Answer: c/2-4
Step-by-step explanation:
This is what I get :
1/2(c-8)
Distributed property
1/2(c) + 1/2(-8)
Do c and -8 as a fraction with denominator 1 solve across
1/2×c/1+1/2×-8/1
Here you can only divide -8by 2
C/2-8/2
To get -4
C/2-4
You can't go further
Answer:
P(4.7≤x≤5.3)≈0.6827≈68.27%
Step-by-step explanation:
Notice that 4.7 is 1 standard deviation below the mean (μ) and 5.3 is 1 standard deviation above the mean. Thus, by the Empirical Rule, ±1σ represents about 68.27% of the data in a normal distribution, meaning that P(4.7≤x≤5.3)≈0.6827
Well, input and output are the functions so look up how to do functions with input and output and you will get the answer or look at this picture
There is a right triangle with the height (4) and half the base (3) as legs and the slant height (x) is the hypotenuse:
4² + 3²= x²
x² = 16+9
x =5 mm
Answer:
We have the matrix ![A=\left[\begin{array}{ccc}-4&-4&-4\\0&-8&-4\\0&8&4\end{array}\right]](https://tex.z-dn.net/?f=A%3D%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D-4%26-4%26-4%5C%5C0%26-8%26-4%5C%5C0%268%264%5Cend%7Barray%7D%5Cright%5D)
To find the eigenvalues of A we need find the zeros of the polynomial characteristic 
Then
![p(\lambda)=det(\left[\begin{array}{ccc}-4-\lambda&-4&-4\\0&-8-\lambda&-4\\0&8&4-\lambda\end{array}\right] )\\=(-4-\lambda)det(\left[\begin{array}{cc}-8-\lambda&-4\\8&4-\lambda\end{array}\right] )\\=(-4-\lambda)((-8-\lambda)(4-\lambda)+32)\\=-\lambda^3-8\lambda^2-16\lambda](https://tex.z-dn.net/?f=p%28%5Clambda%29%3Ddet%28%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D-4-%5Clambda%26-4%26-4%5C%5C0%26-8-%5Clambda%26-4%5C%5C0%268%264-%5Clambda%5Cend%7Barray%7D%5Cright%5D%20%29%5C%5C%3D%28-4-%5Clambda%29det%28%5Cleft%5B%5Cbegin%7Barray%7D%7Bcc%7D-8-%5Clambda%26-4%5C%5C8%264-%5Clambda%5Cend%7Barray%7D%5Cright%5D%20%29%5C%5C%3D%28-4-%5Clambda%29%28%28-8-%5Clambda%29%284-%5Clambda%29%2B32%29%5C%5C%3D-%5Clambda%5E3-8%5Clambda%5E2-16%5Clambda)
Now, we fin the zeros of
.

Then, the eigenvalues of A are
of multiplicity 1 and
of multiplicity 2.
Let's find the eigenspaces of A. For
:
.Then, we use row operations to find the echelon form of the matrix
![A=\left[\begin{array}{ccc}-4&-4&-4\\0&-8&-4\\0&8&4\end{array}\right]\rightarrow\left[\begin{array}{ccc}-4&-4&-4\\0&-8&-4\\0&0&0\end{array}\right]](https://tex.z-dn.net/?f=A%3D%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D-4%26-4%26-4%5C%5C0%26-8%26-4%5C%5C0%268%264%5Cend%7Barray%7D%5Cright%5D%5Crightarrow%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D-4%26-4%26-4%5C%5C0%26-8%26-4%5C%5C0%260%260%5Cend%7Barray%7D%5Cright%5D)
We use backward substitution and we obtain
1.

2.

Therefore,

For
:
.Then, we use row operations to find the echelon form of the matrix
![A+4I_3=\left[\begin{array}{ccc}0&-4&-4\\0&-4&-4\\0&8&8\end{array}\right] \rightarrow\left[\begin{array}{ccc}0&-4&-4\\0&0&0\\0&0&0\end{array}\right]](https://tex.z-dn.net/?f=A%2B4I_3%3D%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D0%26-4%26-4%5C%5C0%26-4%26-4%5C%5C0%268%268%5Cend%7Barray%7D%5Cright%5D%20%5Crightarrow%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D0%26-4%26-4%5C%5C0%260%260%5C%5C0%260%260%5Cend%7Barray%7D%5Cright%5D)
We use backward substitution and we obtain
1.

Then,
