Answer:
Therefore,
Step-by-step explanation:
Given:
![A=\left[\begin{array}{ccc}3&6&9\\2&4&8\\\end{array}\right]](https://tex.z-dn.net/?f=A%3D%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D3%266%269%5C%5C2%264%268%5C%5C%5Cend%7Barray%7D%5Cright%5D)
To Find:
a₂₁ = ?
Solution:
Let,
![A=\left[\begin{array}{ccc}a_{11}&a_{12}&a_{13}\\a_{21}&a_{22}&a_{23}\\\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%26a_%7B23%7D%5C%5C%5Cend%7Barray%7D%5Cright%5D)
We require ' a₂₁ ' i.e Second Row First Column Element
So on Comparing we get
∴
Therefore,
Answer:
1= 102°,3=78°,4=102°
5=102°,6=78°,7=78°,8=102°
Step-by-step explanation:
Answer:
The general plan is to find BM and from that CM. You need 2 equations to do that.
Step One
Set up the two equations.
(7 - BM)^2 + CM^2 = (4*sqrt(2) ) ^ 2 = 32
BM^2 + CM^2 = 5^2 = 25
Step Two
Subtract the two equations.
(7 - BM)^2 + CM^2 = 32
BM^2 + CM^2 = 25
(7 - BM)^2 - BM^2 = 7 (3)
Step three
Expand the left side of the new equation labeled (3)
49 - 14BM + BM^2 - BM^2 = 7
Step 4
Simplify And Solve
49 - 14BM = 7 Subtract 49 from both sides.
-49 - 14BM = 7 - 49
- 14BM = - 42 Divide by - 14
BM = -42 / - 14
BM = 3
Step Five
Find CM
CM^2 + BM^2 = 5^2
CM^2 + 3^2 = 5^2 Subtract 3^2 from both sides.
CM^2 = 25 - 9
CM^2 = 16 Take the square root of both sides.
sqrt(CM^2) = sqrt(16)
CM = 4 < Answer
Step-by-step explanation:
How do linear, quadratic, and exponential functions compare?
Answer:
How can all the solutions to an equation in two variables be represented?
<u><em>The solution to a system of linear equations in two variables is any ordered pair x,y which satisfies each equation independently. U can Graph, solutions are points at which the lines intersect.</em></u>
<u><em /></u>
<u><em>How can all the solutions to an equation in two variables be represented?</em></u>
<u><em>you can solve it by Iterative method and Newton Raphson's method.</em></u>
<u><em /></u>
<u><em>How are solutions to a system of nonlinear equations found?
</em></u>
Solve the linear equation for one variable.
Substitute the value of the variable into the nonlinear equation.
Solve the nonlinear equation for the variable.
Substitute the solution(s) into either equation to solve for the other variable.
<u><em>
</em></u>
<u><em>How can solutions to a system of nonlinear equations be approximated? U can find the solutions to a system of nonlinear equations by finding the points of intersection. The points of intersection give us an x value and a y value. Using the example system of nonlinear equations, let's look at how u can find approximate solutions.</em></u>
Well the equation is (A^2)+(B^2)=(C^2)
A= x
B= 2x
C will always be the longest side because it is the Hypotenuse = 25
So if you plug in those numbers into the equation...
(x^2) + (2x^2) = (25^2)
x^2 + 4x^2 = 625
Combine like terms
5x^2 = 625
Divide by five to both sides
x^2 = 125
Then Square root,
x = sqrt(125)
x = sqrt(25* 5)
x = 5sqrt(5)
