I assume this is supposed to be:
6x^2 - 13x + 5 = 0
(3x - 5) (2x - 1) = 0
Setting each factor to 0 and solving
3x - 5 = 0 2x - 1 = 0
add 5 to both sides add 1 to both sides
3x = 5 2x = 1
divide both sides by 3 divide both sides by 2
x = 5/3 or x = 1/2
We can find the horizontal distance, h, from the plane to the ship as follows :
tan (40) = 1300 / h .....rearrange as.....
h = 1300/tan(40) = about 1549.3 m
We need to find how many terms we have, as follows
1701 = 7(3)n-1 divide both sides by 7
243 = 3n-1 and 35 = 243 so n = 6
So we have
Sn = a1 [ (1-rn) / (1 - r) ]
S6 = 7 [ (1 - 36) / (1 - 3) ] = 7 [ 1 - 729] [ 1 - 3] = 7 [-728] / [-2] = 2548
25=286.93(9.738^x) divide both sides by 286.93
25/286.93 = 9.738^x take the log of each side
log [ 25/286.93] = log [9.738]^x and we can write
log [ 25/286.93] = x * log [9.738] divide both sides by log [9.738]
log [ 25/286.93] / log [9.738] = x = about -1.072
We need to know how many terms are to be summed......obviously, this sequence would have an unbounded infinite sum because each successive term is three times the previous one....
We need to solve this, first
192 = 3(2)n-1 divide both sides by 3
64 = 2n-1 and 26 = 64, so n = 7
So we have the sum of 7 terms
Sn = 3 [ (1 - 27) / (1 - 2) ] = 3 [ (1 - 128) / (-1) ] = 3* 127 = 381
This was answered here..... http://web2.0calc.com/questions/two-geometric-means-are-inserted-between-6-and-384-so-the-four-numbers-form-a-geometric-sequence-what-are-these-two-numbers
384/6 = r3
64 = r3 take the cube root of each side
4 = r
So the second number = 6(4) = 24
And the third = 24(4) = 96
2x+3y=9 (1)
x+5y=8 [ multiply this eq. by -2] = -2x - 10y = -16 ..... add this to the first eq.
0 - 7y = -7
-7y = -7 divide both sides by -7 and y = 1
Using (1) to find x
2x + 3(1) = 9
2x + 3 = 9 subtract 3 from both sides
2x = 6 divide both sides by 2
x = 3
So (x , y) = (3, 1)
