Answer:
-18
Step-by-step explanation:
Let x be our number. First we have to write our equation.
The product of a number and -8 is represented as -8x (-8 multiplied by x). Remember that 'product' signifies multiplication.
The result is eight times the sum of x and 36. This is represented as 8(x + 36). We add x and 36 first because 8 has been multiplied by the sum of those two numbers.
So our equation is:
-8x = 8(x + 36)
We can now solve for x. First, expand the bracket.
-8x = 8x + 288
Next, subtract 8x from both sides. This will give us only one term containing x, which will make it easier to solve.
-8x - 8x = 8x + 288 - 8x
-16x = 288
Divide both sides by -16
x = -18
Instead of X put 5 and instead of y put 2
5+6(2)
5+12= 17
<em>z</em> = 3<em>i</em> / (-1 - <em>i</em> )
<em>z</em> = 3<em>i</em> / (-1 - <em>i</em> ) × (-1 + <em>i</em> ) / (-1 + <em>i</em> )
<em>z</em> = (3<em>i</em> × (-1 + <em>i</em> )) / ((-1)² - <em>i</em> ²)
<em>z</em> = (-3<em>i</em> + 3<em>i</em> ²) / ((-1)² - <em>i</em> ²)
<em>z</em> = (-3 - 3<em>i </em>) / (1 - (-1))
<em>z</em> = (-3 - 3<em>i </em>) / 2
Note that this number lies in the third quadrant of the complex plane, where both Re(<em>z</em>) and Im(<em>z</em>) are negative. But arctan only returns angles between -<em>π</em>/2 and <em>π</em>/2. So we have
arg(<em>z</em>) = arctan((-3/2)/(-3/2)) - <em>π</em>
arg(<em>z</em>) = arctan(1) - <em>π</em>
arg(<em>z</em>) = <em>π</em>/4 - <em>π</em>
arg(<em>z</em>) = -3<em>π</em>/4
where I'm taking arg(<em>z</em>) to have a range of -<em>π</em> < arg(<em>z</em>) ≤ <em>π</em>.
Answer:
n ≥ -3
Step-by-step explanation:
If n can be <em>no less than</em> -3... then we know that n has to be bigger than or equal to -3.
This could also be said as <em>greater than or equal to</em> -3
Conveniently enough... there is a sign for <em>greater than or equal to</em> and it is ≥
