Can someone please help me solve this: solve for x, x/6 - x/10=1/6?
x/6 − x/10 = 1/6 Find LCM 5x/30 − 3x/30 = 1/6 2x/30 = 1/6 x/15 = 1/6 x = 1/6 × 15 x = 15/6 [ x = 5/2 = 2.5 ] Hope this helps Bye !!
Can someone please help me with my math homework. T_T?
Bucket problems! That's what my teacher referred to them as. So you need 50mL. Let's just say you need x mL of the 5%. That means for the 10% you have 50-x mL. Now make an equation x(.05)+(50-X)(.1)=50(.09) solve for x. You get 10. That's how many mL of the 5% you need so you need 40 for the 10% For the second. it works the same. Just plug in the numbers they give you where they belong and solve for whatever you have to solve for. You need x of the 60% but have 100 of the 30% so altogether you have 100+x. Here the equation is x(.6)+100(.3)=(100+x)(.36) and x is 25
Can you solve this? 0+50*1-60-60*0+10?
0 + (50 x 1) - 60 - (60 x 0) + 10 Do the multiplications first. 50 x 1 = 50. 60 x 0 = 0 0 + 50 - 60 - 0 + 10 Then do the adding & subtracting. The result is ZERO. Your friend is not using the correct process of doing the multiplication first before doing the adding and subtracting. This is what your friend is doing: He's working from left to right and doing the operations as they are encountered. 0 + 50 * 1 - 60 - 60 * 0 + 10. 0 + 50 = 50 50 * 1 = 50 50 - 60 = -10 - 10 - 60 = -70 - 70 * 0 = 0 0 + 10 = +10
How do I solve for x, cos2x = sin (x+30)?
How do I solve for x, [math]\cos2x=\sin\left(x+30\right)[/math]?Answer: [math]x=\frac{360^{\circ}n+60^{\circ}}{1\pm2}[/math].The question doesn't say, but from the 30 in the equation I'd guess that it's in degrees and not radians, so I’ll go with that. If it is in radians then the basic process would be the same but the answers would be different.It's well known that [math]\sin\theta\equiv\cos\left(90^{\circ}-\theta\right)[/math]. This gives us[math]\qquad\cos2x=\cos\left(90^{\circ}-x-30^{\circ}\right)=\cos\left(60^{\circ}-x\right)[/math].Now, we also know that [math]\cos\left(\theta\right)\equiv\cos\left(-\theta\right)[/math] and that the cosine function repeats every [math]2\pi[/math] radians, which is equivalent to [math]360^{\circ}[/math]. Both of these must be accommodated but they need to be accommodated only on one side. If we did so on both sides, we would get the same answer.This now gives us[math]\qquad\cos\left(\pm2x\right)=\cos\left(360^{\circ}n+60^{\circ}-x\right)[/math], where [math]n\in\mathbb{I}[/math].Taking the arc cosine of both sides gives us[math]\qquad\pm2x=360^{\circ}n+60^{\circ}-x[/math].Now we simply solve for x.[math]\qquad x=\frac{360^{\circ}n+60^{\circ}}{1\pm2}[/math]Finally, we split the formula as per the [math]\pm[/math].[math]\qquad\boxed{\begin{array}{rl} x_{+}= & 120^{\circ}n+20\\ x_{-}= & 360^{\circ}n+60 \end{array}}[/math]We know that the answer has a period of [math]360^{\circ}[/math], so to test we need only find all of the x-values within any period. We'll choose from [math]0^{\circ}[/math] to [math]360^{\circ}[/math]. As long as we have all of the x-values in this period then the answer must be correct.We can see that[math]\qquad\left\{ 20^{\circ},140^{\circ},260^{\circ}\right\} \subset x_{+}[/math] and[math]\qquad\left\{ 300^{\circ}\right\} \subset x_{-}[/math] are the only x-values between [math]0^{\circ}[/math] and [math]360^{\circ}[/math] and the plot below shows that they are the only ones.
Please help me solve this equation. I am getting confused about the order of operations.?
Well the Order of Operations, the order in which we do arithmetic, is called PEDMAS, which stands for Parenthisis, Exponents, Division/Multiplication, Addition/Subtraction. And to help you remember, the usual phrases that replace it is Please Excuse My Dear Aunt Sally. (But you can make up your own, anything that is helpful and preferable)... So for this equation, you might want to keep "FOIL" in mind. It helps for equations like these. It stands for First, Outer(Outside), Inner(Inside), Last. It's pretty much telling you the basic way of doing it. (and it does end up with the right answer if you do the equation correctly) So for this equation, you want to get the variable alone. Before you can do anything, you should simplify it. You can really simplify anything on the right side, but you can on the left side. By using distributive property, -6(6x-5) equals to -36x+30 as well as +6(6x+5) being +36x+30...It's pretty much the same but you're just making the equation easier by simplfying it. So now the equation is -36x+30+36x+30=x-10...Now, you have to put the "like terms" together(again for simplifying)...So -36x+36X cancels each other out. Then 30+30 equals 60. So the equation is now 60=x-10...In the beginning, I said, you want to get the variable alone. To get rid of then on the right side, you would have to do the opposite of it. Since it is negative 10, you want to put positve 10, other words add 10 to the right side. Though, when you do that, you add 10 to the left side, too. So now you should have your answer- 70=x Here's the work (My explanation might be confusing^^;;) -6(6x-5)+6(6x+5)=x-10 -36x+30+36x+30-=x-10 (-36x+36+30+30=x-10) 30+30=x-10 60=x-10 60+10=x-10+10 70=x And yea
How can I solve this math problem? (2403/38) 9
Are you allowed to use a calculator?If not, then simply start by trying to figure out in the simplest way how many “38’s” are in 2403.Start with the simple, 38 x 10. It equals 380. Not nearly enough. 38 x 100. That equals 3800. Too much, although just from that we can infer that it's closer to 38 x 100 than 38 x 10.Next let's try to break it down a bit more, let's try 38 x 80. We know it's closer to 38 x 100 than 38 x 10. So instead of doing the slightly harder math for 38 x 80, let's just do 38 x 100 - 38 x 20. To do 38 x 20, all you do us find 38 x 10 (which we got earlier) and multiply it by 2. 38 x 10=380. 380 x 2= 300 x 2 + 80 x 2= 600 + 160 = 760. So now let's come back to 38 x 100 - 38 x 20. We now know that it's 3800 - 760. Do a basic subtraction and fund that it's 3040. Still way too big.Next we can try 38 x 100 - 38 x 40 (to see if the answer is 38 x 60). Again, solve for 38 x 60. 38 x 10 x 6= 380 x 6= 300 x 6 + 80 x 6= 3 x 6 x 100 + 8 x 6 x 10= 18 x 100 + 58 x 10= 1800+ 580= 2380. Wow that's really close. Now we can try 38 x 61. We know 38 x 60 is 2380, so we just add 38. That's 2418. Ouch too much. So let's divide 38 by 2. 38/2= 19. Now let's try 38 x 60.5. 2380 + 19= 2399. Still missing 4.. 19 + 4 = 23. We want to divide 38 to find 23. So we are doing 38/x = 23 (the x in this case being a variable and not a multiple). Multiply both sides by x and have 38= 23x. Divide both sides by 23 and find that x = 38/23. Do the division.This is a number with an infinite amount of decimals, so I'm not sure how the teacher expects you to find this without a calculator.. You can do the long division, but it will be hard, just an FYI.After a quick use of the calculator on my phone, I found that 38/23=1.65217391. For simplicity’s sake let's round that to 1.65. Now add that to 38 x (60 + 1.65)= 38 x 61.65= 2403(The number might be off because of rounding), which means 2403 / 38 = 61.65. For 61.65 x 9 you simply do 60 x 9 + 1 x 9 + 0. 6 x 9 + 0.05 x 9=540 + 9 + 5.4 + 0.45== 554. 95!Sorry for the long post, just wanted to explain some simple reasoning for a seemingly complicated problem! Good luck!
Differential Equations: Can someone help me with this (Easy) Cat and Mouse Pursuit Problem?
Let northwest corner be origin. Let the cat's distance from the north wall be y and its distance from the west wall be x at time t. Let Θ be the angle cat makes with the west wall at time t. Then, dy/dt=-10cosΘ dx/dt=10sinΘ tanΘ=(5t-x)/y Let dx/dy be x'(y). Then, (5t-x)/y=-x' So, x'y-x+5t=0. Substitute 5t from the path traveled by the cat, since 10t= - ∫ (from b to y) √(1+x'(z)^2)dz. x'y - x - 0.5* ∫ (from b to y) √(1+x'(z)^2)dz = 0 Take derivative wrt y. x''y - 0.5*√(1+x'^2)=0 You can solve for x(y) analytically. I leave this to you. The boundary conditions would be x(60)=0 since x=0 at the beginning and y=60 x'(60)=0 since the initial angle is zero when cat is facing the mouse. This turns out to be a well-known "Pursuit curve" problem. Note that the differential equation (14) below is different than I derived here (see the extra 0.5), so their solution (15) does not work here... Edit: Corrected some typos...
A simple math problem that I can’t seem to solve…?
Write down these quantities with units, and then multiply / divide so that the units cancel. 45 miles / hour We want to cancel the miles, and have the hours end up on top, so we need to divide by this (since that flips the fraction). (8 miles) / (45 miles / hour) = 8 miles * (1/45) hours / mile = (8)*(1/45) hours = 8/45 hours That's a fraction of an hour, so it is probably better to write it as minutes. We know 60 minutes = 1 hour, so: 60 minutes / 1 hour = 1 We can multiply anything by one, so we'll do that now. The hours cancel, leaving minutes. 8/45 hours * (60 minutes / 1 hour) = 60*8/45 minutes = 12*8/9 minutes = 4*8/3 minutes = 32/3 minutes 32/3 = 10 with a remainder of 2, so that's 10 and 2/3 minutes. A minute is 60 seconds, so 1/3 of a minute is 20 seconds, and 2/3 of a minute is 40 seconds. The answer is 10 minutes and 40 seconds. Hope this helps.
Geometry Help?
Can someone help me please with these questions? I posted a bulletin before this one on the same problems but they did lack information. So I have scanned the page they assigned us for homework today. I really want to learn this but unfortunately I have a teacher who just rushes through things, and this is not my forte. (math that is) THANKS IN ADVANCE!!! We have to find the value of X. I uploaded the scanned picture of the book to tinypic.com. Can someone please help me? THANK YOU SO MUCH! http://i16.tinypic.com/48mi8pi.jpg