Please help with these two fairly easy math problems?
1.) Since sqrt(12)/4 = sqrt(4*3)/4 = 2*sqrt(3)/4 = sqrt(3)/2, we have that sin^-1(sqrt(12)/4) = sin^-1(sqrt(3)/2). The range of sin^-1 is [-pi/2, pi/2], and the angle theta on this interval such that sin(theta) = sqrt(3)/2 is theta = pi/3. Thus sin^-1(sqrt(3)/2) = pi/3. 2.) Let x = log(base 36)(6^(1/3)). Then 36^x = 6^(1/3) -----> (6^2)^x = 6^(1/3) -----> 6^(2x) = 6^(1/3). Now equate exponents to find that 2x = 1/3 ------> x = 1/6 as expected.
WORD PROBLEM anyone? I need help solving a fairly easy one...?
If you were to take just one button out, it could be any color (for example, yellow). If you were to take out two buttons, there's still a chance you could get, say, one yellow and one blue. The same concept applies to taking out three buttons. You could still get one of each color. In other words, if you take out a number of buttons that is less than the number of button color types, you're always gonna run into the chance of picking out all different colors. For example, if you take out four buttons, you might get 1 yellow, 1 brown, 1 green, and 1 blue (one of each color). What if you took out five buttons? There are only four different colors. The fifth button will HAVE TO BE a repeat of one of the colors. So if you take out five buttons, you're bound to get two buttons that are the same color.
Please help solve this math word problem, with working? It's an algebraic question.?
Your two consecutive numbers are x, and x+1 We know that the sum of their squares is equal to 113 so we equate this as: x² + (x + 1)² = 113 x² + x² + 2x + 1 = 113 Combine like terms and bring all of the terms over to the left side to form a quadratic equation 2x² + 2x - 112 = 0 Divide both sides by 2 to get rid of the nasty non-monic quadratic x² + x - 56 = 0 Now to find two numbers such that their sum is +1 and their product is -56, since the product is negative, exactly one of them has to be negative, and if their sum is +1 and one of them is negative then it's fairly obvious the numbers are -7 and +8. (x-7)(x+8) = 0 the product of these two is 0 so: x = 7, 8 (it would normally be negative 8 but that isn't consecutive with 7, if you used the quadratic formula I'm fairly sure you would get + or - 8, but in this case it's 8 and we saved time too) TEST: 7² + 8² = 49 + 64 = 113 Hence, 7 and 8 are the correct values for the two consecutive numbers whose sum of squares are equal to 113.
If you could solve any one world problem, what would it be?
Global Warming.If we leave the global warming to continue on the path its currently moving there are going to be major crisis in about 30 - 40 years. Crisis of proportion we haven’t seen that will lead to wars between the nations over resources especially water.One exampleOver half billion people depend on river Ganga which is fed through glacier in Himalayas. This glacier is retrieving at an alarming rate due to warming temperatures. If this continuous the glacier can disappear in the next 30 years making river Ganga a seasonal river which means the river will dry up during summer and other times. Imagine half a billion people without water for crops, cattle, human consumption. That will lead to unrest that we haven’t seen, pushing people to migrate only to find such problems everywhere. Its an extremely scary situation!!
Can somebody help me in this math word problem?
Ah, well, you see, they're using sea-level as zero, or "0", which is quite common, for measuring altitudes and depths, rather customary, you see. And so, Sasha started off at a position already positive, +212 ft., then went to a new position, negative with respect to the sea-level, you see, and so that's -80. The change in position, would be some number (call it "n", for "number), that you can add to the first position, to change it into the second position, so that would come out as +212 + n = -80 which, rearranging it algebraicly, and solving for n, by subtracting 212 (or adding -212) to both sides (212, to get "n" alone by itself on one side; done to both sides, to keep the = equal, of course), would then be n = -212 + -80, or n = -212 - 80, which amounts to the same thing: a total of -292 feet. That's a change in position, not the altitude. As an altitude, it would be awfully deep, and Sasha didn't go that deep. But suppose it was as you supposed, 212 - 80, to descend 80 feet from the starting altitude of 212 feet: that would get Sasha 80 feet lower, ending up at altitude +132 feet, 132 feet above sea-level, if my arithmetic is any good (I do make mistakes, alas), and unfortunately, that way we'd have Sasha scuba diving up in the air, high above the water, which simply won't do. (Not enough bubbles!) I think the key to this, is that 80 feet is an altitude below sea level, a position, the second position, but they wanted the *change* in position: do you suppose that might be what it is? And thank you for writing the question so clearly, it really helps. :-)
Solve this word problem.?
One end of a wire that is 26 meters long is attached to the top of a tower. The other end of the wire is staked to the ground so that the wire is taut. The distance from the top of the tower to the bottom of the tower is 14 meters greater than the distance from the bottom of the tower to where the wire is staked down. How far is the stake from the base of the tower?
Can anyone solve the precalculus problem explained below?
Trick question. No speed limit is mentioned, so it's impossible to actually determine whether the Smiths were violating it (although it is evident that they were going fairly fast). But the question was whether they can fight the ticket.The answer is that they can and should fight the ticket. In many jurisdictions, speeding tickets require an actual observation of speeding by an officer; it's not legal for the officer to issue a ticket by deducing that a driver must have been speeding.
Am I the only one that finds math word problems very difficult?
No, everyone finds them challenging. The challenge isa.) to classify the problem so we know how to approach the problem with a method andb.) determine what you know or can derive from the given information by translation of english to mathematics, inference, and establishing relationships among variables.Once classified, you have an approach or “method” that you can follow to work through its solution.The algebra or calculus or whatever usually isn’t the challenge. The challenge is the approach, the set up, based on the problem classification and sussing out the information that’s contained in the problem definition.Here is a good source that provides a method and an organized tabular approach to determining what is known and unknown from the problem definition (Algebra problems).Translating Word Problems: ExamplesIn this approach, common problems presented in Algebra can often be classified as one of the following kinds of problems:"Age" problems, involving figuring out how old people are, were, or will be"Area/volume/perimeter" problems, involving very basic geometric formulas"Coin" problems, involving figuring out how many of each type of coin you have"Distance" problems, involving speeds ("uniform rates"), distance, time, and the formula "d = rt"."Investment" problems, involving investments, interest rates, and the formula "I = Prt"."Mixture" problems, involving combining elements and find prices (of the mixure) or percentages (of, say, acid or salt)."Number" problems, involving "Three more than two times the smaller number...""Percent of" problems, involving finding percents, increase/decrease, discounts, etc.Quadratic word problems, such as projectile motion and max/min questions."Work" problems, involving two or more people or things working together to complete a task, and finding how long they took.
Differential Equation help! Word Problem!?
So I've done 2 of the 3 parts to the problem; I need help on part C. Here is the question: As you know, when a course ends, students start to forget the material they have learned. One model (called the Ebbinghaus model) assumes that the rate at which a student forgets material is proportional to the difference between the material currently remembered and some positive constant, a. A) Let Y=F(T) be be the fraction of the original material remembered weeks after the course has ended. Set up a differential equation for Y, using K as any constant of proportionality you may need (let K>0). Your equation will contain two constants; the constant A (also positive) is less than Y for all T. For this part I got dy/dt= -k(y-a) (this is correct) What is the initial condition for your equation? For this part, I got y(0)=1 B) Solve the differential equation. For this part, I got: y= a+e^(-kt+lnabs(1-a)) (this is correct) C) What are the practical meaning (in terms of the amount remembered) of the constants in the solution Y=F(T)? If after one week the student remembers 75 percent of the material learned in the semester, and after two weeks remembers 66 percent, how much will she or he remember after summer vacation (about 14 weeks)? At this point, I don't know how to solve it... Help!