Answer:
$15,059
Step-by-step explanation:
4.1 x 9 = 36.9
36.9% of 11,000 = 4059
11,000 + 4,059 = 15,059
Good morning ☕️
Answer:
<h3>i¹ + i² + i³ +. . .+ i⁹⁹ + i¹⁰⁰ =
0</h3>
Step-by-step explanation:
Consider the sum S = i¹ + i² + i³ +. . .+ i⁹⁹ + i¹⁰⁰
S = i¹ + i² + i³ + . . . + i⁹⁹ + i¹⁰⁰
S = a₁ + a₂ + a₃ +. . . + a₉₉ + a₁₀₀
then, S is the sum of 100 consecutive terms of a geometric sequence (an)
where the first term a1 = i¹ = i and the common ratio = i
FORMULA:______________________
![S=(term1)*\frac{1-(common.ratio)^{number.of.terms}}{1-(common.ratio)}](https://tex.z-dn.net/?f=S%3D%28term1%29%2A%5Cfrac%7B1-%28common.ratio%29%5E%7Bnumber.of.terms%7D%7D%7B1-%28common.ratio%29%7D)
_______________________________
then
![S=i*\frac{1-i^{100} }{1-i}](https://tex.z-dn.net/?f=S%3Di%2A%5Cfrac%7B1-i%5E%7B100%7D%20%7D%7B1-i%7D)
or i¹⁰⁰ = (i⁴)²⁵ = 1²⁵ = 1 (we know that i⁴ = 1)
Hence
S = 0
Answer:
25
Step-by-step explanation:
ratio 5:6
![\frac{5}{6} = \frac{x}{30}](https://tex.z-dn.net/?f=%5Cfrac%7B5%7D%7B6%7D%20%3D%20%5Cfrac%7Bx%7D%7B30%7D)
6x = 5(30)
6x = 150
x = 25
Answer:
Step-by-step explanation:
First, in order to get rid of the parenthesis, you need to multiply 2.5 with whatever is in the parenthesis.
Multiply 2.5 on each side of the equation like this:
(12.75 x 2.5) (24.50 x 2.5) =188.75
Once you're finished, you should get 61.25 and 31.875x =188.75
Now in order to find 'x' you will need to subtract 61.25 to each side of the equation like this:
61.25 - 61.25 = 0
188.75 - 61.25 = 127.5
So now your equation should look like this:
31.875x = 127.5
You have to divide 31.875 by each side now.
Once you divide them, your final answer should be 4.
so x = 4.
I really hope this helps! ^^
In other words, what is <em>y</em> if <em>x</em> is 0.
Look on the graph, a line passes through point, ![A(x,y)\longrightarrow A(0,-3)](https://tex.z-dn.net/?f=A%28x%2Cy%29%5Clongrightarrow%20A%280%2C-3%29)
So we can conclude that at that very point the input <em>x</em> was 0 and the output <em>y</em> was -3 therefore,
![\boxed{f(-3)=0}](https://tex.z-dn.net/?f=%5Cboxed%7Bf%28-3%29%3D0%7D)
Hope this helps.
<u>r3t40</u>