The answer is D.
350 multiplied by 6 is equal to 2100
350 * 0.5 (30 minutes) is equal to 175
2100 + 175 = 2275
First, convert the 25% to a real mathematical number. For percents, this is always done by dividing the 25% by 100%, or 25% / 100% = 0.250.
Second, find out what 25% of $100 is. This is the amount of the sale discount. This is always found by mulitplying 0.250 by the item's cost $100, like this:
0.250 x $100 = $25.00.
So for this sale, you'll save $25.00 on this item.
This means, the cost of the item to you is
$100 - $25.00 = $75.00.
Alternatively, you can think about it this way. The item is 25% off. This means you'll pay 75.000% of the total cost (100% - 25% = 75.000%).
Now what's 75.000% of the total cost?
0.750 x $100 = $75.00.
Just like the result above, the sale price on the item is $75.00.
Hope this helps you on your assignment! :)
The function, g(x), has a constant rate of change and will increase at a faster rate than the function f(x) for all the values of x.
Given:
g(x) = 5/2 x -3 ..... (1)
f(x) = - 3.5 at x = 0
So, putting the value of x=0 in equation (1) for comparison. We get,
g(x) at x = 0
=> g(x) = 5/2 x (0) - 3
=> g(x) = -3
In this value of x function g(x) is faster than function f(x) having a value equal to -3.5.
Similarly, put x = 1 in equation (1) for comparison. We get,
=> g(x) = 5/2 x (1) - 3
=> g(x) = (5-6)/2
=> g(x) = -1/2
In this value of x function g(x) is faster than function f(x) having a value equal to -1.
Similarly, put x = 2 in equation (1) for comparison. We get,
=> g(x) = 5/2 x (2) - 3
=> g(x) = (5-3)
=> g(x) = 2
In this value of x function g(x) is faster than function f(x) having a value equal to 1.5.
Similarly, put x = 3 in equation (1) for comparison. We get,
=> g(x) = 5/2 x (3) - 3
=> g(x) = (15/2 - 3)
=> g(x) = 7.5 - 3
=> g(x) = 4.5
In this value of x function g(x) is faster than function f(x) having a value equal to 4.
Therefore, for all values of x function g(x) is faster than function f(x).
function f(x).
To learn more about the function visit: brainly.com/question/14996787
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