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4 years ago

Solve using matrices. Show your work. x+y=5 2x-y=1

1 answer:

4 0

To solve using matrices, we want to get an identity matrix in the variable columns and see what we get on the far right columns
The coefficients for the problem in matrix form:
1 1 5
2 -1 1
Add -2(row 1) to row 2:
1 1 5
0 -3 -9
Add (1/3) row 2 to row 1
1 0 2
0 -3 -9
Divide Row 2 by -3
1 0 2
0 1 3
We now have an identity matrix inf the first two columns. Row 1 gives us the value of x, and row 2 gives us the value of y:
x = 2, y = 3

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HELP ASAP (Geometry)

Andrei [34K]

1) Parallel line: $y=-2x-3$

2) Rectangle

3) Perpendicular line: $y = 0.5x + 2.5$

4) x-coordinate: 2.7

5) Distance: $d=\sqrt{(4-3)^2+(7-1)^2}$

6) 3/8

7) Perimeter: 12.4 units

8) Area: 8 square units

9) Two slopes of triangle ABC are opposite reciprocals

10) Perpendicular line: $y-5=-4(x-(-1))$

Step-by-step explanation:

The equation of a line is in the form

$y=mx+q$

where m is the slope and q is the y-intercept.

Two lines are parallel to each other if they have same slope m.

The line given in this problem is

$y=-2x+7$

So its slope is $m=-2$ . Therefore, the only line parallel to this one is the line which have the same slope, which is:

$y=-2x-3$

Since it also has $m=-2$

We can verify that this is a rectangle by checking that the two diagonals are congruent. We have:

- First diagonal: $d_1 = \sqrt{(-3-(-1))^2+(4-(-2))^2}=\sqrt{(-2)^2+(6)^2}=6.32$

- Second diagonal: $d_2 = \sqrt{(1-(-5))^2+(0-2)^2}=\sqrt{6^2+(-2)^2}=6.32$

The diagonals are congruent, so this is a rectangle.

Given points A (0,1) and B (-2,5), the slope of the line is:

$m=\frac{5-1}{-2-0}=-2$

The slope of a line perpendicular to AB is equal to the inverse reciprocal of the slope of AB, so:

$m'=\frac{1}{2}$

And using the slope-intercept for,

$y-y_0 = m(x-x_0)$

Using the point $(x_0,y_0)=(7,1)$ we find:

$y-1=\frac{1}{2}(x-7)$

And re-arranging,

$y-1 = \frac{1}{2}x-\frac{7}{2}\\y=\frac{1}{2}x-\frac{5}{2}\\y=0.5x-2.5$

The endpoints of the segment are X(1,2) and Y(6,7).

We have to divide the sgment into 1/3 and 2/3 parts from X to Y, so for the x-coordinate we get:

$x' = x_0 + \frac{1}{3}(x_1 - x_0) = 1+\frac{1}{3}(6-1)=2.7$

The distance between two points $A(x_A,y_A)$ and $B(x_B,y_B)$ is given by

$d=\sqrt{(x_B-x_A)^2+(y_B-y_A)^2}$

In this problem, the two points are

E(3,1)

F(4,7)

So the distance is given by

$d=\sqrt{(4-3)^2+(7-1)^2}$

We have:

A(3,4)

B(11,3)

Point C divides the segment into two parts with 3:5 ratio.

The distance between the x-coordinates of A and B is 8 units: this means that the x-coordinate of C falls 3 units to the right of the x-coordinate of A and 5 units to the left of the x-coordinate of B, so overall, the x-coordinate of C falls at

$\frac{3}{3+5}=\frac{3}{8}$

of the distance between A and B.

To find the perimeter, we have to calculate the length of each side:

$d_{EF}=\sqrt{(x_E-x_F)^2+(y_E-y_F)^2}=\sqrt{(-1-2)^2+(6-4)^2}=3.6$

$d_{FG}=\sqrt{(x_G-x_F)^2+(y_G-y_F)^2}=\sqrt{(-1-2)^2+(3-4)^2}=3.2$

$d_{GH}=\sqrt{(x_G-x_H)^2+(y_G-y_H)^2}=\sqrt{(-1-(-3))^2+(3-3)^2}=2$

$d_{EH}=\sqrt{(x_E-x_H)^2+(y_E-y_H)^2}=\sqrt{(-1-(-3))^2+(6-3)^2}=3.6$

So the perimeter is

p = 3.6 + 3.2 + 2 + 3.6 = 12.4

The area of a triangle is

$A=\frac{1}{2}(base)(height)$

For this triangle,

Base = XW

Height = YZ

We calculate the length of the base and of the height:

$Base =XW=\sqrt{(x_X-x_W)^2+(y_X-y_W)^2}=\sqrt{(6-2)^2+(3-(-1))^2}=5.7$

$Height =YZ=\sqrt{(x_Y-x_Z)^2+(y_Y-y_Z)^2}=\sqrt{(7-5)^2+(0-2)^2}=2.8$

So the area is

$A=\frac{1}{2}(XW)(YZ)=\frac{1}{2}(5.7)(2.8)=8$

A triangle is a right triangle when there is one right angle. This means that two sides of the triangle are perpendicular to each other: however, two lines are perpendicular when their slopes are opposite reciprocals. Therefore, this means that the true statement is

"Two slopes of triangle ABC are opposite reciprocals"

10)

The initial line is

$y=\frac{1}{4}x-6$

A line perpendicular to this one must have a slope which is the opposite reciprocal, so

$m'=-4$

Using the slope-intercept form,

$y-y_0 = m'(x-x_0)$

And using the point

$(x_0,y_0)=(-1,5)$

we find:

$y-5=-4(x-(-1))$

Learn more about parallel and perpendicular lines:

brainly.com/question/3414323

brainly.com/question/3569195

#LearnwithBrainly

8 0

3 years ago

A square and a rectangle have the same perimeter. Find the side lengths of each figure.

aliina [53]

X + 1 + X + 1 + X +1+ X +1 = 2X -3+ 2X -3+ X -1+ X -1
4x + 4 = 4x-6+2x-2
4x+4=6x-8
12=2x
6=x

Side lengths:
6-1=5
6+1=7
2(6)-3=9

Hope that helps! Stay safe!

4 0

3 years ago

Read 2 more answers

What is the answer of −7x+1=−20

Pie

Answer:

Step-by-step explanation:

3 0

4 years ago

Read 2 more answers

Simplify 0.95x-9-1/2x(36-18y)-0.25(-3x+4.8)

iris [78.8K]

Answer: 9xy−16.3x−10.2

Step-by-step explanation:

Distribute:

=0.95x+−9+9xy+−18x+(−0.25)(−3x)+(−0.25)(4.8)

=0.95x+−9+9xy+−18x+0.75x+−1.2

Combine Like Terms:

=0.95x+−9+9xy+−18x+0.75x+−1.2

=(9xy)+(0.95x+−18x+0.75x)+(−9+−1.2)

=9xy+−16.3x+−10.2

7 0

3 years ago

two linear functions have the same slope,same x-intercept, and same y-intercept. Must these functions be identical? Explain your

Leviafan [203]

Yes, because only one line can go through two set points and because they have the same slope and the same x and y-intercepts they have two set points that they must both go through making them identical.

4 0

3 years ago