Answer:
1: 7/12 2:1/3 3:1/2 4:1/3 5:5/8 6:1/2 7:6/10 8:1/1 9:1/4
Step-by-step explanation:
lo saque de otro trabajo igual.
Answer:
b
Step-by-step explanation:
From the given the option moved left 1 unit; reflected across the x axis is the correct.
what is function?
A function from a set X to a set Y in mathematics assigns each element of X exactly one element of Y. The set X is known as the function's domain, and the set Y is known as the function's codomain.
We have to describe the position of the function
relative to the basic function
.
Let the parent function is
.
After shifting one unit right it becomes,
.
After reflection in x axis it becomes,
.
Hence, from the given the option moved left 1 unit; reflected across the x axis is the correct.
To know more about function, click on the link
brainly.com/question/25638609
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Answer:
Neither.
Step-by-step explanation:
I am not completely sure, so make sure to also let other answer this before you trust me lol. I really hope this helps!!
Answers:
- <u>24000 dollars</u> invested at 4%
- <u>18000 dollars</u> was invested at 7%
======================================================
Work Shown:
x = amount invested at 4%
If she invests x dollars at 4%, then the rest (42000-x) must be invested at the other rate of 7%
She earns 0.04x dollars from that first account and 0.07(42000-x) dollars from the second account
This means we have
0.04x+0.07(42000-x)
0.04x+0.07*42000-0.07x
0.04x+2940-0.07x
-0.03x+2940
This represents the total amount of money earned after 1 year.
We're told the amount earned in interest is $2220, so we can say,
-0.03x+2940 = 2220
-0.03x = 2220-2940
-0.03x = -720
x = -720/(-0.03)
x = 24000 dollars is the amount invested at 4%
42000-x = 42000-24000 = 18000 dollars was invested at 7%
----------------------
As a check, we can see that
18000+24000 = 42000
and also
0.04x = 0.04*24000 = 960 earned from the first account
0.07*18000 = 1260 earned from the second account
1260+960 = 2220 is the total interest earned from both accounts combined
This confirms our answers.