Answer: 2 hr.
Explanation: Think back to rate of change. <em>d</em> = <em>rt</em>, <em>r</em> = <em>d/t</em>, <em>t</em> = <em>d/r</em>. In this case, we will be using <em>d</em> = <em>rt</em>. Mph would be <em>r</em>, rate, so you would categorize 15 mph and 8 mph under rate. <em>t</em> should represent the time each cyclist traveled. Tracey's and Emma's distance, <em>d</em>, would be the same as their mph, hence Tracey's being 15<em>t</em> and Emma's would be 8<em>t</em>. When you add Tracey's distance plus Emma's distance, you end up with 46 mi. Now, you need to combine like terms, which should look like 15<em>t</em> + 8<em>t </em> = 46. Add 15 and 8 to get 23, so it should be 23<em>t</em> = 46 now. Then, divide both sides of the equation by 23 and now you should have your answer, <em>t</em> = 2 hr.
Answer:
0.0065 divided by 10
Step-by-step explanation:
Here you go.
The slopes of the original function y = |x| are m = 1 and m = -1 (m is the variable used to represent slope).
when you add a coefficient (number) in front of |x|, it will either make the slopes steeper or more flat. the larger the value of the coefficient, the steeper the slope will be (vice versa for a coefficient smaller than 1, which would make the slope more flat than the parent(original) function).
because these are absolute value functions, they will have two slopes. one slope for the end going up from left to right, and one for the end going down from left to right. this means that one slope must be positive and the other slope must be negative for each function.
with this in mind, the slopes of y = 2|x| are m = 2 and m = -2. the coefficient of 2 narrows the function by a factor of 2 (it is twice as narrow as the parent function). the same rules apply to y = 4|x| with the slopes of this function as m = -4 and m = 4 (it is 4 times narrower than the parent function).
with the fraction coefficients, the function is being widened. therefore, the slopes of y = 1/2 |x| are m = -1/2 and m = 1/2. the slopes of y = 1/5 |x| are m = -1/5 and m = 1/5.
1.567 rounded to the nearest tenths place is 1.6
Y(2)-y(1)/x(2)-x(1)
9-7/-2–(-2)
0
It is a vertical line
