Answer:

------------------------------------
Add up the corresponding elements
Row1,Column1: 6.1+(-7.2) = -1.1
Row1,Column2: 1.9+7.6 = 9.5
Row2,Column1: -7.5+1.8 = -5.7
Row2,Column2: 2.2+(-1.4) = 0.8
Row3,Column1: 3.8+2.3 = 6.1
Row3,Column2: -1.4+2.1 = 0.7
So that's why the answer is
<span>Let x equal your repeating decimal. Call this equation x = 1.273
</span><span>Multiply both sides of your previous expressions by 100. Call this equation 100x = 127.273
</span><span>Subtract 1 from both sides of the equation to get 99x = 126.
</span><span>Divide both sides of the equation by 99 to get 126/99.
</span><span>126/99 reduced is to its lowest terms is 1 and 3/11.
</span>
Answer:
(a) The maximum height of the ball is 16 feet at 1 seconds.
(b) The ball hits the ground at 2 seconds.
PART (A) :
Given function: h(t) = -16t² + 32t
Comparing to quadratic function: ax² + bx + c
In this function: a = -16, b = 32, c = 0
<u>To find the maximum height</u>, use the vertex formula:

<u>Insert values</u>

Then find h(t) = -16(1)² + 32(1) = 16 feet
Conclusion: The maximum height of the ball is 16 feet at 1 seconds.
PART (B) :
When the ball hits the ground, the height [h(t)] will be 0 ft
-16t² + 32t = 0
-16t(t - 2) = 0
-16t = 0, t - 2 = 0
t = 0, t = 2
The ball hits the ground at 2 seconds. Note: the other 0 seconds is for when the ball was launched at the beginning.
Answer:
See explanation
Step-by-step explanation:
You have to graph the system of inequalities presented as

Part A:
1. Draw a dotted line
(dotted because the sign < is without notion "or equal to"). Select one of two regions by substituting the coordinates of origin into inequality:

Since this inequlity is false, the origin doesn't belong to the shaded region, so you have to shade that part which doesn't contain origin (red part in attached diagram).
2. Draw a solid line
(solid because the sign ≥ is with notion "or equal to"). Select one of two regions by substituting the coordinates of origin into inequality:

Since this inequlity is true, the origin belongs to the shaded region, so you have to shade that part which contains origin (blue part in attached diagram).
The intersection of these two regions is the solution area.
Part B:
Plot point (-2,-2). Since this point doesn't belong to the solution area, this is not a solution of the system of two inequalities. You can check it mathematically - substitute x=-2 and y=-2 into the system:

Both inequalities are false, so (-2,-2) doesn't belong to the solution area.