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
Answer:
The SF is 3
Step-by-step explanation:
The sides are 3x the length of the sides of ∆ABC
-3x + 6 + 12 y. I am afraid that there are no like terms to combine, and it is neither an equation nor an inequality.
Hope this Helps!
Answer:
You could find the unit rate by dividing the first term of the ratio by the second term.
A(n)=ar^(n-1) and we can find the rate upon using the ratio of two points...
50/1250=1250r^2/1250r^0
1/25=r^2
r=1/5 so
a(n)=1250(1/5)^1=250
...
You could have also found the geometric mean which is actually quite efficient too...
The geometric mean is equal to the product of a set of elements raised to the 1/n the power where n is the number of multiplicands...in this case:
gm=(1250*50)^(1/2)=250
