Can someone help me with this GRE math question?
In the GRE examination, you are not supposed to make assumptions based on graphical or pictorial illustrations/diagrams.In other words, even if triangle POF seems like 1/3rd the area of PST by visual observation, you cannot conclude that it indeed is.You have neither provided the image or the options given for the question. So, I will go with the option which closely aligns with the idea “Insufficient information/cannot be determined”Edit: Now that the picture is up, the correct answer is Option 4: “Need more info.”But, if we know that O is the mid-point of PS and OF is perpendicular to PT where PS=ST, then the area of POF is less than 1/3rd the area of PST. So in that case, the answer would be Option 2: “B is bigger”.
Math question. Can someone help?
65=13mutiplied by5 and similarly 91=13 multiplied by7. according to the question,13multiplied by5/13multiplied by7. both 13 and 13 will get cancelled and answer is5/7
Can someone help me with a math question?
2 equations: 1 for money: 12.50L + 9.00S = $415.00 1 for number of pizzas: L + S = 38 Two equations in two variables are solvable. Here is how: Multiply the second equation by 9: 9 L + 9 S = 342 Now subtract this equation from the first equation: 12.5 L + 9.00 S = $415.00 9 L + 9 S = 342 -------------------------------------- 3.5L = 73 L = 20.857142857142857142857142857143 Oops, that doesn't look right. The equations have the solutions L = 20.857142857142857142857142857143 S= 17.142857142857142857142857142857 The two numbers add up to 38, 17.142857142857142857142857142857 * 9.00 = $154.29 20.857142857142857142857142857143 * 12.50 = $260.71 $154.29 + $260.71 = $415.00 So these numbers are the solution to the problem as stated. Perhaps you have a typo. Grandpa
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.
Can someone help me with the following math question?
Let's take the speed of the turtoise, who is the slowest among all to be X kmph. Now we know that the Mongoose over takes the turtoise 4 times per round with the 4th overtake at the starting point. Hence we can conclude that the speed of the turtoise is 4 times that of the turtoise. So the speed of the Mongoose is 4X kmph.Now we know that the Rebirth overtakes the Mongoose two times per round with the second overtake taking place at the starting point. So we can conclude that the speed of the Rabbit is two times that of the Mongoose. Thus the speed of the Rabbit is 8X kmph.So the ratio of the speed of the rabbit to that of the tortoise is :-8X/X=8:1.Therefore the ratio of the speed of the Rabbit to that of the turtoise is 8:1.
Can someone help me with Math questions please?
1. the tangent line at x =3. the gradient of tangent line is the derivative of f(x) f'(x) = (-7/(x^2)) - 2: x = 3 f'(x) = -7/9 - 2 = -25/9 (gradient) straight line general formula. when x = 3 y =?? f(x) = 7/3 - 2 =1/3 y = mx + c y = (-25/9)x + c x =3 y =1/3 1/3 = (-25/9)*3 + c c = 26/3 y = -25/9x + 26/3 2. rate change f'(x) f'(x) = 6x^2 - 10^x; where x = 2 f'(x) = 6(2)^2 -10(2) = 24 - 20 = 4 3. f'(x) = -9x^2 + 4x - 4 when x = 0 f'(x) = -4 when x = 2 f'(x) = -32 average = (-4 + -32)/2 = -18 4. f(x) = 2x + 3 , divide the whole thing with x lim x -> 0, 2 + 3/x = 2 f'(x) = 2 5. product rule is given by d(uv) = udv + vdu u = 2x^2+4x-3 du = 4x + 4 v = 5x^3+2x+2 dv = 15x^2 + 2 g'(x)=(2x^2+4x-3)(15x^2+2)+(5x^3+2x+2)... just complete the multiplication and u got it. Have fun.
Math question, can someone help me plizzz?
Let the speed of boat in still water be x km/hr. To reach the campsite, he paddled the boat with the current. So the relative speed of boat is (x+6) km/hr While coming back, against the same current, the relative speed of the boat is (x-6) km/hr. The distance travelled in both directions is same. Distance travelled with the current is speed x time = (x+6)*2 Distance travelled against the current is speed x time = (x-6)*8 distance = same 2x+12 = 8x - 48 6x = 60 x = 10 The speed of the boat in still water is 10 km/hr. Enjoy!! the mathguru!!
Can I get some help on this math question?
All you have to do to work out c is ask yourself how many times the wave repeats over 360 degrees. The answer is 2. So c is 2.As for a and b...Imagine it as: -bcos2x + aThe reason the cos2x graph is flipped upside down is because of the minus sign.Now you just have to find numbers for a and b which slide the graph up and down the y axis so that it looks like the graph. If b is 4, then the peak and troughs are going to be at y = 4 and y = -4. If you then add a = 2, the peak and trough moves up to y = 6 and y = -2. Which fits your graph.So the answer is: y = 2 - 4cos2x
Can someone help me with a math question?! please?!?
We'll label the two bikers A and B. A is going at speed A B is going at speed A + 6km/hr Because they're traveling in opposite directions, the 42.5km distance is going to be the sum of each of their individual travels: distance = velocity * time The time is 1 hour and 15 minutes, which converted to hours is 1.25 hours. 42.5km = (A + B) km/hours * 1.25 hours 42.5km = (A + A + 6) * 1.25 34km/hr = 2A + 6 28km/hr = 2A A = 14km/hr B = A + 6km/hr B = 14 + 6 B = 20km/hr