Answer:
12.73m
Step-by-step explanation:
using Pythagoras theorem
a = x (i change x to a for easy reference due to multiplication)
18 x 18 = a x a + a x a
324 = 2a^2
a^2 = 324 / 2= 162
a =
= 12.7279 = 12.73 (2 decimal place)
We will solve the problem for both cases:
Case 1:
If the amount increases by 7% we have the number for which we are going to multiply is given by:


Case 2:
If the amount decreases by 7% we have the number for which we are going to multiply is given by:


Answer:
single multiplier you would use for each case is:
x = 1.07 (increase)
x = 0.93 (decrease)
Answer:
(-1,5)
Step-by-step explanation:
To find this, you have to add both "x" coordinates, then divide them by two.
Same thing goes for the "y" coordinates, add them, then divide them by two.
x-coordinates ---> 2+(-4)= -2
y-coordinates ---> 1+9= 10
-2÷2= -1
10÷2= 5
That gives you, (-1,5).
Answer:
Domain → (-∞, ∞)
Range → (-2, ∞)
Step-by-step explanation:
1). Domain of a function is defined by the input values (set of x-values) and Range of the function is defined by the output values (set of y-values).
From the graph attached,
Function is defined for all real values of x.
Therefore, domain of the function will be (-∞, ∞).
On y-axis values of the function vary from -2 (excluding 2) to positive infinity.
Therefore, range of the function will be (-2, ∞).
2). Average rate of change of a function in the interval (a, b) is defined by,
Average rate of change = 
By using this expression we can find the average rate of change in the given interval.
Please give the correct interval for which the average rate of change is to be calculated.
Answer:
A. Infinitely many solutions.
Step-by-step explanation:
The row and column numbers are equal in the echelon. This rows have 0001 numbers which indicates that there is free variable at the end. The reduced row echelon has equal number of rows and columns. There are infinitely many solution as the numbers in the rows are zero ending of the free variable 1. If there is no free variable then there will be no solution.