(>_ means greater than or equal to, and <_ means less than or equal to)
Answer:
9: Domain: 0<_x<_10
10: Range: 0<_y<_32
11: Domain: 0<_x<_∞ Range: 10<_y<_∞
12: Domain: 0<_x<_∞ Range: y=25
13: Domain: 0<_x<_15 Range: 0<_y<_15
Step-by-step explanation:
Just look at the highest and lowest points for the x axis and the y axis, and that's your domain and range respectively
Answer:
58.33333%
Step-by-step explanation:
Exam ≥ 80% Exam < 80%
Studied 4+ hours 18 6
Studied < 4 hours 10 14
The number of students who took the exam is 18+10+6+14 =48
The number of students who got an 80 or above is 18+10 = 28
The percent with 80 or above is 80 or above over total
28/48
.5833333
Answer:
12
Step-by-step explanation:
As there are 7 values in the given set, and one is missing, then the full set comprises 8 values.
2,7,8,14,15,20,29.
The middle value is 14 so we need another value, v such that, ½(14 + v) = 13.
This solves as v = 12.
The full set now is,2,7,8,12,14,15,20,29.
Answer:
2-[1, 2]
Step-by-step explanation:
From the graph, we can conclude the following things:
1. The graph is of degree 4 as it intersects the x axis at 4 points.
2. The graph tends to infinity for increasing the value of 'x' along positive or negative x-axis.
3. The graph has 3 turning points between the intervals [-1, 0], [1, 2] and [2, 3]
4. Local maximum: The top of mountain of the graph represents local maximum. So, during the interval [1, 2], there is a local maximum.
5. Local minimum: The lowest point or the valley of the graph represents local minimum.
So, during the intervals [-1, 0] and [2, 3], there are local minimums.
Thus, there is only one local maximum during the interval [1, 2].
Answer:
There are two pairs of solutions: (2,7) and (-1,4)
Step-by-step explanation:
We will use substitution.
y = x^2 + 3
y = x +5
Since the second equation is equal to y, replace y in the first equation with the second equation.
y = x^2 + 3
x + 5 = x^2 + 3
Rearrange so that one side is equal to 0.
5 - 3 = x^2 - x
2 = x^2 - x
0 = x^2 - x - 2
You may use quadratic formula or any form of factoring to find the zeros (x values that make the equation equal to 0).
a = 1, b = -1, c = -2
Zeros = 
and 
Zeros = 2 and -1
Now that you have your x values, plug them into the equations to find their corresponding y values.
y = x^2 + 3
y = (2)^2 + 3
y = 7
Pair #1: (2,7)
y = x^2 + 3
y = (-1)^2 + 3
y = 4
Pair #2: (-1,4)
Therefore, there are two pairs of solutions: (2,7) and (-1,4).
