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AlladinOne [14]
2 years ago
13

The Point-Slope Form of a line perpendicular to y = 2x - 10. and passing through the point (3.1)

Mathematics
1 answer:
valkas [14]2 years ago
3 0

Answer:

y-1=2(x-3)

Point slope form is Y-y1=m(X-x1)

M= slope

y1= the y co-ordinate of your point

x1= the x co-ordinate of your point                                  

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The boundary of the lawn in front of a building is represented by the parabola . The parabola is represented on the coordinate p
Colt1911 [192]
The boundary of the lawn in front of a building is represented by the parabola

y = (x^2) /16 + x - 2

And you have three questions which require to find the focus, the vertex and the directrix of the parabola.

Note that it is a regular parabola (its symmetry axis is paralell to the y-axis).

1) Focus:

It is a point on the symmetry axis => x = the x-component of the vertex) at a distance equal to the distance between the directrix and the vertex).

In a regular parabola, the y - coordinate of the focus is p units from the y-coordinate of the focus, and p is equal to 1/(4a), where a is the coefficient that appears in this form of the parabola's equation: y = a(x - h)^2 + k (this is called the vertex form)

Then we will rearrange the standard form, (x^2)/16 + x - 2 fo find the vertex form y = a(x-h)^2 + k

What we need is to complete a square. You can follow these steps.

1) Extract common factor 1/16 => (1/16) [ (x^2) + 16x - 32]

2) Add (and subtract) the square of the half value of the coefficent ot the term on x =>

16/2 = 8 => add and subtract 8^2 => (1/16) [ (x^2) + 16 x + 8^2 - 32 - 8^2]

3) The three first terms inside the square brackets are a perfect square trinomial: =>

(1/16) [ (x+8)^2 - 32 - 64] = (1/16) [ (x+8)^2  - 96] =>

(1/16) [(x+8)^2 ] - 96/16 =>

(1/16) (x +8)^2 -  6

Which is now in the form a(x - h)^2 + k,
where:

a = 1/16 , h = - 8, and k = -6

(h,k) is the vertex: h is the x-coordinate of the vertex, and k is the y-coordinate of the vertex.

=> a = 1/16 => p =1/4a = 16/4 = 4
 
y-componente of the focus = -6 + 4 = -2

x-component of the focus = h = - 8

=> focus = (-8, -2)

2) Vertex

We found it above, vertex = (h,k) = (-8,-6)


3) Directrix

It is the line y = p units below the vertex = > y = -6 - 4 = -10

y = -10   

 


4 0
3 years ago
A class has 6 boys and 15 girls. What is the ratio of boys to girls
Over [174]
The answer is 6:15, but if u need to simplify the answer would be 2:5.
4 0
3 years ago
Read 2 more answers
Can someone help me with this
aleksandrvk [35]

Answer:

the answer to the second question is h^9

Step-by-step explanation:

you are subtracting 14-5=9

5 0
2 years ago
I need the answer plzz now
Verizon [17]

Answer:

Ea and St

Step-by-step explanation:

6 0
3 years ago
Please help me <br> Show your work <br> 10 points
Svet_ta [14]
<h2>Answer</h2>

After the dilation \frac{5}{3} around the center of dilation (2, -2), our triangle will have coordinates:

R'=(2,3)

S'=(2,-2)

T'=(-3,-2)

<h2>Explanation</h2>

First, we are going to translate the center of dilation to the origin. Since the center of dilation is (2, -2) we need to move two units to the left (-2) and two units up (2) to get to the origin. Therefore, our first partial rule will be:

(x,y)→(x-2, y+2)

Next, we are going to perform our dilation, so we are going to multiply our resulting point by the dilation factor \frac{5}{3}. Therefore our second partial rule will be:

(x,y)→\frac{5}{3} (x-2,y+2)

(x,y)→(\frac{5}{3} x-\frac{10}{3} ,\frac{5}{3} y+\frac{10}{3} )

Now, the only thing left to create our actual rule is going back from the origin to the original center of dilation, so we need to move two units to the right (2) and two units down (-2)

(x,y)→(\frac{5}{3} x-\frac{10}{3}+2,\frac{5}{3} y+\frac{10}{3}-2)

(x,y)→(\frac{5}{3} x-\frac{4}{3} ,\frac{5}{3}y+ \frac{4}{3})

Now that we have our rule, we just need to apply it to each point of our triangle to perform the required dilation:

R=(2,1)

R'=(\frac{5}{3} x-\frac{4}{3} ,\frac{5}{3}y+ \frac{4}{3})

R'=(\frac{5}{3} (2)-\frac{4}{3} ,\frac{5}{3}(1)+ \frac{4}{3})

R'=(\frac{10}{3} -\frac{4}{3} ,\frac{5}{3}+ \frac{4}{3})

R'=(2,3)

S=(2,-2)

S'=(\frac{5}{3} (2)-\frac{4}{3} ,\frac{5}{3}(-2)+ \frac{4}{3})

S'=(\frac{10}{3} -\frac{4}{3} ,-\frac{10}{3}+ \frac{4}{3})

S'=(2,-2)

T=(-1,-2)

T'=(\frac{5}{3} (-1)-\frac{4}{3} ,\frac{5}{3}(-2)+ \frac{4}{3})

T'=(-\frac{5}{3} -\frac{4}{3} ,-\frac{10}{3}+ \frac{4}{3})

T'=(-3,-2)

Now we can finally draw our triangle:

8 0
3 years ago
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