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MA_775_DIABLO [31]
3 years ago
8

Write the slope-intercept form of the equation of each line.

Mathematics
1 answer:
miskamm [114]3 years ago
7 0

Answer:

y = 3x + 2

Step-by-step explanation:

Let's identify two clear points on this line. I can see (0, 2) and (-1, -1)

First you want to find the slope of the line that passes through these points. To find the slope of the line, we use the slope formula: (y₂ - y₁) / (x₂ - x₁)

Plug in these values:

(-1 - 2) / (-1 - 0)

Simplify the parentheses.

= (-3) / (-1)

Simplify the fraction.

-3/-1

= 3

This is your slope. Plug this value into the standard slope-intercept equation of y = mx + b.

y = 3x + b

To find b, we want to plug in a value that we know is on this line: in this case, I will use the first point (0, 2). Plug in the x and y values into the x and y of the standard equation.

2 = 3(0) + b

To find b, multiply the slope and the input of x(0)

2 = 0 + b

Now, we are left with 0 + b.

2 = b

Plug this into your standard equation.

y = 3x + 2

This is your equation.

Hope this helps!

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1. M is the midpoint of LN and O is the midpoint of NP.
Reptile [31]
1. M is the midpoint of LN and O is the midpoint of NP. This makes the triangle MNO equal to half of LNP. Then you can get this equation
MO= (1/2) LP

If you insert MO = 2x +6 and LP = 8x – 20 the calculation would be:
2x+6= (1/2)( 8x-20)
2x+6= 4x-10
2x-4x= -10 - 6
-2x= -16
x=8

2. Centroid is the point that intersects with three median lines of the triangle. The centroid should divide the median lines into 1:2 ratio. In AC lines, A located in the base so A.F:FC would be 1:2

Then, the answer would be:
A.F= 1/(1+2) * AC
A.F= 1/3 * 12= 4

FC= 2/(1+2) * AC
FC= 2/3 * 12= 8

3. Since
∠BAD=∠DAC
∠ABD=∠ACD
AD=AD
The triangle ABD and ACD are similar. You can get this equation
BD=DC
x+8= 3x+12
x-3x= 12-8
-2x=4
x=-2

DC=3x+12= 3(-2) +12= 6

4. Orthocenter made by intersection of triangle altitude
A
BC lines slope would be (-4)-(-1)/1-4= -3/-3= 1. The altitude line slope would be -1, the function would be:
y=-x +a
0= 1+a
a=-1
y=-x-1
B
AC lines slope would be (-4)-(-1)/1-0= -3. The altitude line slope would be 1/3, the function would be:
y=1/3x+a
-1=1/3(4)+a
a=-7/3
y=1/3x - 7/3

C
BC lines slope would be (-1)-(-1)/4 = 0/4. 
The line would be 
0=x+a
a=-1
0=x-1
x=1

y=-x-1 = 1/3x-7/3
-x-(1/3x)=-7/3 +1
-4/3x= -4/3
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y=-x-1
y=-1-1= -2
The orthocenter would be (1,-2)

5. 
a. Circumcenter: the intersection of perpendicular bisector lines<span>
b. Incenter: the intersection of bisector lines
c. Centroid: </span>the intersection of median lines<span>
d. Orthocenter: </span>the intersection of altitude lines
5 0
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1.find the sum. (3^3+5x^2+3x-7)+(8x-6x^2+6)<br> 2. Find the Difference: (8x-4x^2+3\)-(x^3+7x^2+3x-8)
vlada-n [284]

1.

(3^3+5x^2+3x-7)+(8x-6x^2+6)=\\\\=27\ \underline{+\,5x^2}\ \underline{ \underline{+\,3x}}-7\ \underline{\underline{\,+8x}}\ \underline{-\,6x^2}+6=\\\\=-x^2+11x+26

or if you mean (3x^3+5x^2+3x-7)+(8x-6x^2+6):

(3x^3+5x^2+3x-7)+(8x-6x^2+6)=\\\\=3x^3\ \underline{+\,5x^2}\ \underline{ \underline{+\,3x}}-7\ \underline{\underline{\,+8x}}\ \underline{-\,6x^2}+6=\\\\=3x^3-x^2+11x-1

2.

(8x-4x^2+3)-(x^3+7x^2+3x-8)=\\\\=\underline{8x}\ \underline{\underline{-\ 4x^2}}+3-x^3\ \underline{\underline{-\,7x^2}}\ \underline{-\,3x}+8=\\\\=-x^3-11x^2+5x+10

3 0
3 years ago
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