For the first one multiply each term by the LCM of 4 and 6, which is 12:-
15x -36 = 2y..................(1)
and second multiply by 5:-
2x + y = 9..............(2)
solving:-
from equation (2) y = 9-2x, substitute in equation (1):-
15x - 36 = 2(9-2x) = 18 - 4x
19x = 54
x = 54/19 = 2.84
so y = 9 - 2(2.84) = 3.32
Let 'x' be the distance from THE far bank where 700 is the distance to the NEAR bank
boat one has travelled 700 (rate = 700/unit time) boat two has travelled x rate = x / unit time
boat one then travels x + 400 more and boat two travels 700 + (700+x -400) more when they meet
The time is the same rate x time = distance distance/rate = time equate the distances divided by the respective rates
(700 + x + 400)/700 = ( x + 700 + (700+x-400) )/x
1100x + x^2 = 1400x + 700000
x^2-300x -700000 = 0 quadratic formula yields x = 1000
One boat travels 700 the other 1000 whe they first meet.....width of river = 700+ 1000 = 1700 m
Answer:
104°
Step-by-step explanation:
If segments NO and NM are congruent, then angles NMO and NOM are congruent. So, their supplements, angles NML and NOP are congruent. That is ...
∠NML ≅ ∠NOP = 104°
∠NML = 104°
Hello from MrBillDoesMath!
Answer:
-10w - 20
Discussion:
5 ( -2w - 4 ) =
5(-2w) - 5(4) =
-10w - 20
Thank you,
MrB
