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Murrr4er [49]
3 years ago
11

Someone help me with this show work

Mathematics
1 answer:
chubhunter [2.5K]3 years ago
5 0
3x^2 + 11x - 4 = 0
Look for two factors of -12 that add up to 11 -> -1 and 12
(3x - 1)(x + 4) = 0
x = 1/3 or -4
So the answer you want is D

An easy way to answer questions like this is simply to try each option until one works.
3 x 0^2 + 11 x 0 = 0 so not A
3 x (-11/3)^2 + 11 x -11/3 = 0 so not B
3 x 4^2 + 11 x 4 = 92 so not C
3 x (1/3)^2 + 1 x 1/3 = 12/3 = 4 correct
I doubt E would be marked correct but in fact A and E are opposite statements so one of them must be true

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Help me on these questions
mojhsa [17]

Answer:

a) The equation is (y - 1)² = -8 (x - 4)

b) The equation is (x - 1)²/25 + (y - 4)²/16 = 1

c) The equation of the ellipse is (x - 3)²/16 + y²/4 = 1

Step-by-step explanation:

a) Lets revise the standard form of the equation of the parabola with a

   horizontal axis

# (y - k)² = 4p (x - h), (h , k) are the coordinates of its vertex and p ≠ 0

- The focus of it is (h + p , k)

* Lets solve the problem

∵ The focus is (2 , 1)

∵ focus is (h + p , k)

∴ h + p = 2 ⇒ subtract p from both sides

∴ h = 2 - p ⇒ (1)

∴ k = 1

∵ It opens left, then the axis is horizontal and p is negative

∴ Its equation is (y - k)² = 4p (x - h)

∵ k = 1

∴ Its equation is (y - 1)² = 4p (x - h)

- The parabola contains point (2 , 5), substitute the coordinates of the

 point in the equation of the parabola

∴ (5 - 1)² = 4p (2 - h)

∴ (4)² = 4p (2 - h)

∴ 16 = 4p (2 - h) ⇒ divide both sides by 4

∴ 4 = p (2 - h) ⇒ (2)

- Use equation (1) to substitute h in equation (2)

∴ 4 = p (2 - [2 - p]) ⇒ open the inside bracket

∴ 4 = p (2 - 2 + p) ⇒ simplify

∴ 4 = p (p)

∴ 4 = p² ⇒ take √ for both sides

∴ p = ± 2, we will chose p = -2 because the parabola opens left

- Substitute the value of p in (1) to find h

∵ h = 2 - p

∵ p = -2

∴ h = 2 - (-2) = 2 + 2 = 4

∴ The equation of the parabola in standard form is

  (y - 1)² = 4(-2) (x - 4)

∴ The equation is (y - 1)² = -8 (x - 4)

b) Lets revise the equation of the ellipse

- The standard form of the equation of an ellipse with  center (h , k)

 and major axis parallel to x-axis is (x - h)²/a² + (y - k)²/b² = 1  

- The coordinates of the vertices are (h ± a , k )  

- The coordinates of the foci are (h ± c , k), where c² = a² - b²  

* Now lets solve the problem

∵ Its vertices are (-4 , 4) and (6 , 4)

∵ The coordinates of the vertices are (h + a , k ) and (h - a , k)  

∴ k = 4

∴ h + a = 6 ⇒ (1)

∴ h - a = -4 ⇒ (2)

- Add (1) and (2) to find h

∴ 2h = 2 ⇒ divide both sides by 2

∴ h = 1

- Substitute the value of h in (1) or (2) to find a

∴ 1 + a = 6 ⇒subtract 1 from both sides

∴ a = 5

∵ The foci at (-2 , 4) and (4 , 4)

∵ The coordinates of the foci are (h + c , k) , (h - c , k)

∴ h + c = 4

∵ h = 1

∴ 1 + c = 4 ⇒ subtract 1 from both sides

∴ c = 3

∵ c² = a² - b²

∴ 3² = 5² - b²

∴ 9 = 25 - b² ⇒ subtract 25 from both sides

∴ -16 = -b² ⇒ multiply both sides by -1

∴ 16 = b²

∵ a² = 25

∵ The equation of the ellipse is (x - h)²/a² + (y - k)²/b² = 1

∴ The equation is (x - 1)²/25 + (y - 4)²/16 = 1

c) How to identify the type of the conic  

- Rewrite the equation in the general form,  

 Ax² + Bxy + Cy² + Dx + Ey + F = 0  

- Identify the values of A and C from the general form.  

- If A and C are nonzero, have the same sign, and are not equal  

 to each other, then the graph is an ellipse.  

- If A and C are equal and nonzero and have the same sign, then

 the graph is a circle  

- If A and C are nonzero and have opposite signs, and are not equal  

 then the graph is a hyperbola.  

- If either A or C is zero, then the graph is a parabola  

* Now lets solve the problem

∵ x² + 4y² - 6x - 7 = 0

∵ The general form of the conic equation is

   Ax² + Bxy + Cy² + Dx + Ey + F = 0  

∴ A = 1 and C = 4

∵ If A and C are nonzero, have the same sign, and are not equal  to

  each other, then the graph is an ellipse.

∵ x² + 4y² - 6x - 7 = 0 ⇒ re-arrange the terms

∴ (x² - 6x ) + 4y² - 7 = 0

- Lets make x² - 6x completing square

∵ 6x ÷ 2 = 3x

∵ 3x = x × 3

- Lets add and subtract 9 to x² - 6x to make the completing square

 x² - 6x + 9 = (x - 3)²

∴ (x² - 6x + 9) - 9 + 4y² - 7 = 0 ⇒ simplify

∴ (x - 3)² + 4y² - 16 = 0 ⇒ add 16 to both sides

∴ (x - 3)² + 4y² = 16 ⇒ divide all terms by 16

∴ (x - 3)²/16 + 4y²/16 = 1 ⇒ simplify

∴ (x - 3)²/16 + y²/4 = 1

∴ The equation of the ellipse is (x - 3)²/16 + y²/4 = 1

5 0
3 years ago
I need help with 14 and 15
bekas [8.4K]

Answer:

14: 9 + 5 ℎ

15: − 1 6 + 9/ 9

Step-by-step explanation:

14 : ℎ ⋅ 5 + ( 9 )

5ℎ+ (9)

^^ Re-order terms so constants are on the left

5ℎ+ ⋅ 9 also 5 ℎ+9  then after this you would wanna re arange the terms so it would be 9 + 5 ℎ

7 0
3 years ago
How do I convert radians to degrees? Is it pi/180 or 180/pi? I get them confused! Thanks
o-na [289]
It could be either one, depending on whether you intend to multiply or divide.

When you convert, just remember that degrees ==> less radians,
and radians ==> more degrees.  Then you can always check the
results, and make sure that you used the right conversion factor
in the right direction.
5 0
3 years ago
Read 2 more answers
A school raised &4,589 in a sale. Mr Simmons class raised 1/100 of the money. How much did Mr Simmons class raise?
Ksenya-84 [330]

Alright, lets get started.

total amount school raised in sale = $ 4589

Mr Simmons class raised \frac{1}{100} of the money.

So, if we want to find amount of the money Mr. Simmons raised, we sould take one hundered of the total amount.

Hence money raised by Mr. Simmons class = \frac{1}{100} * 4589

Hence money raised by Mr. Simmons class = \frac{4589}{100}

Hence money raised by Mr. Simmons class = $ 45.89 : Answer

Hope it will help :)


4 0
3 years ago
Read 2 more answers
A chemical refinery needs a vat. They want the vat to be a rectangular prism (with square bases) that has a maximum volume of 1,
Alex17521 [72]

Answer:

For least material to be used lengths of square base and sides = 10 units.

Step-by-step explanation:

Let the lengths of the square base and the sides = x feet, x feet and y feet

Area of the square base = x² feet

Volume of the rectangular prism = Area of the square base × Height

                                                      = x²y cubic feet

1000 = x²y

y = \frac{1000}{x^2} -------(1)

Material used in the prism = Surface area of the rectangular prism

                                            = 2(lb + bh + hl)

Here, h =  height of the prism

l = length of the base

w = Width of the base

Material to be used (S) = 2(xy + x² + xy) - Area of lid

                                  S = 2(x² + 2xy) - x²

                                  S = x² + 2xy

Now by substituting the value of y from equation (1),

S = x² + 2x(\frac{1000}{x^{2} })

  = x² + \frac{2000}{x}

For least amount of material used,

We will find the derivative of the given function and equate it to zero.

S' = 2x - \frac{2000}{x^{2} }

2x - \frac{2000}{x^{2} } = 0

2x³ = 2000

x³ = 1000

x = 10 feet

From equation (1),

y = \frac{1000}{(10)^2}

y = 10 feet

Therefore, for least amount of the material used lengths of square base and sides will be 10 feet.

7 0
3 years ago
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