Answer:
Function B has the greater initial value because the initial value for function A is 4 and the initial value for Function B is 5
Step-by-step explanation:
- <em>The initial value of a function is the output value of the function when the input value is 0</em>
Initial value of A is y=4 at x=0,
and
initial value of B is y=0*6+5= 5 at x=0
Function B has the greater initial value because the initial value for function A is 4 and the initial value for Function B is 5
125/12 would be your equivalent improper fraction
<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:
Continuous: g(x) and j(x)
Removable: h(x) and m(x)
Infinite: f(x) and i(x) and k(x)
Jump: l(x)
Step-by-step explanation:
Answer:
x = -1, y = -4
Step-by-step explanation:
Let's solve our system of equations by substitution.
y = 5x + 9y = −x + 3
Step: Solve = 5x + 9 for y:
y = 5x + 9
Step: Substitute 5x + 9 for y in y = −x + 3:
y = −x + 3
5x + 9 = −x + 3
5x + 9 + x = −x + 3 + x (Add x to both sides)
6x + 9 = 3
6x + 9 + −9 = 3 + −9 (Add -9 to both sides)
6x = −6
6x/ 6 = −6/6(Divide both sides by 6)
x = −1
Step: Substitute −1 for x in y = 5x + 9:
y = 5x + 9
y = (5)(−1) + 9
y = 4(Simplify both sides of the equation)
So our answers are y = -4 and x = -1.
