 2 1 0 1 1 0 0 0 0 0 0 (128+64) Get it???
Can someone please please help me solve this? I am verifying trigonometric identities. cot^2y(sec^2y-1)=1?
cot^2y(sec^2y-1)=1 cot^2y(tan^2y) 1/(tan^2y)*(tan^2y) (tan^2y)/(tan^2y) 1