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cos (A) = 2/7, cos(B) = 3/4, cos(C) = 3/13

The answer is x = 118.

There are six non-real roots.

The answer is 32.

The answer is 12/5.

The answer is 9/16

Angle RAC = 68 degrees.

q(15) = -5

Points M, N, and O are the midpoints of sides KL, LJ, and JK, respectively, of triangle JKL. Points P, Q, and R are the midpoints of NO, OM, and MN, respectively. If the area of triangle PQR is 27, then what is the area of triangle LPQ?

Note:

The line joining the midpoints of a two sides of a triangle is parallel to the 3rd side. You can look up the proof if you wish.

So

KJ is parallel to MN is parallel to QP

ans

KL is parallel to ON is parallel to QR

So every triangle in this figure is similar to all the others.

LPQ=27u^2

So the area of MNO = 4*27 =  108u^2

And the area of LJK = 4*108 = 432u^2

Before I think on it any further can you confirm that it really is triangle LPQ that you want the area of ...

Here is the diagram.

Let DEF be an equilateral triangle with side length 3. At random, a point G is chosen inside the triangle. Compute the probability that the length DG  is less than or equal to 2  

Draw DEF, equilateral triangle with side length 3. 

Put the point of your compass on D and draw a circle with radius 2. 

If DG is within the segment of the circle that's inside the triangle, it will be 2 or less. 

If DG is outside the segment of the circle that's inside the triangle, it will be more than 2. 

Figure the area of the segment inside the triangle.  A_Segment     I got 2.09  (I called it 60/360 of the whole circle.)  

Figure the area of the entire triangle.                      A_Triangle       I got 3.90  (I used sin(60^o) to calculate the altitude.)  

Divide A_Segment by A_Triangle.  The quotient is the probability.  2.09 / 3.90  =  0.54  

y=(x+4)(x+1)(x-2)^2

If a normal distribution is graphed then the Highest column would be in the middle.

Then it would go down equally on both sides. It would look symmetrical about the middle.

If more of the scores are on the top side or on the bottom end then the graph/distribution is not normal,  it is skewed.

So is this distribution normal?

In how many seconds will the ball hit the ground?

Hello Guest!

\(h=v_0t-\dfrac{t^2}{2}\cdot g\\ -180ft=\frac{28ft}{s}t-\dfrac{t^2}{2}\cdot \dfrac{9.81\ m}{s^2}\cdot \dfrac{3.28\ ft}{m}\ |\ :\ ft\\ -180=\dfrac{28}{s}t-t^2\cdot \dfrac{9.81}{s^2}\cdot\ 1.64\\ t^2\cdot \dfrac{9.81}{s^2}\cdot\ 1.64-\dfrac{28}{s}t-180=0\)

\(t = {\dfrac{28}{s} \pm \sqrt{\dfrac{784}{s^2} +4\cdot \dfrac{9.81\cdot 1.64}{s^2}\cdot 180} \over 2\cdot \dfrac{9.81}{s^2}\cdot\ 1.64}\\ t= \dfrac{28s\pm \sqrt{784+4\cdot 9.81\cdot 1.64\cdot 180}\ s}{2\cdot9.81\cdot 1.64}\\ \color{blue}t=4.215\ s\)

In 4.215 seconds will the ball hit the ground.

  !

(4, 12, 16, 24, 32, 36, 64, 124, 132, 136, 164, 216, 236, 264, 312, 316, 324, 364, 412, 416, 432, 436, 612, 624, 632, 1236, 1264, 1324, 1364, 1432, 1436, 1624, 1632, 2136, 2164, 2316, 2364, 2416, 2436, 3124, 3164, 3216, 3264, 3412, 3416, 3612, 3624, 4132, 4136, 4216, 4236, 4312, 4316, 4612, 4632, 6124, 6132, 6312, 6324, 6412, 6432, 12364, 12436, 13264, 13624, 14236, 14632, 16324, 16432, 21364, 21436, 23164, 23416, 24136, 24316, 31264, 31624, 32164, 32416, 34216, 34612, 36124, 36412, 41236, 41632, 42136, 42316, 43216, 43612, 46132, 46312, 61324, 61432, 63124, 63412, 64132, 64312) >>Total = 97 integers divisible by 4

The number of solutions is 14.

The smallest possible value is 25/4.

The answer is 3/4.

YES, It is correct ! As follows:

1146 , 1155 , 1164 , 1236 , 1245 , 1254 , 1263 , 1326 , 1335 , 1344 , 1353 , 1362 , 1416 , 1425 , 1434 , 1443 , 1452 , 1461 , 1515 , 1524 , 1533 , 1542 , 1551 , 1614 , 1623 , 1632 , 1641 , 2136 , 2145 , 2154 , 2163 , 2226 , 2235 , 2244 , 2253 , 2262 , 2316 , 2325 , 2334 , 2343 , 2352 , 2361 , 2415 , 2424 , 2433 , 2442 , 2451 , 2514 , 2523 , 2532 , 2541 , 2613 , 2622 , 2631 , 3126 , 3135 , 3144 , 3153 , 3162 , 3216 , 3225 , 3234 , 3243 , 3252 , 3261 , 3315 , 3324 , 3333 , 3342 , 3351 , 3414 , 3423 , 3432 , 3441 , 3513 , 3522 , 3531 , 3612 , 3621 , 4116 , 4125 , 4134 , 4143 , 4152 , 4161 , 4215 , 4224 , 4233 , 4242 , 4251 , 4314 , 4323 , 4332 , 4341 , 4413 , 4422 , 4431 , 4512 , 4521 , 4611 , 5115 , 5124 , 5133 , 5142 , 5151 , 5214 , 5223 , 5232 , 5241 , 5313 , 5322 , 5331 , 5412 , 5421 , 5511 , 6114 , 6123 , 6132 , 6141 , 6213 , 6222 , 6231 , 6312 , 6321 , 6411 , Total =  125

Therefore, the probability is = 125 / 6^4 =125 / 1296

I don't think or know anything that's why i am asking 

The answer is 2/3

Can you explain? I don't understand the problem

\({1 \over a} + {1 \over b} = {b \over ab} + {a \over ab} = {a + b \over ab} = {7 \over ab} \)

\((a+b)^3 = a^3 + b^3 + 3a^2b+3ab^2 = 343\)

\(44 + 3a^2b+3ab^2 = 343\)

\(3a^2b+3ab^2 = 299\)

\(3ab(a+b) = 299\)

\(3ab \times 7 = 299\)

\(21ab = 299\)

\(ab = {299 \over 21}\)

\({7 \over {299 \over 21}} = {{147 \over 7} \over {299 \over 21}} = \color{brown}\boxed{147 \over 299}\)

The answer is 102