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Two cars are 500 miles apart & directly moving towards each other. One car is at a speed of 100 mph and the other is at 70 mph. Supposing that the cars start at the same time how much time it take for the two cars to meet?
Solution
Let's assume t represent the amount of time which the cars are traveling before they meet. Now, we have to sketch a figure for this one. This figure will help us to write the equation which we'll have to solve out.
From this figure we can note that the Distance Car A travels as well as the Distance Car B travels has to equal the total distance separating the two cars, 500 miles.
Following is the word equation for this problem into two separate forms.
We utilized the standard formula here twice, once for each car. We know that the distance a car travels is the rate of the car times the time traveled by the car. In this case we know that Car A travels at 100 mph for t hours & that Car B travel at 70 mph for t hours as well. Plugging these in the word equation and solving out gives us,
100t + 70t = 500
170t = 500
t = 500/170 =2.941176 hrs
Thus, they will travel for approximately 2.94 hours before meeting.
I can find what x means I just cant do the interval notation correctly
(xy)-3/(x^-5y)3
Bills Roast Beef sells 3 times as many sandwiches as Pete''s deli. The difference between their sales is 170 sandwiches. How many sandwiches did each sell?
x2 -x = 7 -3 -5 1 5 -3
#questionfWRWETWE..
-(1-5n)-7n
how to expand when asked to divide, multiply etc
This section doesn't actually have many to do with the rest of this chapter, but since the subject required to be covered and it was a fairly short chapter it appeared like as good
For these properties we will suppose that x > 0 and y > 0 log b ( xy ) = log b x + log b y log b ( x/y) = log b x - log b y log b (x r ) = r log x If log
(m2-3m-10)/(m-5)
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