Two<span> trains </span>leave different<span> cities heading toward each </span>other<span> at </span>different<span> speeds. ... At the </span>same time<span>Train B, </span>traveling 60 mph<span>, leaves Eastford heading toward Westford. ... Since an equation remains true as </span>long<span> as we perform the </span>same<span> operation ... that the train's rate is 40 </span>mph<span>, which means it </span>will travel<span> 40 </span>miles<span> in </span>one<span> hour.</span>
The answer is 42 bc 42-6 is 36
Answer:
8
6
4
2
0
Step-by-step explanation:
Answer:
2+under root 3 is the correct answer
First of all, we can solve this by elimination (adding the 2 equations and eliminating 1 variable).
3x+10y=-47
5x+7y=40
It looks like we can eliminate the x by making one 15x, and the other -15x (15 is the least common multiple [LCM] of 3 and 5).
5 (3x+10y=-47) --> 15x+50y=-235
-3 (5x+7y=40) --> -15x-21y=-120
15x-15x = 0, 50y-21y = 29y, -235-120 = -355
0 + 29y = -355
29y/29 = -355/29
y = 12.24
then plugged that into 3x+10y=-47
3x+122.4=-47
3x+122.4-122.4=-47-122.4
3x = -169.4
x = 56.47
Something is wrong... are you sure the 2nd equation wasn't equal to NEGATIVE 40??
Or the 1st one equal to POSITIVE 40??
