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dimaraw [331]
3 years ago
13

Write a system of linear equations.Write two-variable equations in slope-intercept form.

Mathematics
1 answer:
kolbaska11 [484]3 years ago
4 0

Answer:

A linear relation can be written in the slope-intercept form as:

y = a*x + b

Where a is the slope, and b is the y-intercept.

And if a line passes through the points (x₁, y₁) and (x₂, y₂), the slope can be written as:

a = (y₂ - y₁)/(x₂ - x₁)

Now let's look at our lines.

The left one passes through the points (0, -3) and (-2, 0)

Then the slope of this line is:

a = (0 - (-3))/(-2 - 0) = 3/-2 = -(3/2)

Then the line will be something like:

y = -(3/2)*x + b

Knowing that this line passes through the point (0, -3), we know that when x = 0, we must have y = -3

Then:

-3 = (-3/2)*0 + b

-3 = b

So the first linear equation is:

y = -(3/2)*x - 3

Now let's look at the second line, this one passes through the points (0, 2) and (1, 0)

Then the slope of this line is:

a = (0 - 2)/(1 - 0) = -2

And we can write the line as:

y = -2*x + b

Knowing that this line passes through the point (0, 2), we know that when x = 0 we must have y = 0, replacing that we get:

2 = -2*0 + b

2 = b

Then the equation of this line is:

y = -2*x + 2

Now that we know both equations we can write our system as:

y = (-3/2)*x - 3

y = -2*x + 2

To solve it, we need to remember that y = y

then:

(-3/2)*x - 3 = y = -2*x + 2

(-3/2)*x - 3  = -2*x + 2

We can solve this for x.

2*x - (3/2)*x = 2 + 3

(1/2)*x = 5

x = 5((1/2) = 5*2 = 10

And to find the y-value, we need to input this x-value in one of the equations:

y = -2*10 + 2 = -20 + 2 = -18

Then the solution of the system is the point (10, -18)

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Nine million nine hundred thousand five thousand four hundred and eighty two.
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What is the fourth term of a number pattern that begins with 250 and subtracts 13? A. 211 B. 224 C. 198 D. 302
Alina [70]

Create the equation:

250-13x

Plug in 4 for the equation and solve:

250-13(4)

250-52

198

The answer is C. 198

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Triangle ABC has vertices A(-5, -2), B(7, -5), and C(3, 1). Find the coordinates of the intersection of the three altitudes
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Answer:

Orthocentre (intersection of altitudes) is at (37/10, 19/5)

Step-by-step explanation:

Given three vertices of a triangle

A(-5, -2)

B(7, -5)

C(3, 1)

Solution A by geometry

Slope AB = (yb-ya) / (xb-xa) = (-5-(-2)) / (7-(-5)) = -3/12 = -1/4

Slope of line normal to AB, nab = -1/(-1/4) = 4

Altitude of AB = line through C normal to AB

(y-yc) = nab(x-xc)

y-1 = (4)(x-3)

y = 4x-11           .........................(1)

Slope BC = (yc-yb) / (xc-yb) = (1-(-5) / (3-7)= 6 / (-4) = -3/2

Slope of line normal to BC, nbc = -1 / (-3/2) = 2/3

Altitude of BC

(y-ya) = nbc(x-xa)

y-(-2) = (2/3)(x-(-5)

y = 2x/3 + 10/3 - 2

y = (2/3)(x+2)    ........................(2)

Orthocentre is at the intersection of (1) & (2)

Equate right-hand sides

4x-11 = (2/3)(x+2)

Cross multiply and simplify

12x-33 = 2x+4

10x = 37

x = 37/10  ...................(3)

substitute (3) in (2)

y = (2/3)(37/10+2)

y=(2/3)(57/10)

y = 19/5  ......................(4)

Therefore the orthocentre is at (37/10, 19/5)

Alternative Solution B using vectors

Let the position vectors of the vertices represented by

a = <-5, -2>

b = <7, -5>

c = <3, 1>

and the position vector of the orthocentre, to be found

d = <x,y>

the line perpendicular to BC through A

(a-d).(b-c) = 0                          "." is the dot product

expanding

<-5-x,-2-y>.<4,-6> = 0

simplifying

6y-4x-8 = 0 ...................(5)

Similarly, line perpendicular to CA through B

<b-d>.<c-a> = 0

<7-x,-5-y>.<8,3> = 0

Expand and simplify

-3y-8x+41 = 0 ..............(6)

Solve for x, (5) + 2(6)

-20x + 74 = 0

x = 37/10  .............(7)

Substitute (7) in (6)

-3y - 8(37/10) + 41 =0

3y = 114/10

y = 19/5  .............(8)

So orthocentre is at (37/10, 19/5)  as in part A.

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Answer:

The distance around a circle on the other hand is called the circumference (c).

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