1. The line containing points(-1,3)&(3,8)has slope? a.5/2 b.5/4 c.2/5 d.4/5???????????\n2. The midpoint of the segment joining points (a,b)&(j,k)is? a.(j-a,k-b) b.(j-a/2, k-b/2) c.(j+a,k+b) d.j+a/2, k+b/2)\n3.The altitude of an equilateral triangle is 7squareroots of 3 units long. The length of one side of the triangle is? a.7 b.14 c.14 squareroots of 3 d.7 squareroots of 3/2\n4.The area of a sqaure is 36. The length of the diagonal of the square is? a.36squareroots of 2 b.6squareroots of2 c.3squareroots of 2 d.6\n5.The only defined term of those listed is? a.line b.angle c.plane d.point\n6.The intersection os two planes is a? a.line b.segment c.point d.ray\n7. Which of the following items can be measured? a.plane b.line c.ray d.segment\n8. Ray OX bisects AOC and m<AOX=42*(DEGREES). m,AOC=? a.42* b.84* c.21* d.68*\n9. In triangle ABC, m,A=47*, m,B=62*, m,c=? a.81* b.61* c.71* d.51*\n\n\nplease help me this is just about the last of my homeschool work...yes...i can see the light..almost
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1.B\n2.D\n3.A\n4.b\n5.D\n6.c\n7.d\n8.B\n9.C
31867
sigh...you shouldn't wait this late to do your homework...and you should study more, but i'll show you how to do this one - hopefully it'll help you with the other ones you posted lol.\n\nyou know the line is 2x-3y=18\n\nfind the slope of this line since you'll use it later:\n\n2x-3y=18\n-3y=18-2x\ny=-6+(2/3)x\n\nthus the slope is 2/3, since in that equation, y=b+mx, m=2/3.\n\nnow the line you want to find is perpendicular to this.  meaning the slope is the negative reciprocal.\n\nnew slope = -1/m = -1/(2/3) = -3/2\n\nnow you found the new m.  plug this into the given slope-interceot form equation and you get:\n\ny=mx+b\ny=-(3/2)x+b\n\nnow since you know that this line passes through point (-1,8), plug these y and x values into the equation and solve for b:\n\ny=-(3/2)x+b\n8=-(3/2)(-1)+b\n8=3/2+b\nb=8-3/2\nb=16/2-3/2\nb=13/2\n\nnow plug b back into your equation and you get:\n\ny=-(3/2)x+b\ny=-(3/2)x+13/2\n\nwhich is your answer :)  now go try and solve the rest of your problems by yourself.\n\n[[edit]]\n\nand don't listen to the guy above me - he's wrong and i don't like beer.
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>>>> what is the altitude of a right triangle?\n\nI have several here, and I am at 12 metres above sea level.\nSeriously, you haven’t provided enough information. You need at minimum the length of 2 sides or a side and an angle as well as the orientation to be able to calculate height (altitude).\n\n>>>>>>if the equation is d=2r what is the d and the r what exectly \n>>>>>>>am I finding with this equation \n\nd = diameter. r= radius. You are finding either how far it is from one side of a circle to another (d), or how far it is form the edge of a circle to the centre (r).
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1)One leg of a right triangle is 12 cm long. The diference between the other two sides is 2 cm.\nFind the sides of the triangle. (Pythagorean Theorem)\n\nWhat we can take from this problem is that one LEG is 12 cm long, while the other LEG is unknown (we will call it X) and the HYPOTENUSE -- the longest side of a right triangle -- is also unknown (we will designate it X+2, since it differs 2 cm from the unknown leg).\n\nSetting up Pythagorean Theorem as always:\n\n "Leg squared" + "Leg squared" = "Hypotenuse squared"\n\n  12^2 + x^2 = (x+2)^2\n\nExpanding:\n\n  144 + x^2 = x^2 + 4x + 4\n\nCollecting like terms:\n\n  4x = 140\n\nDivide by 4:\n\n   x = 35\n\nThe unknown leg is 35 cm, while the hypotenuse is 37.  Thus, we can also call the triplet (12,35,37) a "Pythagorean Triple" (of which there are infinitely many!).\n\n\n2) The hypotenuse of a right triangle is 58 cm long. The diference between the other two sides\nis 2 cm. Find the sides of the triangle. (Pythagorean Theorem\n\nIn this case, we know the hypotenuse, but do not know the lengths of the legs.  For convenience, we will again use the expressions X and (X+2) for each side's length.\n\nBy P.T.:\n\n x^2 + (x+2)^2 = 58^2\n\nExpanding:\n\n x^2 + x^2 + 4x + 4 = 3364\n\nCollect like terms:\n\n 2 x^2 + 4x - 3360 = 0\n\nDivide by the GCF (2):\n\n  x^2 + 2x - 1680 = 0\n\nCan we factor this trinomial easily?  We could, if we knew integer factors of -1680 (one positive, one negative) that combine to make 2.\n\nLuckily, -1680 can be factored into -2*2*2*2*3*5*7, and these can be grouped into two sets:\n      2*3*7 = 42\n -2*2*2*5 = -40\n\nFactor:\n\n (x+42) (x-40) = 0\n\nAlthough two solutions come from these factors (-42 and 40), we may only consider the positive answer, since lengths of triangle's legs are always positive!\n\n Therefore, x=40 (Leg 1), and Leg 2 is 42 cm.
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Your question is not well posed, but I'm going to restate it in the way that I'm guessing you're thinking.  \n\nSuppose you have a 20 inch diameter circle.  Take two points on the circle and draw a straight line segment between them that is precisely 8 inches in length (this is termed a "secant", or a "chord").  What is now the arc-length of the segment of circle between those two points?\n\nIf that is your question, then here is how I work the problem:\n\nYou have defined an equilateral triangle, two edges of which are the radii of the circle, 10 inches each, and one which is the chord of 8 inches.  We can now determine the ANGLE between the two 10 inch edges using basic trigonometry.  The easiest way for me to think about this, offhand is to divide the equilateral triangle in half, making two right triangles.  The hypotenuse is 10 inches, the opposite leg is 4 inches, and therefore the sin function of the angle is equal to 4/10.  the angle is therefore (approximately) 23.6 degrees.  Since we "trigged" a right triangle that was half of the original equilateral triangle, we need to double the result, yielding 47.2 degrees.\n\nThe angle between the two radii that subtend the arc who's chord is 8 inches (on a 20 inch circle) is 47.2 degrees.\n\n47.2 degrees represents 47.2/360 of the circle (or 0.13 of the circle).  Therefore the arc length of the segment in question is given by 0.13 x the circumfrence.\n\nThe circumfrence is pi times the diameter or 3.14 x 20 = 62.8 inches, therefore the arc length is given by 0.13 x 62.8 =\n8.2 inches\n\nThe answer here isn't terribly surprising.  The 20 inch diameter circle is quite large compared to the short segment of an 8 inch straight line.  The "gentle curvature" of the circle compared to the line means that the arc length of the circular segment isn't going to be much longer than the straight line.  If  you took it to the extreme and compared the circular segment from a circle who's DIAMETER was about 8 inches, then the arc length would become nearly half the circle, 3.14 x 8 / 2 =12.5 inches.  On the other hand, if the diameter of the circle got really really big (like the diameter of the earth for example) then you'd not be able to tell the difference between the straight 8 inch rod and the curve of the circle at all.  The answer would approach exactly 8 inches.\n\nI hope that answered your question!
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16.72 inches squared.\n\n\nDivide an equilateral triangle into two equal right triangles (meaning, draw the height).  The right triangles will have angles of 30, 60, 90, right (an equilateral triangle has 3 60 degree angles)?\n\nThe ratio of the sides in such a right triangle (the smallest leg being a) is a:a(sq(3)):2a, where 2a is the hypotenuse and a(sq(3)) is the other leg.\n\nThe "other leg" in this case is the height.  So a, half of the length of the side of the whole equilateral triangle, is 5.382/sq(3).\n\nMultiply this by 2 to get the whole length of the side.  We have our base and height.\n\n\nThe area of an equilateral triangle is 1/2bh.  b=2(5.382)/sq(3)\nh=5.382\n\nPlug those in and you get 16.72.
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Angle of elevation should be that raised from the ground until it is vertical (90 degrees), correct?  The wall which the ladder is placed against will have a vertical face (90 degrees).  The ladder, therefore, represents the hypotenuse of the triangle so formed.\nCosine is the length of the adjacent sided divided by the hypotenuse.  So cos 70 = X/40 or 40*(cos 70) = X.\nWith me so far?\nOnce you have X, you have the hypotenuse and 1 side of a right angle triangle and finally a use for Pythagoras's theorem the square on the hypotenuse equals the sum of the squares on the other 2 sides.\nCan you do the rest of the arithmetic?  I hope this problem was in Trigonometry, otherwise I have given you the wrong way to solve it. But it is the easier way.
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here you assume there is almost no point of intersection...or at least that it is exactly a point therefor no measureable space so you must determine the two circumfrences...use the equation\nc=pi * diameter\nc1 = pi *6\nc2 = pi * 18\n\ntotal distance is 24pi\n\nhope that helps!\n\nactually, that is a good point, when i first read it i assumed that we would want to put the rope around the circle...meaning that we would have the rope touch the circle at all times...the problem with stating that there is a line from one circle to the other will be slightly inaccurate, the line from the two circles will be different from the horizontal by several degrees, that would be the need for teh trig function.  I would say that for this point in the school year for a high school geometry class they probably havnt gotten to trig yet...\n\nhope this helps!!
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1.find the midpoint of BC usingthe midpoint formula\n(x1+x2/2,y1+y2/2) say (x,y)\n2.find the slope os BC using theformula \ny2/y1/x2-x1 say m\n3.the slope of the perpendicular bisector will be the negative inverse of this -1/m\n4. now use the point (x,y) and slope (-1/m) and find the equation of the perpendicular bisector
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Consider the triangle with one vertex at the center and two vertexes on the ends of the bent steel, one at each end. This is an isosceles triangle with an angle in radians equal to 3/19 (remember that the degree in radians is simply the length of the arc subtended by the angle divided by the length of the radius). Now consider the altitude from the center to the short side of the triangle. Clearly, this line will bisect the angle and divide the triangle into two equal right triangles, both having a hypotenuse equal in length to the radius and an angle of one half the small angle in the isosceles triangle (3/38). The altitude itself is the side adjacent to this angle and also the distance between the center and the midpoint of the straight line between the two ends of the bent steel. The distance between the midpoint of the bent steel, and the midpoint of this line, is equal to the radius minus the length of the altitude. Since the cosine of an angle is equal to the length of the adjacent side is equal to the proportion of the adjacent side to the hypotenuse (which in this case is the altitude and the radius, respectively), it follows that the length of the altitude is equal to the radius times the cosine of 3/38. Thus the distance you are looking for is:\n\n190 ft - 190 ft * cos(3/38)\n190(1-cos(3/38)) ft\napprox. 7.1 in
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1(A) 1 L = 1000 cm^3 therefore 2L = 2000 cm^3\n1(B) 1 L = 1 Kg therefore 2L = 2Kg\n\n*******************\n\n2)\nTwo thirds less than half of what number is five sixths\n(1/2)X - (2/3) = (5/6)\n\nmultiply every term by 6\n3x - 4 = 5\n\nadd 4 to both sides\n3x = 9\nx = 3\n\n*******************\n\n3)\nvolume of a cube = side^3 = 12^3 = 1728 cubic inches\nvolume of square pyramid = (1/3)(b^2)(h)\n where b = 12 and h = 8 (because the total height = 20)\n\n v = (1/3)(144)(8) = 384 cubic inches\n\n total volume = 2112 cubic inches\n\n*******************\n\n4)\nwe do not know the relation ship between the segments, if they form a line, if they form a right triangle, or if they lie on top of each other....\n\nif they form a line segment, where "b" is between "a" and d", then the total meximum length is 12 + c. if they lie on top of each, then they have a minimum distance of |12 - c| (absolute value of 12 - "c")\n\n*******************\n\n5)\nyou would need (1 x 10^3 or 1000) to form 1 meter multiplied by 1000 to form 1 KM\n1000 x 1000 = 1,000,000 or 1 x 10^ 6\n\n*****************\n6)\nx^2 + x  : GCF = x\nx(x + 1)\n\n\n12m^2n^3 + 18mn^2 - 24m^2n^3 \n12m^2n^3 and 24m^2n^3 are like terms\ntherefore\n\n= -12m^2n^3 + 18mn^2\n\nGCF = 6mn^2\n\n6mn^2(-2mn + 3) or 6mn^2(3 - 2mn)\n\n*****************\n\n7)\n-2 2/3w - 1 1/3 = 4 \n\nadd 1 1/3 to both sides\n\n-2 2/3w = 5 1/3\n\n= -(8/3)w = (16/3)\n\nmultiply both sides by -(3/8)\n\nw = -2
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Its been a long time since I took Geometry but maybe I can give you a small pointer.\n\nTry dividing the rectangle into 3 segments lengthwise and 3 segments widthwise\n\nA--B--C--D\nE--F--G--H\nJ---K--L--M\nN--O--P--Q\n\nDraw the diagonal through NKGD\n\nProve that the segments NK, KG, and DG are equal (prove that the triangles are the same)\n\nNow draw the segments from the vertex Q to the mid point of AD and AN.  Prove that the mid point of AD is the same as the mid point of BC and the same for AN and EJ.\n\nNow we have to prove that the line segment we drew in the last step instersect points K and G. You can probably do this with similar triangles (the triangle NQ and mid point of EJ is similar to JK and mid point of EJ therefore the point K is in the triangle)\n\nSo if you prove that the lines drawn from Q to the midpoints intersect K and G and we proved that NK KG and GD are congruent we have proved the original hypothesis.\n\nYou can turn this gabably gook into a real mathematical proof.\n\nGood Luck
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The midpoint of that segment is (3, -2) - all you do is take the average of the x coordinates and the average of the y coordinates.\n\nThe slope of that segment is found by\n\nm = (y2 - y1)/(x2 - x1)\n= (1 - (-5))/(4 - 2)\n= 6/2\n= 3\n\nSince you're looking for a line perpendicular to this one, take the slope, invert it, and change the sign.  So the slope of the new line will be -1/3.\n\nNow you have a point and a slope for the new point: m = -1/3, and point (3, -2).  Use the point-slope formula to find the equation of the line:\n\ny - y1 = m(x - x1)\ny - (-2) = (-1/3)(x - 3)\ny + 2 = (-1/3)x + 1\ny = (-1/3)x - 1
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15: First you cut the triangle in half so the hypotenuse is 17 and the bottom leg is 8 (half of 16=8) and then use Pythagorean theorem so it would be 17^2 - 8^2 (=225) and take the square root of that (=15).\nGood luck on your midterm!
235319
Look at the triangle formed by the line segment AB.  The diagonal side(I'm assuming it's AB) is the hypotenuse.  Label the side across from θ as D.  Label the straight side next to θ as E.  \ncscθ equals the hypotenuse divided by D\ncscθ=(AB)/D\nI don't know what m is supposed to be in the equation, so I can't help you there.\n\nEdit: To find m, take the inverse tangent of D/E.
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If the bases are in the ratio of k:1, then we can say they have lengths of kx and x (kx and x have ratio k:1). Let's say the trapezoid has height h.\n\nThe mid-segment of the trapezoid cuts the trapezoid into two similar trapezoids, and has length (1/2)(kx + x)\n\nThe area of a trapezoid is (1/2)(height)(b1 + b2).\n\nThe top trapezoid has height h/2, and bases kx and (1/2)(kx + x).  It has an area \nA1 = (1/2)(h/2)(kx + (1/2)(kx + x))\n     = (h/4)(kx + kx/2 + x/2)\n     = (h/4)(3kx/2 + x/2)\n     = (hx/8)(3k + 1)\n\nThe bottom trapezoid has height h/2 and bases (1/2)(kx + x) and x.  It has an area\nA2 = (1/2)(h/2)((1/2)(kx + x) + x)\n     = (h/4)(kx/2 + x/2 + x)\n     = (h/4)(kx/2 + 3x/2)\n     = (hx/8)(k + 3)\n\nIf A1:A2 = 5:2, then \n[(hx/8)(3k + 1)]/[(hx/8)(k + 3)] = 5/2\n(3k+1)/(k+3) = 5/2\n2(3k+1) = 5(k+3)\n6k + 2 = 5k + 15\nk = 13\n(C)
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If M and N are midpoints of RT and ST, then MN is parallel to RS.  There is a property of triangles that says that if you have a base and a segment inside the triangle whose endpoints are midpoints of the other two sides, the segment is half the length of the base.\n\nSo in your case, we need to find the length of the base RS.  Since the points are on the x-axis, the length is simply |2a-2b|.  Thus, the length of the segment MN is (|2a-2b|)/2, or |a-b|.