This answer is 2/11
I used my brain lol !!!
Part A: x = -5/4, 3 || (-5/4, 0) (3, 0)
To find the x-intercepts, we need to know where y is equal to 0. So, we will set the function equal to 0 and solve for x.
4x^2 - 7x - 15 = 0
4 x 15 = 60 || -12 x 5 = 60 || -12 + 5 = -7
4x^2 - 12x + 5x - 15 = 0
4x(x - 3) + 5(x - 3) = 0
(4x + 5)(x - 3) = 0
4x + 5 = 0
x = -5/4
x - 3 = 0
x = 3
Part B: minimum, (7/8, -289/16)
The vertex of the graph will be a minimum. This is because the parabola is positive, meaning that it opens to the top.
To find the coordinates of the parabola, we start with the x-coordinate. The x-coordinate can be found using the equation -b/2a.
b = -7
a = 4
x = -(-7) / 2(4) = 7/8
Now that we know the x-value, we can plug it into the function and solve for the y-value.
y = 4(7/8)^2 - 7(7/8) - 15
y = 4(49/64) - 49/8 - 15
y = 196/64 - 392/64 - 960/64
y = -1156/64 = -289/16 = -18 1/16
Part C:
First, start by graphing the vertex. Then, use the x-intercepts and graph those. At this point we should have three points in a sort of triangle shape. If we did it right, each of the x-values will be an equal distance from the vertex. After we have those points graphed, it is time to draw in the parabola. Knowing that the parabola is positive, we draw in a U shape that passes through each of the three points and opens toward the top of the coordinate grid.
Hope this helps!
Answer: 1/9
Step-by-step explanation:
60 + 48 = 108 minutes
12 / 108 = 1/9
Answer:
Explanation:
There are only two mutually exclusive possibilities for each complex electronic system: being <em>defective</em> or being <em>not defective</em>:
- Number of complex electronic systems: 7
- Nmber of defective systems: 2
- Number of not defective systems: 5
<em>Probability </em>that no defective systems are found in the sample is the equal to the probability of selecting them consecutively without replacement and finding each one is not defective:
- Porbability the first is not defective × probability the second is not defective
- 5/7 × 4/6 = 20/42 = 10/21
Answer:
C. No. The sum of the dimensions of the eigenspaces equals nothing and the matrix has 3 columns. The sum of the dimensions of the eigenspace and the number of columns must be equal.
Step-by-step explanation:
Here the sum of dimensions of eigenspace is not equal to the number of columns, so therefore A is not diagonalizable.
