Answer:
Step-by-step explanation:
Given
5 tuples implies that:
implies that:
Required
How many 5-tuples of integers are there such that
From the question, the order of the integers h, i, j, k and m does not matter. This implies that, we make use of combination to solve this problem.
Also considering that repetition is allowed: This implies that, a number can be repeated in more than 1 location
So, there are n + 4 items to make selection from
The selection becomes:
Expand the numerator
<u><em>Solved</em></u>
Answer:
60 %
Step-by-step explanation:
multiply .90 * 150
X= 41
-(12)/(1+5x)=-(3)/(x-10)
Multiply the numerator of the first fraction by the denominator of the second fraction. Set this equal to the product of the denominator of the first fraction and the numerator of the second fraction.
−12⋅(x−10)=(1+5x)•-3
-12⋅(x-10)=(1+5x)•-3
Solve the equation for x.
Simplify -12• (x-10)
-12x-12•-10=(1+5x) •-3
Multiply
-12 by -10
-12x+120=(1•5x)•-3
Simplify (1+5x)•-3
Apply the distributive property.
− 12x + 120 = 1• -3 + 5x •-3
Multiply
-12x+120=1⋅-3+5x⋅-3
Multiply -3 by 1
-12x+ 120=-3+5x•-3
Multiply -3 by 5
-12x +120=-3-15x
Move all terms containing x
to the left side of the equation.
Add 15x to both sides
-12x+120+15x=-3
Add
3x+120=-3
Move all terms not containing x
to the right side of the equation.
Subtract 120 from -3
3x=-123
Divide each term by 3 and simplify
Divide each term in 3x = -123 by 3.
3x / 3 = -123 / 3
Cancel the common factor of 3.
-123/3 divid by 1
X = -123/3
Divid -123/3 = 41
So X= 41
A.The first Quadrant
B.The fourth Quadrant and the coordinates are (5,-3)
C. The third Quadrant and the coordinates are (-5,-3)