The estimated population, if the rate is a growth rate is 34696 populations
The estimated population, if the rate is a decline rate is 5657 populations
<h3> Exponential functions</h3>
The standard exponential equation is expressed as ![y = ab^x](https://tex.z-dn.net/?f=y%20%3D%20ab%5Ex)
where;
- a is the intercept
- b is the growth rate
- x is the time taken
If the initial population is 15,000 with a rate of 15%, then;
a) If the rate is a growth rate, the estimated population will be expressed as;
![y = 15000(1+0.15)^{6}\\y=15000(1.15)^6\\y = 15000(2.313)\\y \approx 34,696](https://tex.z-dn.net/?f=y%20%3D%2015000%281%2B0.15%29%5E%7B6%7D%5C%5Cy%3D15000%281.15%29%5E6%5C%5Cy%20%3D%2015000%282.313%29%5C%5Cy%20%5Capprox%2034%2C696)
The estimated population, if the rate is a decline rate is 34696 populations
b) If the rate is a decline rate, the estimated population will be expressed as;
![y = 15000(1-0.15)^{6}\\y=15000(0.85)^6\\y = 15000(0.37715)\\y \approx 5,657](https://tex.z-dn.net/?f=y%20%3D%2015000%281-0.15%29%5E%7B6%7D%5C%5Cy%3D15000%280.85%29%5E6%5C%5Cy%20%3D%2015000%280.37715%29%5C%5Cy%20%5Capprox%205%2C657)
The estimated population, if the rate is a decline rate is 5657 populations
Learn more on exponential function here: brainly.com/question/12940982
The values of the function when x = 1 are y = 1 and y = 1
<h3>How to choose one value that is in the domain of both y(x) = x^2 and y^2 = x?</h3>
The functions are given as:
y(x) = x^2
y^2 = x
Also, the graphs of the functions are given.
From the given graph, we have:
- The domain of y(x) = x^2 is the set of all real numbers
- The domain of y^2 = x is the set of all real numbers greater than or equal to 0
The common numbers in both domain are represented by x >= 0
An example of such number is
x = 1
So, we have:
y(1) = 1^2 ⇒ y(1) = 1
y^2 = 1 ⇒ y = 1
Hence, the values of the function when x = 1 are y = 1 and y = 1
Read more about domain and range at:
brainly.com/question/2264373
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Answer: 17
Step-by-step explanation: There are 34 squares in the blue figure each one is 0.5 cm. So what we do is multiply 0.5 x 34 which gives us the answer of 17.
Answer:
1.81 inch by 8.38 inch by 6.38 inch
Smallest Value=1.81 inch
Largest Value=8.38 inch
Step-by-step explanation:
The cardboard is 12 in. long and 10 in. wide
Let the length of the square cut off=x
Length of the box=12-2x
Width of the box=10-2x
Height of the box=x
Volume of the box=lwh
V=x(12-2x)(10-2x)
The dimensions of the box that will yield maximum volume occurs at the point where the derivative of V=0.
![V^{'}=4\,\left( 30 - 22\,x + 3\,x^{2}\right)](https://tex.z-dn.net/?f=%20V%5E%7B%27%7D%3D4%5C%2C%5Cleft%28%2030%20-%2022%5C%2Cx%20%2B%203%5C%2Cx%5E%7B2%7D%5Cright%29%20)
Thus:
4(30-22x+3x²)=0
Since 4≠0
3x²-22x+30=0
Solving for x using a calculator gives:
x=1.81 or x=5.52
x cannot be 5.52 inch since the width is 10 inch and removing 2(5.52) from the width gives a negative result.
When x=1.81 inch
Length of the box=12-2x=12-2(1.81)=8.38inch
Width of the box=10-2x=10-2(1.81)=6.38 inch
Therefore, the dimensions at which the Volume is maximum are: 1.81 inch by 8.38 inch by 6.38 inch