You need to graph it so I guess
It seems that some the work is already here, but I'd be glad to!! So for #3 which is 9x^2+15x, we can factor out both a 3 and an x (3x) so we know that 3x * 3x =9x^2 and 3x * 5 = 15x so once we take the 3x out of the equation, we are left with 3x(3x+5) and that's as far as you can factor.
For #4, we see that the common factor is 10m because 10m * 2n = 20mn and 10m * 3 = 30m so once we take 10m out of the original, it becomes 10m(2n-3)
For #5, this one the common factor is 4xy because 4xy * 2xy=8x^2y^2 and 4xy*x= 4x^2y and 4xy*3=12xy so once we take the 4xy out of the equation, it becomes 4xy(2xy-x-3)
Hope this helps!
Answer:
The number lines are shown below.
Step-by-step explanation:
In the given problem the sign of inequality is missing.
We know, that there are 4 signs of inequality ">", "<", "≥" and "≤".
The possible inequalityes are
In , all points on the right side of 23 are included in the solution set.
In , all points on the left side of 23 are included in the solution set.
In , 23 and all points on the right side of 23 are included in the solution set.
In , 23 all points on the left side of 23 are included in the solution set.
Step-by-step explanation:
For the quadratic equation to have 1 repeated real solution, the discriminant b² - 4ac must be zero.
=> (-z)² - 4(z - 5)(5) = 0
=> z² - 20(z - 5) = 0
=> z² - 20z + 100 = 0
=> (z - 10)² = 0
Therefore z = 10.
The value of the year at the end of 2025 will be given by:
A=P(1+r/100)^n
where:
A=future value
P=principle
r=rate
n=number of terms
hence for the data given;
p=35000
R=5.5
n=(2025-2017)*2=16
Thus
A=35000(1+5.5/100)^16
A=$82, 434. 20
