Consider the following.f(x)=74 – x2Find the x-values at which f is not continuous. Which of the discontinuities are removable?
1 answer:
Answer
given,
f(x) = 74 - x²
A function is continuous when it is differential at all the points.
so, the given function is continuous for all the value of x.
We should know that if a function is quadratic then it is differential at all the point. And if it is differential then it is definitely continuous.
Hence, there is no discontinuous point present in the equation.
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Answer:
The correct answer is two designs are used 12 times each and other four designs are used for 13 times each.
Step-by-step explanation:
Width of the quilt Lu is making is 18 rows of 20 squares each.
Thus the perimeter of the quilt is 20 + 20 + 18 + 18 = 76.
We can also say there are 76 square as the boundary which are to be bordered with toy designs.
Each square can hold 1 design of toy and there are 6 different designs of toys.
Thus number of times each design is used = 12
.
Thus there are two designs used 12 times each and other four designs are used for 13 times each.
The first year, the amount is 40,000
the second year is 40000 + 4.2% of 40000, or 0.042 * 4000, so 40000+(0.042*4000)
common factoring that we get 40000(1 + 0.042), or just 40000(1.042)
in short, the starting amount is 40000, and to get the next term's value you'd use the "common ratio" of 1.042, namely the multiplier of 1.042.
for the third year it'll be 40000(1.042) + (0.042 *40000(1.042) ), again, common factoring that
40000(1.042)(1 + 0.042) or 40000(1.042)(1.042) or 40000(1.042)²
therefore,
the second year is 40000 + 4.2% of 40000, or 0.042 * 4000, so 40000+(0.042*4000)
common factoring that we get 40000(1 + 0.042), or just 40000(1.042)
in short, the starting amount is 40000, and to get the next term's value you'd use the "common ratio" of 1.042, namely the multiplier of 1.042.
for the third year it'll be 40000(1.042) + (0.042 *40000(1.042) ), again, common factoring that
40000(1.042)(1 + 0.042) or 40000(1.042)(1.042) or 40000(1.042)²
therefore,