Mathematics question! ?
First, subtract 32 from both sides: F-32=9/5C Then, divide both sides by 9/5 (i.e. multiply by 5/9): 5/9(F-32)=C
Please Help with a few mathematics questions?
Hi, I'm currently doing year 10 and have fallen a little behind in maths. I have a assignment which is due tomorrow and most of the question i know how to do apart from a couple The assignment is 6 pages long and i really need help with one of the pages... 1. A mobile company has only two plans for monthly payment. PLAN a-- a fixed charge of $30 and 50 cents a call after 150 calls, there is no extra charge for the first 150 calls PLAN b -- No fixed charge and 30 cents a call (A) Write down a rule for the charge, $Cb, the monthly cost of n calls using plan b (B) The rule for the cost ,$Ca, of n calls using plan a is Ca= (c If n is less or equal to 150 (an+b If n>150 Find the value of c,a and b (C) For how many calls is it better to use PLAN A? and PLAN B? 2 Find the equation of the straight line taht passes through the point (-3,6) and is perpendicular to the line with eqaution y = 2x + 6 3 Solve the following for x 5 - 2x = 6x +3 5x over 4 -8 =12
Mathematics Question...HELP?
fourfold
Mathematics question help?
Given: s^3 = 500 Take the cube root of both sides: s = cube root(500) Using a calculator (or Google): s ≈ 7.93700526 (Round as appropriate)
Who can help me with these maths questions?
Hey buddy, I can help you in Question 41, but Sorry not in 42. I haven't studied 3D yet, but will soon finish it. :-)Now in Question 41, First of all, You must know what is an odd function, its a function which is symmetric about the origin. For example, y = sinx. If you find the area of sinx from -90° to +90°, it would be zero. "This is the main property which is we have to apply in this question."Now in the limits, if you put x = 1 in F(x), you will get F(1)=0. Because F(x) is actually defined as the area of f(x), which is an odd function, from -1 to any arbitrary value of x. Now, since you are getting F(1) =0, apply the L'Hospital Rule, Since you are getting a finite limit, Numerator must not reduce to zero. Therefore, Equation upgrades to:Lt(x->1) F'(x)/G'(x). Now, you can apply the Newton-Leibnitz Formula to easily obtain F'(x) and G'(x) and then put x=1. After all substitutions, you will get f(1/2) = 7PS;Please Read about L'Hospital Rule and the Newton-Leibnitz Theorem immediately if you haven't heard any thing about them till now, They are the master keys to any Problems in Functions, Limits, Continuity, Differentiability and much more......Hope this helped,Have Fun :-)
Quick Mathematics Question! Please Help! Thanks!?
Let C(n,r) stand for the combination of n objects taken r at a time. We know that C(n,r) = n!/[r!(n-r)!] 1. (2a-3b)^5 = C(5,0)(2a)^5 (-3b)^0 + C(5,1)(2a)^4 (-3b)^1 + C(5,2)(2a)^3 (-3b)^2 + C(5,3)(2a)^2 (-3b)^3 + C(5,4)(2a)^1 (-3b)^4 + C(5,5)(2a)^0 (-3b)^5 = 1(32a^5)(1) + 5(16a^4)(-3b) + 10(8a^3)(9b^2) + 10(4a^2)(-27b^3) + 5(2a)(81b^4) + 1(1)(-243b^5) = 32a^5 - 240a^4b + 720a^3b^2 - 1080a^2b^3 + 8100ab^4 -243b^5 Notice that the sum of the variables on each term is 5. 2. The (k+1)th term of (x+y)^n is C(n,k)x^k y^(n-k). For the 6th term of (a+2)^10, we would put k = 5, n = 10, x = a, and y = 2 C(10,5)(a)^5 (2)^(10-5) You can simplify it. 3. For the fifth term of (3x-y)^8, replace n with 8, k with 4, x with 3x and y with -y.
Please...help me with this additional mathematics question??
first set: 32 = p*2^n second: 160 = p*4^n divide eqn 2 by eqn1 , p gets cancelled >>5 = (4/2)^n >>5 = 2^n take log: >>log5 = nlog2 >>n = log5 / log2 = 2.321928 now substitute n in eqn 1 32 = p * 2^(2.321928) >>p = 32 / (2^(2.321928)) >>p = 6.4
Can someone help me solve this maths question?
Let’s visualize the problem and clarify definitions.The first drawing is the container, with its six sides. Only three sides or “faces” are visible. The container is closed.The second picture removes the faces that were in the way and allows us to imagine the inside of the container.Read your word problem and label the dimensions, the lengths of the parts of the container, with their specified values.Now, imagine placing a box in the container, shoved snug in the corner marked by the arrow. If we place another box just to the right, touching and pressed against the back wall, what is the combined width? Keep doing this. How many boxes can you line up against the back wall before you reach the right side?Now, imagine adding boxes against the left wall, on the floor of the container, toward the front. How many boxes fit, counting the one in the corner?With those two answers, we can figure out how many boxes are needed to form a single layer on the base of the container. Repeated addition: the number of boxes across the back, repeated for each row in front until the front edge of the container.Go back and think of the empty container. Stack boxes on top of each other. How many boxes tall is the stack that reaches the top of the container. Now we know how many layers or tiers of boxes - how many rectangles of boxes laid on top of each other can fill the container. Repeated addition of the boxes in one tier, times the boxes high the tiers can go.That’s the first answer.The second answer is calculating the area of the outside of the container. That’s the area of the six faces. I gave them all names. List them. Note the lengths of the two adjacent sides of each of those rectangles. The area of a face is the product of those sides. You have six faces. Add their separate areas together. The faces come in three pairs of matching dimensions.[Edit]The other answers so far do not show a crucial piece of work. In the real world, boxes cannot be broken to fit in spaces that are too narrow for the box dimensions. In this case, the numbers worked out evenly. But it might have been the case that space would go unused in the front, down the side or at the top of the container. To divide the volume of the container by the volume of a box only calculates the ideal capacity. If any dimension was not a multiple of five, the actual number of boxes that could be packed would be reduced by the wasted space.