1) Vertex: (0,2)
AOS x=0
table : x = -1, 0, 1
x^2+2 = 3, 2, 3
(x,y) = (-1,3), (0,2), (1,3)
2) vertex: (-5, -1)
AOS x= -5
table : x = -7, -6, -5, -4, -3
x^2+10x+24 = 3, 0, -1, 0, 3
(x,y) = (-7,3), (-6,0), (-5,-1), (4,0), (-3,3)
3) vertex: (4, -1)
AOS x=4
table : x = 2, 3, 4, 5, 6
y = 3, 0, -1, 0, 3
(x,y) = (2, 3), (3, 0), (4, -1), (5, 0), (6, 3)
if the picture didn’t attach along with this then just use the points i gave (x,y) to plot a graph. always remember that whenever your equation has x^2 in it, it is a parabola. and do not forget to write down “y=“. my teacher takes off a lot of points if i don’t. have a great day!
Answer: 8(8+x)=80
Step-by-step explanation: 8 times is shown in parenthesis, 8+x is the sum, and 80 is the answer
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Add sides 2

Simplification

Divide sides by 9




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Answer:
The balance in her lunch account at the end of the day on Friday is -$1.15.
Step-by-step explanation:
The balance at the end of the week will be the result of adding up the initial balance with the amount deposited on Monday minus the total amount spent that week.
Initial balance=$2.60
Amount deposited on Monday=$20
Total amount spent that week=$4.75*5=$23.75
Balance=$2.60+$20-$23.75
Balance=$22.60-$23.75
Balance=-$1.15
According to this, the answer is that the balance in her lunch account at the end of the day on Friday is -$1.15.
Answer:
The ball is at a height of 120 feet after 1.8 and 4.1 seconds.
Step-by-step explanation:
The statement is incomplete. The complete statement is perhaps the following "A ball is thrown vertically in the air with a velocity of 95 ft/s. What time in seconds is the ball at a height of 120ft. Round to the nearest tenth of a second."
Since the ball is launched upwards, gravity decelerates it up to rest and moves downwards. The position of the ball can be determined as a function of time by using this expression:

Where:
- Initial height of the ball, measured in feet.
- Initial speed of the ball, measured in feet per second.
- Gravitational constant, equal to
.
- Time, measured in seconds.
Given that
,
,
and
, the following second-order polynomial is found:


The roots of this polynomial are, respectively:
and
.
Both roots solutions are physically reasonable, since
represents the instant when the ball reaches a height of 120 ft before reaching maximum height, whereas
represents the instant when the ball the same height after reaching maximum height.
In nutshell, the ball is at a height of 120 feet after 1.8 and 4.1 seconds.