
√8 is between 2 and 3, because 2²=4<8, but 3²=9>8. Also, our value is closer to 3 than to 2, so it is more than 2.5 and we have C and D options left.
Among these two numbers we find the one which is closer to √8.
C. 27=√729 ⇒ 2.7=√7.29
D. 28=√784 ⇒ <u>2.8=√7.84</u>
Hence our answer is D) 2.8
5^-3 is 1/125
The negative in the exponent makes it so there is a 1 over the problem if there was no negative. When you have a negative in the exponent, the answer will be the reciprocal (opposite) of the question with no negative.
For example,
9^2 = 81
9^-2 = 1/81
When ur polynomial has more then one variable....and this one does...the degree is the highest term degree
7a^3b^2 ....this term has a degree of 5
2a^4...this term has a degree of 4
4b....this term has a degree of 1
15....this term has a degree of 0
so the highest degree term has a degree of 5....so that is the degree of this polynomial.
now if u just had 1 variable, the degree would be the highest exponent
Answer:
5
Step-by-step explanation:
Answer:
A function f(x) is said to be periodic, if there exists a positive real number T such that f(x+T) = f(x).
You can also just say: A periodic function is one that repeats itself in regular intervals.
Step-by-step explanation:
The smallest value of T is called the period of the function.
Note: If the value of T is independent of x then f(x) is periodic, and if T is dependent, then f(x) is non-periodic.
For example, here's the graph of sin x. [REFER TO PICTURE BELOW]
Sin x is a periodic function with period 2π because sin(x+2π)=sinx
Other examples of periodic functions are all trigonometric ratios, fractional x (Denoted by {x} which has period 1) and others.
In order to determine the period of the determined graph however, just know that the period of the sine curve is the length of one cycle of the curve. The natural period of the sine curve is 2π. So, a coefficient of b=1 is equivalent to a period of 2π. To get the period of the sine curve for any coefficient b, just divide 2π by the coefficient b to get the new period of the curve.
Hopefully this helped a bit.