The amount of Tina money can be expressed in an exponent function like this:
an= $1100(1.0725)^n
The variable an represent the total money and variable n is the years needed to achieve that amount.
Then, the time needed for the money to reach $6,600 would be:
an= $1100(1.0725)^n
$6,600= $1100(1.0725)^n
$1,100(6)= $1100(1.0725)^n
6= (1.0725)^n
n= log1.075 6
n= 24.78
Answer:
7.722 cm^2
Step-by-step explanation:
Step 1: Find the area of the square.
- Area of square = s^2, where s = side
- 6^2 = 36 cm^2
Step 2: Find the area of the circle.
- We are given the diameter of the circle as 6 cm, so the radius must be 3 cm.
- Area of circle is , so now we plug in and simplify.
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Step 3: Subtract area of circle from area of square.
Therefore, the area of the shaded region is 7.722 cm^2.
Have a lovely rest of your day/night, and good luck with your assignments! ♡
Correct Options:- A) x-9 and E) x+6
Step by Step Solution:
Answer:
10.3125 (NOT ROUNDED)
Step-by-step explanation:
1. I changed it to decimals. 1/4 can easily be changed into a decimal.
2. I set 1/4 = x/100. Then cross multiplied. That would give you 4x = 100.
3. Now you would simplify that. x would be 25. So its 25/100 wich then turns to .25.
4. Now you are going to replace them so it would be 1.25 and 8.25.
5. To find out how much milk chocolate there was you would multiply 1.25 and 8.25.
6. That's your final answer
Answer:
f(-3) = -20
Step-by-step explanation:
We observe that the given x-values are 3 units apart, and that the x-value we're concerned with is also 3 units from the first of those given. So, a simple way to work this is to consider the sequence for x = 6, 3, 0, -3. The corresponding sequence of f(x) values is ...
34, 10, -8, ?
The first differences of these numbers are ...
10 -34 = -24
-8 -10 = -18
And the second difference is ...
-18 -(-24) = 6
For a quadratic function, second differences are constant. This means the next first-difference will be ...
? -(-8) = -18 +6
? = -12 -8 = -20
The value of the function at x=-3 is -20.
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The attachment shows using a graphing calculator to do a quadratic regression on the given points. The graph can then be used to find the point of interest. There are algebraic ways to do this, too, but they are somewhat more complicated than the 5 addition/subtraction operations we needed to find the solution. (Had the required x-value been different, we might have chosen a different approach.)