Answer:
4
Step-by-step explanation:
Let's say her speed was x miles/hour during the first 3 miles runThen, time = distance/speedt1 = 3/x eq1 In the next 4 miles run, her speed = x-1 miles/hourTime taken:t2 = 4/(x-1) eq2 Now, total time:t1 + t2 = 1 3/5 hourssubstitute t1 and t2 from eqs. 1 and 2 3/x + 4/(x-1) = 1 3/5=> 3/x + 4/(x-1) = 8/5
=> 3(x-1) + 4x = 8x(x-1)/5=> 35x - 15 = 8x2 - 8x=> 8x2 - 43x + 15 = 0=> (8x-3)*(x-5) = 0=> x = 3/8 or 5 miles/hourx can not be 3/8 miles/hour because in that case, the speed during 4 miles run would be 3/8-1 = negative numberi.e. speed during 3 miles segment = 5 miles/hourand speed during 4 miles segment = 5-1 = 4 miles/hour
The answer is -3x^2+10xy+8y^2
Answer:
We have:
x < 4 AND x > a.
if a = 4 and we use an "or" instead of the "and" we have:
x < 4 or x > 4.
This is:
"x is larger than 4 or smaller than 4."
Then the solution of this is all the real numbers except the value x = 4.
The set of solutions can be written as:
{xI x ∈ R \ [4]}
Where this reads:
"x belongs to the set of the reals minus the number 4".
Or we also could write it as:
x ∈ (-∞, 4) ∪ (4, ∞)
Where we have two open ends in the "4" side, so the value x = 4 does not belong to that set.
Answer:
add them togethr then subract the diffrece to get x
Step-by-step explanation:
