9514 1404 393
Answer:
2) 1/3
Step-by-step explanation:
The repeating decimal can be converted to a fraction by multiplying it by 10^n and subtracting the original value. n = number of repeating digits. This process cancels all of the repeating digits, so you can find the equivalent fraction.

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Your calculator can help you answer this:
1/5 = 0.2
1/3 = 0.333.... (repeating) . . . . the number you're looking for
3/10 = 0.3
3/5 = 0.6
Answer:
y = 1
Step-by-step explanation:
Passes through (6,-1)(6,-1)
Find the slope (m)
m=(y2 - y1) / (x2 - x1)
m=(-1 - (-1) ) / (6-6)
m= 0
Parallel to x - 3y = 3
-3y = -x +3
y = 1/3 x -1
Equation of the line equation is y - y1 = m ( x - x1 )
y - ( -1 ) = 0 ( x - 6 )
y + 1 = 0
y = 1
-12 because surface is usually considered as "0" so 12 ft below the surface means -12.
Answer:
(A) -3 ≤ x ≤ 1
Step-by-step explanation:
The given function is presented as follows;
h(x) = x² - 1
From the given function, the coefficient of the quadratic term is positive, and therefore, the function is U shaped and has a minimum value, with the slope on the interval to the left of <em>h</em> having a negative rate of change;
The minimum value of h(x) is found as follows;
At the minimum of h(x), h'(x) = d(h(x)/dx = d(x² - 1)/dx = 2·x = 0
∴ x = 0/2 = 0 at the minimum
Therefore, the function is symmetrical about the point where x = 0
The average rate of change over an interval is given by the change in 'y' and x-values over the end-point in the interval, which is the slope of a straight line drawn between the points
The average rate of change will be negative where the y-value of the left boundary of the interval is higher than the y-value of the right boundary of the interval, such that the line formed by joining the endpoints of the interval slope downwards from left to right
The distance from the x-value of left boundary of the interval that would have a negative slope from x = 0 will be more than the distance of the x-value of the right boundary of the interval
Therefore, the interval over which <em>h</em> has a negative rate of change is -3 ≤ x ≤ 1