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Crazy boy [7]
2 years ago
9

The pre-image is

Mathematics
2 answers:
yKpoI14uk [10]2 years ago
8 0
It is the last one.

A pre-image is the original before another shape is created from it, either dilated, translated, reflected, or rotated.
bezimeni [28]2 years ago
6 0

Answer

Step-by-step explanation:

It's D

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Anyone know this Geometry problem?
Yuliya22 [10]

Answer:

ST = 20

Step-by-step explanation:

RT is the sum of RS and ST

Replacing with length you get:

17 + x + 6 = 3x - 56

17 + 6 + 5 = 3x - x

28 = 2x

14 = x

ST = x + 6 = 14 + 6 = 20

6 0
3 years ago
Whats the answer to this problem
vlabodo [156]
The answer would be to the right since after you factor the equation out the 9 is positive
8 0
3 years ago
Please solve this with explanations!
lyudmila [28]

Answer:

D. -2x - 8

Step-by-step explanation:

1) Simplify 1/3(6x+15) to 6x+15/3

-6x+15/3 - 3

2) Factor out the common term 3.

-3(2x+5)/3 - 3

3) Cancel 3.

-(2x+5)-3

4) Remove parentheses.

-2x-5-3

5) Collect like terms.

-2x+(-5-3)

6) Simplify

-2x-8

3 0
2 years ago
Please answer these questions! will mark brainlist!
timofeeve [1]

<u><em>1</em><em>s</em><em>t</em><em> </em><em>photo....</em></u>

<em>Yes</em>

<em>Explanation</em>

All when divided becomes 1/3 , so it is proportional...

<em>Pls </em><em>mark</em><em> </em><em>brainliest</em>

<u><em>2</em><em>n</em><em>d</em><em> </em><em>photo</em><em>.</em><em>.</em><em>.</em><em>.</em></u>

Yes<em> </em>

<em>Explai</em><em>n</em><em>ation</em>

<em>Also </em><em>when </em><em>divided </em><em>becomes </em><em>1</em><em>/</em><em>3</em><em>.</em><em>.</em><em>.</em><em> </em><em>Same </em><em>proportional</em>

7 0
2 years ago
Read 2 more answers
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
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