Answer:
all real numbers
Step-by-step explanation:
The function f(x) = 3|x +4| +1 is defined for all values of x. Its domain is <em>all real numbers</em>.
P = a + b + c
p = 2400
a = longest = 2c - 200
b = middle = a - 200 = 2c - 200 - 200 = 2c - 400
2400 = (2c - 200) + (2c - 400) + c
2400 = 5c - 600
2400 + 600 = 5c
3000 = 5c
3000/5 = c
600 = c <== shortest side
a = 2c - 200.....= 2(600) - 200 = 1200 - 200 = 1000<== longest side
b = 2c - 400...= 2(600) - 400 = 1200 - 400 = 800 <== middle side
Answer:
One of the obvious non-trivial solutions is
.
Step-by-step explanation:
Suppose the matrix A is as follows:
![A=\left[\begin{array}{ccc}a_{11}&a_{12}&a_{13}\\a_{21}&a_{22}&3_{23}\\a_{31}&a_{32}&a_{33}\end{array}\right]](https://tex.z-dn.net/?f=A%3D%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7Da_%7B11%7D%26a_%7B12%7D%26a_%7B13%7D%5C%5Ca_%7B21%7D%26a_%7B22%7D%263_%7B23%7D%5C%5Ca_%7B31%7D%26a_%7B32%7D%26a_%7B33%7D%5Cend%7Barray%7D%5Cright%5D)
The observed system
after multiplying looks like this
![Ax=0 \iff \left[\begin{array}{ccc}a_{11}&a_{12}&a_{13}\\a_{21}&a_{22}&a_{23}\\a_{31}&a_{32}&a_{33}\end{array}\right] \cdot \left[\begin{array}{ccc}x_1\\x_2\\x_3\end{array}\right] =0 \iff \\ \\a_{11}x_1+a_{12}x_2+a_{13}x_3=0\\a_{21}x_1+a_{22}x_2+a_{23}x_3=0\\a_{31}x_1+a_{32}x_2+a_{33}x_3=0\\\\](https://tex.z-dn.net/?f=Ax%3D0%20%5Ciff%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7Da_%7B11%7D%26a_%7B12%7D%26a_%7B13%7D%5C%5Ca_%7B21%7D%26a_%7B22%7D%26a_%7B23%7D%5C%5Ca_%7B31%7D%26a_%7B32%7D%26a_%7B33%7D%5Cend%7Barray%7D%5Cright%5D%20%5Ccdot%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7Dx_1%5C%5Cx_2%5C%5Cx_3%5Cend%7Barray%7D%5Cright%5D%20%3D0%20%5Ciff%20%5C%5C%20%5C%5Ca_%7B11%7Dx_1%2Ba_%7B12%7Dx_2%2Ba_%7B13%7Dx_3%3D0%5C%5Ca_%7B21%7Dx_1%2Ba_%7B22%7Dx_2%2Ba_%7B23%7Dx_3%3D0%5C%5Ca_%7B31%7Dx_1%2Ba_%7B32%7Dx_2%2Ba_%7B33%7Dx_3%3D0%5C%5C%5C%5C)
Since we now that
, where
are the columns of the matrix A, we actually know this:
![-2\cdot \left[\begin{array}{ccc}a_{11}\\a_{21}\\a_{31}\end{array}\right] +3\cdot \left[\begin{array}{ccc}a_{12}\\a_{22}\\a_{32}\end{array}\right] -5\cdot \left[\begin{array}{ccc}a_{13}\\a_{23}\\a_{33}\end{array}\right] =\left[\begin{array}{ccc}0\\0\\0\end{array}\right]](https://tex.z-dn.net/?f=-2%5Ccdot%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7Da_%7B11%7D%5C%5Ca_%7B21%7D%5C%5Ca_%7B31%7D%5Cend%7Barray%7D%5Cright%5D%20%2B3%5Ccdot%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7Da_%7B12%7D%5C%5Ca_%7B22%7D%5C%5Ca_%7B32%7D%5Cend%7Barray%7D%5Cright%5D%20-5%5Ccdot%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7Da_%7B13%7D%5C%5Ca_%7B23%7D%5C%5Ca_%7B33%7D%5Cend%7Barray%7D%5Cright%5D%20%3D%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D0%5C%5C0%5C%5C0%5Cend%7Barray%7D%5Cright%5D)
Once we multiply and sum up these 3 by 1 matrices, we get that these equations hold:

This actually means that the solution to the previously observed system of equations (or equivalently, our system
) has a non-trivial solution
.
Answer:
Hello some parts of your question is missing below is the missing part
c. If you randomly select a navel orange, what is the probability that it weighs between6.2 and 7 ounces
Answer: A) 0.0099
B) 0.6796
C) 0.13956
Step-by-step explanation:
weight of Navel oranges evenly distributed
mean ( u ) = 8 ounces
std ( б )= 1.5
navel oranges = X
A ) percentage of oranges weighing more than 11.5 ounces
P( x > 11.5 ) = 
= P ( Z > 2.33 ) = 0.0099
= 0.9%
B) percentage of oranges weighing less than 8.7 ounces
P( x < 8.7 ) = 
= P ( Z < 0.4667 ) = 0.6796
= 67.96%
C ) probability of orange selected weighing between 6.2 and 7 ounces?
P ( 6.2 < X < 7 ) = 
= P ( -1.2 < Z < -0.66 )
= Ф ( -0.66 ) - Ф(-1.2) = 0.13956
The formula for the area of a circle is pi times the radius squared. Here are the steps to solving this problem.
1. First, find the radius. Since the radius is allows half of the diameter, then we divide the diameter by two. 50/2= 25
2. Next, we square the radius. 25^2= 625
3. After that we multiply 625 ft by pi(3.14).
625 times 3.14= 1962.5
4. The last step is to round the the answer to the nearest square foot. Since 5 is at the end, we round up to 1,963.
So the answer is 1,963 ft^2.