Answer:
17 nickles !
Step-by-step explanation:
First, identify the variables:
n = amount of nickels
d = amount of dimes
Next, setup the equations based on what you know. The first equation is:
n + d = 28
For the second equation, we know that a dime is worth 10¢ and a nickel is 5¢, so it should be:
0.05n + 0.10d = 1.95
This a three-step answer:
In one formula (you can use any of them; most people use the simplest one), single out the variable on one side
Apply the first formula into the second formula, and solve it to get the value of one variable
Apply the answer from the second formula into the first formula, and solve it to get the value of the other variable
======
Step One:
n + d = 28
n + d - d = 28 - d
n = 28 - d
Step Two:
0.05n + 0.10d = 1.95
(0.05 * (28 - d)) + 0.10d = 1.95
1.40 - 0.05d + 0.10d = 1.95
1.40 + 0.05d = 1.95
1.40 - 1.40 + 0.05d = 1.95 - 1.40
0.05d = 0.55
d = 11
Step Three:
n = 28 - d
n = 28 - 11
n = 17
======
Your answer should be 17 nickels and 11 dimes.
You can double check by applying the variables into both formulas.
n + d = 28
17 + 11 = 28
28 = 28
0.05n + 0.10d = 1.95
(0.05 * 17) + (0.10 * 11) = 1.95
0.85 + 1.10 = 1.95
1.95 = 1.95
I hope this helped.
Answer:
(
3
x
+
1
)
(
3
x
−
1
)
is the factorised form of the expression
idk if its right so dont come at me and i also think that it didnt paste right.
<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:
Step-by-step explanation:
Find common difference from subtracting any term in the sequence with the previous term.
Apply nth term formula.
