Maths Help - A computer screen is 18 cm wide and 12cm high. The x is the horizontal and y is the vertical.?
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1a)\n\n18 / 7.5 = 36 / 15 = 12 / 5\n12t - 5t^2 = t(12 - 5t) = 0 at 0, 12 / 5\n\nSo at time T = 0 the target is at the origin at the lower left hand corner, and at time T = 12 / 5 = 2.4 the target is at the lower right hand corner (18, 0).\n\n2b)\n\nAt t = 0, rM = (14, 12) is where the missile is fired from.\n\n2a)\n\nThe x coordinate of the missile is a constanr at 14, i.e., the missile is fired straight up.  So it must the target at (14, ?).\n\nThe target moves right according to 7.5t.\n\n14 / 7.5 = 28 / 15 is the time when the target would pass through x = 14.\n\ny = (12t - 5t^2) at time 28 / 15 is\n(12 * 28 / 15) - (5 * 28 * 28 / (15 * 15))\n = (1 / (15 * 15)) (12 * 15 * 28 - 5 * 28 * 28)\n = (5 * 28 / (15 * 15)) (12 * 3 - 28)\n = (5 * 8 * 28) / (15 * 15)\n = (8 * 28) / (3 * 15) = 224 / 45\n\nSo 12 - ct = 224 / 45 at time t = 28 / 15,\nc = (12 - (224 / 45)) / (28 / 15)\n = ((12 * 45) - 224) / (28 * 3)\n = (540 - 224) / (28 * 3)\n = 316 / (28 * 3) = 79 / 21\n\nwith c = 79 / 21\nat time 28 / 15\nthe target is hit at (14, 224 / 45)\n\n3)\n\nI'll let someone else graph something in proportional font ASCII where multiple whitespace is shown collapsed into a single space.
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There will be a vertical asymptote at x=0. When x values get increasingly closer to zero from the right, the graph approaches infinity. When the x values approach 0 from the left, the graph approaches negative infinity.\n\nSimilarly, there is also a horizontal asymptote at y=4. As x approaches infinity, the graph approaches 4 (1/infinity + 4 = 4). Also, as x approaches negative infinity, the graph also approaches 4 (1/negative infinitive + 4 = 4).
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It is similar to calculating the area. \n\nIt is a product of Length and Width (L x W).
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w1=width of 1st rect.\nL1=length of 1st rect\nA1=area of 1st rect\nw2=width of 2nd rect\nL2=length of 2nd rect\nA2=area of 2nd rect\nw1=x\nL1=4x\nA1=length*width\n=4*(square of x)\nw2=x+2\nL2=4x+5\nA2=(x+2)(4x+5)\n=x(4x+5)+2(4x+5)\n=4*(square of x)+5x+8x+10\n=4*(square of x)+13x+10\n\nAccording to the given condition\nA1+270=A2\n4*(square of x)+270=4*(square of x)+13x+10\n270=13x+10                //4*(square of x) gets cancelled\n13x=260\nx=20\ntherefore width of 1st rect=20cm\nlenght of 1st rect=20*4=80cm
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A = bh\nb = base and h = height
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There are 12 inches in a foot.\n\nSo divide by 12.
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Well,\n\nhere you can find comparison table:\n\nhttp://michaelbluejay.com/electricity/computers.html\n\nThe main factor is width of screen
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Why is this under Laptops & Notebooks? it should be under Help with Homework, if there is such a thing. Measure it if you're an adult, but if you are a kid (or just need help), you need to use algebra to find it. This might help you:\n\nArea = Length X Width\nLength = Width +11\nArea = 80\n\nTherefore\n\n80 = Width X Width + 11\n\nThat's called substitution
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9:3 simplifies to 3:1\n\ndivide both by 3 to get the answer
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let l be the length and w be the width\n lw=111\n\nbut it has a margin so Area= (l-11cm)(w-4)\ndifferntiate\nbut we have 3 unknowns so we use the first ea lw=111\nl=111/w\n\nA=(111/w-11)(w-4)\n\ndA/dw= (111/w-11)+(w-4)(-111/w^2)=0\n\n111/w-444/w^2= 111/w-11\ncancel 111/w\n-444/w^2=-11\ncrossmultiply\nw^2=-444/-11\nw=6.4 cm\nl=111/6.4=17.3
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this looks fairly accurate to me\n\n x = b*h/(b + h)\n\nx=side of the square\nwhere b is the base of the triangle\nh is the height of the triangle
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If the question is to simplify:\n\nSqrt(144(x^10)(y^12)(z^18)) then the first answer is correct:\n    \n12(x^5)(y^6)(z^9)
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Use "ray tracing" and ratio and proportion. First draw a horizontal line from the point source to the screen that is 2.20 + 1.40 m long. At a distance of 2.20 m from the point source let the horizontal line be a perpendicular bisector to a line that represents the aperture (unknown height as yet). At the screen let the horizontal line be a perpendicular bisector of a line that is 14.0 cm (0.14 m) high. Now draw two lines from the point source to the ends of the line representing the spot of light on the screen. Where these two lines cross the line of unknown length, 2.20 m from the point source, is the aperture.\n\nYou now have two isosceles triangles with a common apex at the point source, the base of the smaller one being the aperture and the base of the larger being the line representing the spot of light on the screen. Since these are similar triangles, you know the ratio of their heights is the same as the ratio of their bases. Plug in the numbers and solve for the aperture width.
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you need to identify that there're two parts of the force, one acting horizontally, the other acting vertically.\nSo:\n1) the vertical part is:\n70sin30\n2) tje horizontal part is:\n70cos30
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mass = (side) x (side) x (side) or (height) x (depth) x (width)
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H*W= area\nheight times width\n\none side has a height of 8 and its width is 9 thus the area would be 72
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I think  you mean ,you have:\nBase x height = 1x 1/2 \nthan by comparaison base =1\n                    height =1/2