Answer:
Each golf ball costs about <u>$2</u>.
Step-by-step explanation:
Given:
Donald bought a box of golf balls for $9.27. There were 18 golf balls in the box.
Rename the decimal dividend as a whole number that is compatible with the divisor to estimate the quotient.
Now, to find the cost of each golf ball.
<u>As given:</u>
<em>Rename the decimal dividend as a whole number that is compatible with the divisor to estimate the quotient.</em>
As, the cost of golf ball box = $9.27.
So, 9.27 nearest to whole number is 9.
Thus to get the cost of each ball:
18 golf balls cost = $9.
So, 1 golf ball cost = 
Therefore, each golf ball costs about $2.
Answer:
D is the answers for the question
Step-by-step explanation:
please mark me as brainlest
1/32 maybe................
Answer:
Anything in the form x = pi+k*pi, for any integer k
These are not removable discontinuities.
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Explanation:
Recall that tan(x) = sin(x)/cos(x).
The discontinuities occur whenever cos(x) is equal to zero.
Solving cos(x) = 0 will yield the locations when we have discontinuities.
This all applies to tan(x), but we want to work with tan(x/2) instead.
Simply replace x with x/2 and solve for x like so
cos(x/2) = 0
x/2 = arccos(0)
x/2 = (pi/2) + 2pi*k or x/2 = (-pi/2) + 2pi*k
x = pi + 4pi*k or x = -pi + 4pi*k
Where k is any integer.
If we make a table of some example k values, then we'll find that we could get the following outputs:
- x = -3pi
- x = -pi
- x = pi
- x = 3pi
- x = 5pi
and so on. These are the odd multiples of pi.
So we can effectively condense those x equations into the single equation x = pi+k*pi
That equation is the same as x = (k+1)pi
The graph is below. It shows we have jump discontinuities. These are <u>not</u> removable discontinuities (since we're not removing a single point).
1st one that I can possibly think behind my brain: 326/1000 or simplified is 163/500
2nd: Three hundred and twenty six thousandths
Hope it helps! Happy Halloween!