The number of integer values of n for which the value 4000 × (0.4)ⁿ is an integer is; 9
<h3>How to find the integer values in algebra?</h3>
We are given the expression;
4000 × (0.4)ⁿ
This can also be expressed as;
4000 × (²/₅)ⁿ which can further be expressed in exponent form as;
(2⁵ × 5³) × (²/₅)ⁿ = 2⁽⁵ ⁺ ⁿ⁾ × 5⁽³ ⁻ ⁿ⁾
Since this expression is an integer, we need:
1) 5 + n ≥ 0 for which n ≥ -5
2) 3 - n ≥ 0 for which n ≤ 3
Taking the intersection gives;
-5 ≤ n ≤ 3
Thus;
Number of Integer Values = 3 - (-5) + 1 = 9
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Answer:
see below
Step-by-step explanation:
x^-2 = 1/x^2 Let x = -2 1/ (-2)^2 = 1/4
x^-1 = 1/x^1 Let x = -2 1/ (-2)^1 = 1/(-2) = -1/2
x^0 = 1
x^1 = x Let x = -2 (-2)
x^2 = Let x = -2 (-2)^2 = 4
Answer:
1.78260245797
Step-by-step explanation:
<u><em>PLEASE MARK BRAINLIEST</em></u>
Answer:(13,-8)
Step-by-step explanation:
Answer: 22j
Explanation:
1) From the graph you can calculate the vector v as folllows:
Translation of the extreme (-5,0) to the origin (0,0) implies add 5 units to - 5.
Then, add 5 units to the x-coordinate of the other extreme: 7 + 5 = 12; and keep the y-coordinate, which is 2.
Therefore, the coordinates of the vector v are (12,2)
2) Now you can find the vector v - 4u applying the operations to the coordinates of the two vectors:
v - 4u = (12,2) - 4(3, -5) = (12,2) - (12,-20) = (12 - 12, 2 + 20) = (0,22)
3) Changing the notation in terms to i, j; that is 0i + 22j = 22j
