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Triss [41]
2 years ago
6

ASAP PLEASE! PART TO PART RATIOS!

Mathematics
1 answer:
Bad White [126]2 years ago
8 0

Answer:

b). (0, 4)

if you need an explanation, comment, and I will put one in for you :)

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6 bags of baseball cards

Step-by-step explanation:

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3 years ago
Marcus went to the school gym to practice his 3 point shots
Schach [20]

Answer:

Plz write a proper question and by the way I'm not writing it For <em>points</em><em> </em><em>sake</em>..It's a incomplete question.....

3 0
3 years ago
PLEASE HELP GUYS i am struggling so much, two questions
ankoles [38]

Answer: the equation of the standard parabola

1) (y-6)^2 = 4 (x-1)

The equation of the standard parabola

2) (x+5)^2 = 16(y-2)

Step-by-step explanation:

<u>Explanation </u>

<u>Parabola:-</u>

The set of points in a plane whose distance from a fixed point and a constant ratio to their corresponding perpendicular distance from a fixed straight line is a conic.

Let S be a fixed point and l be a fixed straight line from any point P,the perpendicular PM is drawn to the line 'l'

  • The locus of P such that \frac{SP}{PM} = constant
  • The fixed point  'S' is called the Focus.
  • The fixed line'l 'is called the directrix of the conic
  • The constant ratio is known as the eccentricity, denoted by 'e'
  • If e=1 , the conic is called a parabola

1) <u> Step 1</u> :-

Given the focus   S = (2,6) and directrix is x=0

we know that \frac{SP}{PM}=1

now cross multiplication , we get

SP = PM

squaring on both sides,we get

SP^{2} = PM^2

step 2:-

now using distance formula is

  • \sqrt(((x_{2}-x_{1})^2+(y_{2} -y_{1} )^2)

Given S =(2,6) and P(x,y) be any point on parabola

SP^2 = (x-2)^2+(y-6)^2........(1)

Now using perpendicular distance formula

let P(x , y ) be any point on the parabola

  • \frac{ax_{1}+by_{1}+c   }{\sqrt{a^2+b^2} }

Given the directrix is x =0 and P(x,y) be any point on parabola

PM^2 = \frac{x^2}{\sqrt{1}^2 }......(2)

equating equation(1) and equation (2), on simplification

we get (x-2)^2+(y-6)^2 = x^2.....(3)

  • apply (a-b)^2 = a^2+  b^2+2 ab

now the equation (3) is

(y-6)^2 = 4 x-4

now the standard form of parabola is

(y-k)^2 = 4 a(x-h)

<u>Final answer</u>:-

(y-6)^2 = 4 (x-1)

2) <u> Explanation:-</u>

<u>step 1:</u>

Given vertex of a parabola is A(-5,2) and its focus is S(-5,6)

here the given points of 'x'co- ordinates are equal

  • Therefore the axis AS is parallel to y- axis

now the standard equation of parabola

(x-h)^2 = 4 a (y-k)

now you have to find' a' value

Given vertex of a parabola is A(-5,2) and its focus is S(-5,6)

The distance of AS = \sqrt{(-5-(-5)^2+(2-6)^2}

 on simplification we get a =4

<u>Final answe</u>r :-

the vertex (h,k) = (-5,2) and a=4

(x-h)^2 = 4 a (y-k)

The standard parabola is (x+5)^2 = 16 (y-2)

5 0
3 years ago
What is Y=2x-1 and Y=3x+2
Readme [11.4K]

Answer:

{y,x} = {-7,-3}

Step-by-step explanation:

- 3x = 2

  y - 2x = -1

-3x + y = 2        -2x + y = -1  

y = 2x - 1

(2x-1) - 3x = 2

    - x = 3   x = - 3

y = 2x-1

   x = -3

y = 2(-3)-1 = -7

4 0
4 years ago
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