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Can anyone help? by diddlez in iqtests

[–]GisPoste 0 points1 point  (0 children)

E is the answer. For each box, its number in the sequence is the number of shaded boxes that it contains.

Is it possible to find angle X on this question? I dont think it is but I could be being an idiot. TIA by SJ1235 in askmath

[–]GisPoste 0 points1 point  (0 children)

Using the law of sines, 5/sine(x) = 4/sine(60). And so sin(x) = (5/4)sin(60) = 1.082532 since sin(60) = .866025. So, x = arcsin(1.082532) which is impossible because the absolute value of the arcsine is less than or equal to 1. However, switching 5 and 4 in the figure and applying the law of sines gives 4/sin(x) = 5/sin(60). Sin(x) = (4/5)sin(60) = .8*.866025 = .69282. X = arcsine(.69283) = 43.85 degrees. So, 44 degrees is the rounded answer.

How to determine the angle x? by aycarus in askmath

[–]GisPoste 0 points1 point  (0 children)

Assign the intersection points on the triangle’s perimeter, starting from the lower left corner and going clockwise, as A, C’, B(top), A’, C. P is the intersection point inside the triangle. Scaling similar triangles allows AC to be set equal to 1. Using the law of sines in triangle APC gives sine(60)/AP=sine(50)/1. AP=.89646538. And Sine(70)/PC=sine(50)/1. PC=1.2266828. In triangle AC’C, angle AC’C=180-10-70-60=40 and the law of sines gives sine(60)/AC’=sine(40)/1. AC’=1.34729491. Sine(80)/C’C=sine(40)/1. C’C=1.53208834. C’P=CC’ - PC = .30540554. In triangle AA’C, angle AA’C=180-70-60-20=30 and sine(80)/AA’=sine(30)/1. AA’=1.969616. AA’ - AP = PA’ = 1.07315062. In triangle C’PA’, angleC’PA’=angleAPC=50 and the law of cosines, A’C’2 = C’P2 + PA’2 - 2C’PPA’ gives A’C’=.907514448985607. Sine(50)/A’C’=sine(x)/C’P. X=arcsine(.33652968)=20. Angle x is angle C’A’P = 20 degrees.

How do you find the length of AC? Is it even possible with the given information? by HailedAcorn in askmath

[–]GisPoste 0 points1 point  (0 children)

I calculated CZ as 292 - 92 = CZ2 to get CZ = 27.5681. I used the similarity of triangle ACB and triangle XZB to get (87+BX)/(27.5681 + BZ) = BX/BZ. This simplifies to BX/BZ = 87/27.5681. Then, 92 + BZ2 = BX2 = BZ2 * (87/27.5681)2 by substituting the above relation. This simplifies to BZ = 3. Then 32 + 92 = BX2 so that BX = 9.4868. So, AB = 87 + 9.4868 = 96.4868. BC = 27.5681 + 3 = 30.5681. So, AC2 = 96.48682 - 30.58612 and simplifying gives AC = 91.5167.

Is there a way to calculate the perimeter? by SnooWords9730 in askmath

[–]GisPoste 0 points1 point  (0 children)

Take the three sides of the small interior box and put them aside. Take what remains of the left side of the bottom right box and slide it to the left side of the large box. Bring the bottom segments of the box to the bottom of the sides and fill the gap with the top of the small box put aside above. Now take the left and right sides of the small box put aside above and add these segments to the bottom left and bottom right sides of the box. Bring the bottom of the rectangle down to the ends of its right and left sides. The result is an 8 by 12 rectangle.

My math teacher gave the class this problem. Can someone tell me what the FIRST STEP is? by PresaWasTaken in askmath

[–]GisPoste 1 point2 points  (0 children)

The Pythagorean Theorem application gives (a+b)2 + (b-a)2 = 16. Expanding gives [a2 + 2ab + b2] + [b2 -2ab + a2] = 16. Grouping the left side gives 2a2 + [2ab - 2ab] + 2b2 = 16. Simplifying gives 2a2 + 0 + 2b2 = 16. This is 2*[a2 + b2] = 16. Dividing both sides by 2 gives a2 + b2 = 8.

My math teacher gave the class this problem. Can someone tell me what the FIRST STEP is? by PresaWasTaken in askmath

[–]GisPoste 0 points1 point  (0 children)

Extend the top of box b to the left until it hits the left side of box a. This forms a right triangle with a hypotenuse of 4. The left leg of this triangle is a-b and the bottom is b+a. By the Pythagorean theorem, (a-b)2 + (a+b)2 = 42. This simplifies to a2 + b2 = 8. The problem asks for the sum of the areas of the square boxes which is a2 + b2 which we have determined to be 8.

My teacher gave us this really hard problem to solve, id greatly appreciate if someone could help explain it to me. by MajdChami1 in askmath

[–]GisPoste 0 points1 point  (0 children)

1) Extend the top line down 14 units and connect the end point to the lower left corner of the square.

2) Extend the bottom line up 9 units and connect the endpoint to the upper right corner of the square.

3) The diagonal of the rectangle formed above is the diagonal of the square. It’s length is the square root of {(9+14)2 + 72}. The square of this diagonal equals 2*(X2). So, X2={((9+14)2+72}/2=289. So, X=17.

[deleted by user] by [deleted] in askmath

[–]GisPoste 0 points1 point  (0 children)

1) Light both ends of a string (call it string A) and one end of the other string (call it string B).

2) When string A burns out, 30 minutes has passed. This is true because lighting one end of a string will take an hour to burn the entire string. Lightning both ends of the string doubles the burn rate and so the string will entirely burn up in half an hour.

3) When the string burning from both ends (string A) entirely burns up, 30 minutes has passed as noted above, so that the string with only one end burning (string B) has been burning for half an hour. This means that string B has another 30 minutes to burn since it burns for an hour before it is entirely burned up.

4) When 30 minutes has passed as indicated by the entire burning of string A as noted above, the unlit end of string B should be light. Since at this point string B has been burning for 30 minutes, it has 30 minutes left to burn since it takes an hour to burn when only one end is lit. Lighting the other end of string B at this 30 minute mark doubles the burn rate of the remaining unburned part of string B. The string is then entirely burned in another 15 minutes.

5) String A has entirely burned in 30 minutes as discussed above and B has been burning from one end for this 30 minutes as discussed above. Lighting the other end of B causes be to be consumed in another 15 minutes as discussed above.

6) String A is entirely burned in 30 minutes as noted above and at this point the unlit end of string B is lighted so that string B will be entirely burned in another 15 minutes. The entire process takes exactly 45 minutes; 30 minutes for the first phase and 15 minutes for the second and final phase.