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solve for x. measure of angleAOB=28 measure of angle BOC=3x-2 measure of angle AOD=6x

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Solve for x. The measure of anle AOB=28. The meas... - OpenStudy


Solve for x. The measure of anle AOB=28. The measure of angle BOC= 3x-2. The measure of angle AOD=6x.
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Angle AOB=x+3, Angle AOC=2x+11, Angle BOC=4x-7. Solve for x.


If angle AOB = 28, Angle BOC = 3X-2, and Angle AOD = 6x ... Solve for x: measure of AOC= 7x-2 measure of AOB= 2x+8 measure of BOC= 3x+14 math
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Solve for x. Find the angle measures to check your work. m&lt ...


If angle AOB = 28, Angle BOC = 3X-2, and Angle AOD = 6x ... Solve for x: measure of AOC= 7x-2 measure of AOB= 2x+8 measure of BOC= 3x+14 geometry line BD bisects ...
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If the measure of angle AOC = 7x - 2, the measure of angle ...


If the measure of angle AOC = 7x - 2, the measure of angle AOB = 2x + 8, and the measure of angle BOC = 3x + 14, solve for x. 10 Points? Find answers now! No. 1 ...
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Solve for X. The measure of angle AOB = 4x-2. The... - OpenStudy


Solve for X. The measure of angle AOB = 4x-2. The measure of angle BOC = 5x+10. The measure of angle COD = 2x+14.
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If the measure of angle one is 2 minus nine and the measure of angle 2 is 27 minus 4x. solve for x

If the measure of angle one is 2 minus nine and the measure of angle 2 is 27 minus 4x. solve for x If the measure of angle one is 2x minus 9 and the measure of angle 2 is 27 minus 4x. solve for x You haven't told us the relationship between the two angles. Do they add up to 90 degrees? Do they add up to 180 degrees? Is there some other relationship? Possibility one: (2x - 9) + (27 - 4x) = 90 2x - 4x - 9 + 27 = 90 -2x + 18 = 90 -2x = 72 x = -36 A negative angle is measured clockwise from the x axis, that is, sweeping downward around the origin. Check: (2x - 9) + (27 - 4x) = 90 (2(-36) - 9) + (27 - 4(-36)) = 90 (-72 - 9) + (27 + 144) = 90 -81 + 171 = 90 90 = 90 Possibility two: (2x - 9) + (27 - 4x) = 180 2x - 4x - 9 + 27 = 180 -2x + 18 = 180 -2x = 162 x = -81 Check: (2x - 9) + (27 - 4x) = 180 (2(-81) - 9) + (27 - 4(-81)) = 180 (-162 - 9) + (27 + 324) = 180 -171 + 351 = 180 180 = 180 We need more information.
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If the measure of angle one is "2x minus 9" and the measure of angle 2 is "27 minus 4x", solve for x

Are these angles on a straight line? If so, 2x-9+27-4x=180, so -2x-18=180 and -2x=198, so x=-99. If they're angles in a right-angled triangle, then 2x-9+27-4x=90 and -2x=108, so x=-54.
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In a triangle, the measure of the 2nd angle is twice the 1st angle. The 3rd angle is 15 degrees less than the 2nd angle. If there are 180 degrees in a triangle, find the measure of each angle.

X = first angle, 2x= second angle, 2x-15 . Their sum is 180. Solve for x. I got x= 39.  Plug 39 to x to get 2 (39) to get second angle =78. To get third angle subtract 15 to get 63. First angle=39. Second angle =78. Third angle = 63.
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Flying bearing 325 degrees 800km,turns course 235 degrees flies 950km - find bearing from A.

1. Wendy leaves airport A, flying on a bearing of 325 degrees for 800km. She then turns on a course of 235 degrees and flies for 950km. Find Wendy's bearing from A. frown 2. Wendy decides to turn and fly straight back to A at a speed of 450 km/h. How long will it take her? There are two coordinate systems used to solve this problem. The first system, used in aeronautics, is a system in which angles are measured clockwise from a line that runs from south to north. The second, used to plot graphs, measures angles counter-clockwise from the +X axis, which runs from left to right. It is necessary to convert the angles stated in the problem to equivalent angles in the rectangular coordinate system. Bearing 325 degrees is 35 degrees to the left of north. North converts to the positive Y axis, so the angle we want is the complementary angle measured up from the negative X axis. 90 - 35 = 55 If we could show a graph, we would see a line extending up to the left, from the origin at (0,0) to a point 800Km (scaled for the graph) from the origin. What we want to know is the X and Y coordinates of that point. By dropping a perpendicular line down to the X axis, we form a right-triangle, with the flight path forming the hypotenuse. We'll call the 55 degree angle at the origin angle A, and the angle at the far end of the flight path angle B. Keep in mind that angle B is the complement of angle A, so it is 35 degrees. Because we will be working with more than one triangle, let's make sure we can differentiate between the x and y sides of the triangles by including numbers with the tags. Side y1 is opposite the 55 degree angle, so... y1 / 800 = sin 55 y1 = (sin 55) * 800 y1 = 0.8191 * 800 = 655.32    We'll round that down, to 655 Km Side x1 is opposite the 35 degree angle at the top, so... x1 / 800 = sin 35 x1 = (sin 35) * 800 x1 = 0.5736 * 800 = 458.86     We'll round that up, to 459 Km At this point, the aircraft turns to a heading of 235 degrees. Due west is 270 degrees, so the new heading is 35 degrees south of a line running right to left, which is the negative X axis. Temporarily, we move the origin of the rectangular coordinate system to the point where the turn was made, and proceed as before. We draw a line 950Km down to the left, at a 35 degree angle. From the far endpoint of that line, we drop a perpendicular line to the -X axis ("drop" is the term even though we can see that the axis is above the flight path). Side y2 is opposite this triangle's 35 degree angle at the adjusted origin, so... y2 / 950 = sin 35 y2 = (sin 35) * 950 y2 = 0.5736 * 950 = 544.89    We'll round that up, to 545 Km Side x2 is opposite this triangle's 55 degree angle, so... x2 / 950 = sin 55 x2 = (sin 55) * 950 x2 = 0.8191 * 950 = 778.19    We'll round that down, to 778 Km The problem asks for the bearing to that second endpoint from the beginning point, which is where we set the first origin. We now draw our third triangle with a hypotenuse from the origin to the second endpoint and its own x and y legs. Because the second flight continued going further out on the negative X axis, we can add the two x values we calculated above. x3 = x1 + x2 = 459 Km + 778 Km = 1237 Km The first leg of the flight was in a northerly direction, but the second leg was in a southerly direction, meaning that the final endpoint was closer to our -X axis. For that reason, it is necessary to subtract the second y value from the first y value to obtain the y coordinate for the triangle we are constructing. y3 = y1 - y2 = 655 Km - 545 Km = 110 Km Using x3 and y3, we can determine the angle (let's call it angle D) at the origin by finding the tangent. tan D = y3 / x3 = 110 / 1237 = 0.0889 Feeding that value into the inverse tangent function, we find the angle that it defines. tan^-1 0.0889 = 5.08 degrees The angle we found is based on the rectangular coordinate system. We need to convert that to the corresponding bearing that was asked for in the problem. We know that the second endpoint is still to the left of the origin. The -X axis represents due west, or 270 degrees. We know that the endpoint is above the -X axis, so we must increment the bearing by the size of the angle we calculated. Bearing = 270 + 5.08    approximately 275 degrees The second part of the problem asks how long it will take Wendy (the pilot) to fly straight back to the origin. Distance = sqrt (x3^2 + y3^2) = sqrt (778^2 + 1237^2) = sqrt (605284 + 1530169)              = sqrt (2135453) = 1461.319    We'll round that one, too   1461 Km Wendy will fly 1461 Km at 450 Km/hr. How long will that take? t = d / s = 1461Km / (450Km/hr) = 3.25 hours   << 3hrs 15 mins
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What is the diameter of a spiral coil of .65265 inch diameter pipe 100 feet long?

The equation of a spiral in polar coordinates has the general form r=A+Bø, where A is the starting radius of the spiral and B is a factor governing the growth of the spiral outwards. For example, if B=0, there is no outward growth and we just have a circle of radius A. A horizontal line length A represents the initial r, and the angle ø is the angle between r and this horizontal line. So r increases in length as ø increases (this angle is measured in radians where 2(pi) radians = 360 degrees, so 1 radian is 180/(pi)=57.3 degrees approximately.) If B=1/2 and A=5", for example, the minimum radius would be 5" when ø=0. When ø=2(pi) (360 degrees), r=5+(pi), or about 8.14". This angle would bring r back to the horizontal position, but it would be 8.14" instead of the initial 5". At ø=720 degrees, the horizontal line would increase by a further 3.14". Everywhere on the spiral the spiral arms would be 3.14" apart. What would B be if the spiral arms were 0.65625" apart? 2(pi)B=0.65625, so B=0.65625/(2(pi))=0.10445". The equation of the spiral is r=5+0.10445ø. To calculate the length of the spiral we have two possible ways: an approximate value based on the similarity between concentric circles and a spiral; or an accurate value obtainable through calculus. The approximate way is to add together the circumferences of the concentric circles: L=2(pi)(5+(5+0.65625)+...+(5+0.65625N)) where L=spiral length and N is the number of turns. L=2(pi)(5N+0.65625S) where S=0+1+2+3+...+(N-1)=N(N-1)/2. This formula arises from the fact that the first and last terms (0, N-1) the second and penultimate terms (1, N-2) and so on add up to N-1. So, for example, if N were 10 we would have (0+9)+(1+8)+(2+7)+(3+6)+(4+5)=5*9=45=10*9/2. If N were 5 we would have 0+1+2+3+4=10=(0+4)+(1+3)+2=5*4/2. L=12*100 inches. L=1200=2(pi)(5N+0.65625N(N-1)/2)=(pi)N(10+0.65625(N-1))=(pi)N(9.34375+0.65625N). If the external radius is r1 and the internal radius is r then the thickness of the spiral is r1-r and since 0.65625 is the gap between the spiral arms N=(r1-r)/0.65625. N is an integer, but, since it is unlikely that this equation would actually produce an integer we would settle for the nearest integer. If we solve this equation for N, we can deduce the external radius and diameter of the spiral: N(9.34375+0.65625N)=1200/(pi)=381.97; 0.65625N^2+9.34375N-381.97=0 and N=(-9.34375+sqrt(1089.98))/1.3125=18 (nearest integer). This means that there are 18 turns of the spiral to make the total length about 100 feet. If X is the final external diameter of the coiled pipe and the internal radius is 5" (the minimum allowable) then X/2 is the external radius, so N=((X/2)-5)/0.65625. We found N=18 so we can find X: X=2*(0.65625*18+5)=33.625in. Solution using calculus Using calculus, we can work out the relationship between the length of the spiral and other parameters. We start with any polar equation r(ø) and a picture: draw a line representing a general value of r. At a small angle dø to this line we draw another line a little bit longer, length r+dr. Now we join the ends together to make a narrow-angled triangle AOB where angle AOB=dø and AB=ds, the small section of the curve. In the triangle AO is length r and BO is length r+dr. If we mark the point C along BO so that CO is length r, the same as AO, we have an isosceles triangle COA. Because the apex angle is small, CA=rdø, the length of the arc of the sector. In triangle ABC, CB=dr, AB=ds and CA=rdø. By Pythagoras, AB^2=CB^2+CA^2, that is, ds^2=dr^2+r^2dø^2, because angle BCA is a right angle as dø tends to zero. The length of the curve is the result of adding the tiny ds values together between limits of r or ø. We can write ds=sqrt(dr^2+r^2dø^2). If we divide both sides by dr, we get ds/dr=sqrt(1+(rdø/dr)^2) so s=integral(sqrt(1+(rdø/dr)^2)dr, where s is the length of the curve. The integral is definite if we define the limits of r. For our spiral we have r=A+Bø, making ø=(r-A)/B and B=p/(2(pi)), where p is the diameter of the pipe=0.65625", so we can substitute for ø in the integral and the limits for r are A to X/2, where A is the inner radius (A=5") and X/2 is the outer radius. dø/dr=2(pi)/p, a constant=9.57 approx. s=integral(sqrt(1+(2(pi)r/p)^2)dr) between limits r=A to X/2. After the integral is calculated, we solve for X putting s=1200". The expression (2(pi)r/p)^2 is large compared to 1, so s=integral((2(pi)r/p)dr) approximately and s=[(pi)r^2/p] (r=A to X/2); therefore, since we know s=1200, we can write ((pi)/p)(X^2/4-A^2)=1200. Therefore X=2sqrt(1200p/(pi))+A^2)=33.21". Compare this answer with the one we got before and we can see they are close. [We could get a formal solution to the integral, using hyperbolic trigonometric or other logarithmic functions, but such a solution would make it very difficult or tedious to solve for X, since X would appear in logarithmic expressions and in other expressions making it difficult or impossible to isolate X. For example, the next term in the expansion of the integral would be (p/(4(pi))ln(X/2A), having a value of about 0.06. It is anticipated, therefore, that an approximation would be sufficient in this problem with the given figures.] We can feel justified in using the formula for finding the length of pipe, L, when X=6'=72": L=((pi)/p)(1296-25)=6084.52"=507' approximately. This length of pipe would hold 507/100*0.96 gallons=4.87 gallons.      
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solve for x. measure of angleAOB=28 measure of angle BOC=3x-2 measure of angle AOD=6x


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how to solve triangle algerbra problems

Problem: how to solve triangle algerbra problems Show me how to find the missing number in a triangle. Like one point has 45 degrees another has 14 degrees the last has a b. How do you get the measure ments for the b. The sum of the angles in any triangle is 180 degrees. Since you know two of the angles, add them and subtract that from 180. b = 180 - (45 + 14) b = 180 - 59 b = 121 Answer: b = 121 degrees
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In a triangle, the sum of the interior angles is 180o.

x + y + z = 180 x + 43 + 67 = 180 x + 110 = 180 x = 180 - 110 x = 70 degree
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find the measure of an angle whose compliment is nine times its measure

find the measure of an angle whose compliment is nine times its measure x + 9x = 90 10x = 90 x = 9 9x = 81 The angle is 9 degrees, its complement is 81 degrees
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What is the value of x?

3x + ( 2x + 20 ) +(4x - 20) = 180 angle 1 + angle 2 + angle 3 = 180 9x = 180 x = 180/9 x = 20 degree
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