The point B is located at (3, 0). Therefore option B is the correct answer.
On graph A, B, C, D and E points are given.
Graph is a mathematical representation of a network and it describes the relationship between lines and points. A graph consists of some points and lines between them. The length of the lines and position of the points do not matter.
The numbers on a coordinate grid are used to locate points. Each point can be identified by an ordered pair of numbers; that is, a number on the x-axis called an x-coordinate, and a number on the y-axis called a y-coordinate. Ordered pairs are written in parentheses (x-coordinate, y-coordinate).
Here, on the point B is located at (3, 0).
The point B is located at (3, 0). Therefore option B is the correct answer.
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Point B is located at (0, -3). When both x and y coordinates are positive,it lies in the first quadrant. If both are negative, it lies in the thirdquadrant. If x is negative and y is positive, it is in the second quadrant. If xis positive and y is negative, it is in the fourth quadrant.
what is x?
18
30
26
What is its area?
Answer: 49 feet squared
A square has four equal sides. It the perimeter is 28, then each side is 7 feet. The area is just the side length squared. 7 times 7 equals 49
The balance of Albert is $2159.07; the balance of Marie is $2244.99, the balance of Hans is $2188.35, and the balance of Max is $2147.40. Marie is $10,000 richer at the end of the competition.
Compound interest is defined as interest paid on the original principal and the interest earned on the interest of the principal.
To determine the balance of Albert’s $2000 after 10 years :
If the amount of $1000 at 1.2 % compounded monthly,
A = P(1 +r/n)ⁿ n = 10 years
here P = $1000 and r = 1.2
A = 1000(1 + 0.001)¹²⁰
A = $1127.43
If Albert $500 losing 2%
So 0.98 × 500 = $490
If $500 compounded continuously at 0.8%
So A = P
A = 500
A = 541.6
So the balance of Albert’s $2000 after 10 years :
Total balance = 1127.43 + 490.00+ 541.64 = $2159.07
To determine the balance of Marie’s $2000 after 10 years:
If 1500 at 1.4 % compounded quarterly,
A = 1500(1 + 0.0035)⁴⁰ = $1724.99
If $500 Marie’s gaining 4 %
So 1.04 × 500 = $520.00
So the balance of Marie’s $2000 after 10 years
Total balance = 1724.99 + 520.00 = $2244.99
To determine the balance of Hans’ $2000 after 10 years:
If $2000 compounded continuously at 0.9%
So A = 2000
A = $2188.3
To determine the balance of Max’s $2000 after 10 years :
If $1000 decreasing exponentially at 0.5 % annually
So A = 1000(1 - 0.005)¹⁰= $951.11
If $1000 at 1.8 % compounded bi-annually
So A = 1000(1 + 0.009)²⁰ = $1196.29
So the balance of Max’s $2000 after 10 years
Total balance = 951.11 + 1196.29 = $2147.40
Therefore, Marie is $10,000 richer at the end of the competition.
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Answer:
Step-by-step explanation:
Albert:
$1000 earned 1.2% annual interest compounded monthly
= 1000 (1+.001)120
(periodic interest = .012/12 ,n is periods = 10yr x 12 mos)
$500 lost 2% over the course of the 10 years
= 500 (.98)
$500 grew compounded continuously at rate of 0.8% annually
= 500 e^008(10) 10 years interest .008 (in decimal form)
Add these three to see how Albert did with his investments
Answer:
(3,2)
Step-by-step explanation:
Use the midpoint formula.
Add the x's together and then divide that by 2. Then do the same thing for the y's.
10+(-4)=6; 6/2=3
-2+6=4; 4/2=2
So, m=(3,2)