Answer:
Step-by-step explanation:
we have
we know that
The radicand must be greater than or equal to zero
so
Solve for x
Adds 4 both sides
Divide by 2 both sides
so
The domain is the interval for x ----> [2,∞)
All real numbers greater than or equal to 2
Annually The amount after 10 years = $ 7247.295
quarterly compound after 10 years = $7393.5
Continuously interest =$7,419
Given:
P = the principal amount
r = rate of interest
t = time in years
n = number of times the amount is compounding.
Principal = $4500
time= 10 year
Rate = 5%
To find: The amount after 10 years.
The principal amount is, P = $4500
The rate of interest is, r = 5% =5/100 = 0.05.
The time in years is, t = 10.
Using the quarterly compound interest formula:
A = P (1 + r / 4)4 t
A= 4500(1+.05/4)40
A= 4500(4.05/4)40
A= 4500(1.643)
Answer: The amount after 10 years = $7393.5
Using the Annually compound interest formula:
A = P (1 + r / 100) t
A= 4500(1+5/100)10
A= 4500(105/100)10
Answer: The amount after 10 years = $ 7247.295
Using the Continuously compound interest formula:
e stands for Napier’s number, which is approximately 2.7183
A= $2,919
Answer: The amount after 10 years = $4500+$2,919=$7,419
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Answer:
Area: 3.7 ft²
Ratio: 73.14 : 1
Step-by-step explanation:
Perimeter of an octagon = 8*side
Replacing with perimeter = 7 ft:
7 = 8*side
side = 7/8 ft = 0.875 ft
that is, each side of the model is 7/8 ft length.
Area of an octagon = 2*(1 + √2)*side²
Area of an octagon = 2*(1 + √2)*(7/8)²
Area of an octagon = 3.7 ft²
Perimeter of real gazebo = 8*8 = 64 ft
Then, the ratio of the perimeters (in feet) of the real gazebo floor to the model gazebo floor is 64:0.875. Multiplying each term by 8/7, we get 73.14:1
5
3y=15-4x
y=5-4/3x
so the answer is 5
The answer is B.
The absolute value is the "positive version" of a number. So, the absolute value does nothing to positive numbers, and switches the sign of negative numbers. Here are some examples:
So, as you can see, the result of an absolute value can't be negative, and as such, it can't be less than a negative number.