There are two ways to work this out: normal variables or using "imaginary" numbers.
Normal variables:
![(7+2i)(3-i)\\(7*3)+[7*(-i)]+(3*2i)+[2i*(-i)]\\21-7i+6i-2i^{2}\\\\21-i-2i^{2}](https://tex.z-dn.net/?f=%20%287%2B2i%29%283-i%29%5C%5C%287%2A3%29%2B%5B7%2A%28-i%29%5D%2B%283%2A2i%29%2B%5B2i%2A%28-i%29%5D%5C%5C21-7i%2B6i-2i%5E%7B2%7D%5C%5C%5C%5C21-i-2i%5E%7B2%7D)
Imaginary numbers:
Using the result from earlier:
Now since
, then the expression becomes:
The answer is some number between 9 and 10.
Explanation:
√83
is an irrational number. You won't be able to simplify it any further either, since it doesn't have any perfect square factors.
However, you will be able to tell between which two numbers it lies in.
9^2 is 81 and 10^2 is 100
. Therefore, you can say that a certain number between 9 and 10 is 83 when squared.
If you're looking for an exact answer, then it will be 9.11043357914... (I got that using a calculator).
Answer:
<h3>The correct answer is <u>D: -30</u></h3>
Answer:
\frac{2x^2}{3} =24
Step-by-step explanation:
times both sides by 3
2x²=72
divide both sides by 2
x²=36
sqrt both sides
x=+/-6
x=-6 or 6
-6<6 so -6 is the smallest solution
We take the value of F in the inequality by taking the inequalities in group. Let the first group be:
(1) -20 ≤ 59(F - 32)
Then, the second group would be,
(2) 59(F - 32) ≤ - 15
Calculating for the values of F,
(1) -20 ≤ 59F - 1888
1888 - 20 ≤ 59F
1868 ≤ 59F
F ≥ 31.66
(2) (59)(F - 32) ≤ - 15
59F - 1888 ≤ -15
59F ≤ 1873
F ≤ 31.75
The values of F are therefore,
31.66 ≤ F ≤ 31.75
