Answer:
Below in bold.
Step-by-step explanation:
No repetition.
There are a total of 7 numbers.
For the number to be even it must end in 0, 2 or 8.
The other 3 numbers will be the number of permutations of 3 from the 6 other numbers.
Number of 4 digit numbers ending in 0
= 6P3 = 6! / 6-3!
= 120.
Now the number could also end in 2 or 8,
so there are 3*120 = 360 4-digit even numbers (no repetition).
With repetition.
There are 3 * 7^3 = 1029 with repetition.
Answer:
x=6
Step-by-step explanation:
1/2x+17=20
lets get rid of the 17 first by subtracting
1/2x=3
we have 1/2 x so to get a full x multiply both sides by 2
x=6
Answer:
y=17/3x+2
Step-by-step explanation:
Get 2 points from the graph
(3;19) (-10;2)
Grad=change in y/change in x
=2-19/-10-3
=-17/-13
=17/3
y=mx+c
19=17/13(3)+c
19=17+c
19-17=c
2=c
y=17/13x+2.
Well I think you are supposed to find the equation of the line so I think that's it.
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 .
Hello!
It appears that this is creating a triangle. It gives us leg lengths of 3 and 5. If we assume that this is a right triangle, we can use the Pythagorean Theorem. We add up the square of each of our numbers. Then we find the square root of our answer. Basically, a²+b²=c².
25+9=34
√34≈5.83
Therefore, the eagle is about 5.83 miles from its nest.
I hope this helps!
