Answer:
y = -1/3x + 5
Step-by-step explanation:
-1/3 is the slope
5 is the y-intercept
Answer:
It is C. with love, with patience and with faith
she'll make her way
Step-by-step explanation:
Because i got it right on edge
Step-by-step explanation:
-2×4/5
-8/5
= -1.6
That's the answer
Answer:
6.7
General Formulas and Concepts:
<u>Pre-Algebra</u>
Order of Operations: BPEMDAS
- Brackets
- Parenthesis
- Exponents
- Multiplication
- Division
- Addition
- Subtraction
Equality Properties
<u>Trigonometry</u>
- [Right Triangles Only] Pythagorean Theorem: a² + b² = c²
Step-by-step explanation:
<u>Step 1: Identify Variables</u>
Leg <em>a</em> = <em>x</em>
Leg <em>b</em> = 2
Hypotenuse <em>c</em> = 7
<u>Step 2: Solve for </u><em><u>x</u></em>
- Substitute [PT]: x² + 2² = 7²
- Isolate <em>x </em>term: x² = 7² - 2²
- Exponents: x² = 49 - 4
- Subtract: x² = 45
- Isolate <em>x</em>: x = 3√5
- Evaluate: x = 6.7082
- Round: x ≈ 6.7
Answer with Step-by-step explanation:
We are given that two matrices A and B are square matrices of the same size.
We have to prove that
Tr(C(A+B)=C(Tr(A)+Tr(B))
Where C is constant
We know that tr A=Sum of diagonal elements of A
Therefore,
Tr(A)=Sum of diagonal elements of A
Tr(B)=Sum of diagonal elements of B
C(Tr(A))=
Sum of diagonal elements of A
C(Tr(B))= Sum of diagonal elements of B
Tr(C(A+B)=Sum of diagonal elements of (C(A+B))
Suppose ,A=![\left[\begin{array}{ccc}1&0\\1&1\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D1%260%5C%5C1%261%5Cend%7Barray%7D%5Cright%5D)
B=![\left[\begin{array}{ccc}1&1\\1&1\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D1%261%5C%5C1%261%5Cend%7Barray%7D%5Cright%5D)
Tr(A)=1+1=2
Tr(B)=1+1=2
C(Tr(A)+Tr(B))=C(2+2)=4C
A+B=![\left[\begin{array}{ccc}1&0\\1&1\end{array}\right]+\left[\begin{array}{ccc}1&1\\1&1\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D1%260%5C%5C1%261%5Cend%7Barray%7D%5Cright%5D%2B%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D1%261%5C%5C1%261%5Cend%7Barray%7D%5Cright%5D)
A+B=![\left[\begin{array}{ccc}2&1\\2&2\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D2%261%5C%5C2%262%5Cend%7Barray%7D%5Cright%5D)
C(A+B)=![\left[\begin{array}{ccc}2C&C\\2C&2C\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D2C%26C%5C%5C2C%262C%5Cend%7Barray%7D%5Cright%5D)
Tr(C(A+B))=2C+2C=4C
Hence, Tr(C(A+B)=C(Tr(A)+Tr(B))
Hence, proved.
