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3 years ago

PLEASE ANSWERRRR

Mathematics

1 answer:

QveST [7]3 years ago

8 0

The equation that gives the line is A. $y=10x+45.$

Step-by-step explanation:

Step 1:

First, we need to plot some points on the line on the graph. Some of the points are (1, 55), (3, 65), and (3, 75).

We need to find which of the given equations relating the x values to the values of y in the determined coordinates.

y is the dependent value whereas x is independent i.e. the y value changes according to the x value.

Step 2:

Now we substitute the values in the given equation to see which corresponds correctly.

For $y= 10x + 45$ , when x = 2, $y= 10(2) + 45 = 20 + 45 = 65.$

For $y= x + 45$ , when x = 2, $y= 2 + 45 = 2 + 45 = 47.$

For $y= -10x + 45$ , when x = 2, $y= -10(2) + 45 = -20 + 45 = 25.$

For $y= 45x + 10$ , when x = 2, $y= 45(2) + 10 = 90 + 10 = 100.$

Only the first equation gets a value of y, that corresponds to a coordinate on the line of the graph, so A. $y=10x+45$ represents the line of best fit.

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What is the length of the third side of the window frame below?

sergiy2304 [10]

Answer:

a=15

Step-by-step explanation:

Nice you can take your time now XD

c²-a²=b²

39²-36²=a²

1521-1296=a²

a²=225

a²=√225

a=15

Too easy! :D

6 0

4 years ago

[10 points] Given matrix A =  2 2 3, −6 −7 8 (a) (5 points). Show that A has no LU decomposition. (b) (5 points). Find the dec

ch4aika [34]

Answer:

Both the answers are as in the solution.

Step-by-step explanation:

As the given matrix is not in the readable form, a similar question is found online and the solution of which is attached herewith.

Part a:

Given matrix is : A = $\left[\begin{array}{ccc}0&3&4\\1&2&3\\-3&-7&8\end{array}\right]$

Here,

$det(A) =\left|\begin{array}{ccc}0&3&4\\1&2&3\\-3&-7&8\end{array}\right| = -55 \neq 0.$

Then, A is non-singular matrix.

Here, A₁₁= 0.

If we write A as LU with L lower triangular matrix and U upper triangular matrix, then A₁₁=L₁₁U₁₁.

So, As

A₁₁ = 0 gives L₁₁U₁₁= 0 ,

This indicates that either L₁₁= 0 or U₁₁ = 0.

If L₁₁= 0 or U₁₁ = 0, this would made the corresponding matrix singular, which contradicts the condition as A is non-singular.

Therefore, A has no LU decomposition.

Part b:

By the implementation of the various row operations

interchange R1 and R2

$\left[\begin{array}{ccc}1&2&3\\0&3&4\\-3&-7&8\end{array}\right]$

R3+3R1=R3

$\left[\begin{array}{ccc}1&2&3\\0&3&4\\0&-1&17\end{array}\right]$

R3+(1/3)R2 = R3

$\left[\begin{array}{ccc}1&2&3\\0&3&4\\0&0&55/3\end{array}\right]$

Therefore, U = $\left[\begin{array}{ccc}1&2&3\\0&3&4\\0&0&55/3\end{array}\right]$ .

Here, LP = E₁₂=E₃₁=-3 &E₃₂=-1/3

$LP=\left[\begin{array}{ccc}0&1&0\\1&0&0\\0&0&1\end{array}\right] \left[\begin{array}{ccc}1&0&0\\0&1&0\\-3&0&1\end{array}\right]\left[\begin{array}{ccc}1&0&0\\0&1&0\\0&-1/3&1\end{array}\right]$

$LP=\left[\begin{array}{ccc}0&1&0\\1&0&0\\-3&-1/3&1\end{array}\right]$

$LP=\left[\begin{array}{ccc}1&0&0\\0&1&0\\-1/3&-3&1\end{array}\right]\left[\begin{array}{ccc}0&1&0\\1&0&0\\0&0&1\end{array}\right]$

So now U is given as

$U=\left[\begin{array}{ccc}1&2&3\\0&3&4\\0&0&55/3\end{array}\right]\\L=\left[\begin{array}{ccc}1&0&0\\0&1&0\\-1/3&-3&1\end{array}\right]\\P=\left[\begin{array}{ccc}0&1&0\\1&0&0\\0&0&1\end{array}\right]\\$

4 0

3 years ago

Solve the system of linear equations.

sweet-ann [11.9K]

Answer:

dependent system
x = 2 -a
y = 1 +a
z = a

Step-by-step explanation:

Let's solve this by eliminating z, then we'll go from there.

Add 6 times the second equation to the first.

(3x -3y +6z) +6(x +2y -z) = (3) +6(4)

9x +9y = 27 . . . simplify

x + y = 3 . . . . . . divide by 9 [eq4]

Add 13 times the second equation to the third.

(5x -8y +13z) +13(x +2y -z) = (2) +13(4)

18x +18y = 54

x + y = 3 . . . . . . divide by 18 [eq5]

Equations [eq4] and [eq5] are identical. This tells us the system is dependent, and has an infinite number of solutions. We can find them in terms of z:

y = 3 -x . . . . solve eq5 for y

x +2(3 -x) -z = 4 . . . . substitute into the second equation

-x +6 -z = 4

x = 2 - z . . . . . . add x-4

y = 3 -(2 -z)

y = z +1

So far, we have written the solutions in terms of z. If we use the parameter "a", we can write the solutions as ...

x = 2 -a

y = 1 +a

z = a

_____

Check

First equation:

3(2-a) -3(a+1) +6a = 3

6 -3a -3a -3 +6a = 3 . . . true

Second equation:

(2-a) +2(a+1) -a = 4

2 -a +2a +2 -a = 4 . . . true

Third equation:

5(2-a) -8(a+1) +13a = 2

10 -5a -8a -8 +13a = 2 . . . true

Our solution checks algebraically.

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