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Zolol [24]
3 years ago
14

Laura plans to paint the 8 foot high rectangular walls of her room, and before she buys paint she needs to know the area of the

wall surface to be painted. Two walls are 10 feet wide, and the the other 2 walls are 15 feet wide. The combined area of the 1 window and the 1 door in her room in 60 square feet. What is the area in square feet, of the wall surface Laura plans to paint? A.200 B.340 C.360 D.390 E.400
Mathematics
2 answers:
Mila [183]3 years ago
7 0

Answer:

340

Step-by-step explanation:

2 × (10 × 8) = 160

2 × (15 × 8) = 240

160 + 240 - 60 = 340 square feet

hram777 [196]3 years ago
6 0

Answer:

B: 340

Step-by-step explanation:

We are told that her walls are 8 ft high.

Now, Two walls are 10 feet wide. Thus, area of the first two walls will be;

A1 = 2 × (10 × 8) = 160 sq.ft

Also, the other 2 walls are 15 feet wide.

Thus, Area of other two walls will be;

A2 = 2 × (15 × 8) = 240 sq.ft

We are told that the combined area of the 1 window and the 1 door in her room in 60 square feet.

This means that we will subtract 60 Sq.ft from the sum of the area of the 2 walls earlier calculated to get the total area to be painted.

Thus;

Area to be painted = 160 + 240 - 60 = 340 Sq.ft

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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
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