Answer:
The answer is below
Step-by-step explanation:
Solve this system of equations by graphing. First graph the equations, and then type the solution. y=2x+2 y=x–2
Solution:
The equation of a straight line is in the form y = mx + b, where m is the slope of the graph and b is the y intercept (value of y when x = 0). To graph the line, online geogebra graphing calculator was used for plotting the two graphs by writing the equation of the graphs.
After plotting the two graphs, the intersection of the two lines should be traced to get the x coordinate and y coordinate of the point. This point is the solution of the two graph equations, hence:
The solution = (-4, -6)
Step-by-step explanation:
Answer:
17.106 Teaspoons of fertilizer
Step-by-step explanation:
Joni has a circular garden with a diameter of 16.5. If she uses 2 teaspoons of fertilizer for every 25 square feet of garden how many teaspoons of fertilizer will Joni need for her entire garden?
Step 1
We calculate the area of the garden.
The garden is circular in shape.
Area of a circle = πr²
r = radius = Diameter/2
Diameter = 16.5 ft
r = 16.5/2
r = 8.25 ft
Area of the circle = π × 8.25²
Area of the circle (Garden) = 213.82464998 ft²
Approximately = 212.825ft²
Step 2
If she uses 2 teaspoons of fertilizer for every 25 square feet of garden how many teaspoons of fertilizer will Joni need for her entire garden?
25 ft² = 2 teaspoons of fertilizer
212.825ft² = x
Cross Multiply
25 ft² × x = 212.825ft² × 2
x = 213.825× 2/25 ft²
x = 17.106 Teaspoons of fertilizer
Answer:
c
Step-by-step explanation:
Since in a pass code, the placement of the digits is
important, therefore this means that to solve for the total number of
possibilities we have to make use of the principle of Permutation. The formula
for calculating the total number of possibilities using Permutation is given
as:
P = n! / (n – r)!
where,
n = is the total amount of numbers to choose from = 20
r = is the total number of digits needed in the passcode =
4
Therefore solving for the total possibilities P:
P = 20! / (20 – 4)!
P = 20! / 16!
P = 116,280
<span>Hence there are a total of 116,280 possibilities of pass
codes.</span>
