Answer:
983040
Step-by-step explanation:
Second Year : 15x4=60
Third Year : 60x4=240
Fourth Year : 240x4=960
Fifth Year : 960x4=3840
Sixth Year : 3840x4=15360
Seventh Year : 15360x4 = 61440
Eighth Year: 61440x4=245760
Ninth Year : 245760x4= 983040
Y<-60
hope this helps and good luck;)
First you must have the quadratic equal to zero. In order to do this you must subtract 7 to both sides
x^2 + 20x + (100 - 7) = 7 - 7
x^2 + 20x + 93 = 0
Now you must find two numbers who's sum equals 20 and their multiplication equal 93
Are there any? NO!
This means that you have to use the formula:
In this case:
a = 1
b = 20
c = 93
^^^We must simplify √28
√28 = 2√7
so...
simplify further:
-10 + √7
or
-10 - √7
***plus/minus = ±
Hope this helped!
~Just a girl in love with Shawn Mendes
<h3>
Answer:</h3>
System
Solution
- p = m = 5 — 5 lb peanuts and 5 lb mixture
<h3>
Step-by-step explanation:</h3>
(a) Generally, the equations of interest are one that models the total amount of mixture, and one that models the amount of one of the constituents (or the ratio of constituents). Here, there are two constituents and we are given the desired ratio, so three different equations are possible describing the constituents of the mix.
For the total amount of mix:
... p + m = 10
For the quantity of peanuts in the mix:
... p + 0.2m = 0.6·10
For the quantity of almonds in the mix:
... 0.8m = 0.4·10
For the ratio of peanuts to almonds:
... (p +0.2m)/(0.8m) = 0.60/0.40
Any two (2) of these four (4) equations will serve as a system of equations that can be used to solve for the desired quantities. I like the third one because it is a "one-step" equation.
So, your system of equations could be ...
___
(b) Dividing the second equation by 0.8 gives
... m = 5
Using the first equation to find p, we have ...
... p + 5 = 10
... p = 5
5 lb of peanuts and 5 lb of mixture are required.
Answer:
The answer is the 2nd option
Step-by-step explanation:
