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3 years ago

A homeowner wants to enlarge a rectangular closet that has an area of (x2 + 9x + 20) ft?.

Mathematics

2 answers:

Tresset [83]3 years ago

7 0

Answer:

width: (x+4) ft, length: (x+5) ft

Step-by-step explanation:

If the area was x2 + 9x + 20, and the length was (x+5), we find the width dividing the first polynomial by the second (or find what polynomial, multiplied by the second, will give us the first). We can see that the last value of the first polynomial is 20, and the last of the second is 5, so dividing these two values will give us the last value of the width polynomial, so we have that the width before construction is (x+4) ft.

Basile [38]3 years ago

4 0

Answer:

The dimension of the rectangular closet before construction = (x + 5) ft by (x + 4) ft

Step-by-step explanation:

The homeowner has a rectangular closet and the area of the closet is (x² + 9x + 20) ft². The area of a rectangle is the length multiplied by the width. The length is given as (x + 5) ft. After construction the area change to (x² + 14x + 48) ft². The length became (x + 6) ft.

The dimensions before construction can be calculated as follows:

Mathematically,

area of rectangle = Length × width

area = (x² + 9x + 20) ft².

Length = (x + 5) ft

Width = unknown

area = LW

(x² + 9x + 20) = (x + 5) W

W = (x² + 9x + 20) / x + 5

W = (x + 4) ft

The dimension of the rectangular closet before construction = (x + 5) ft by (x + 4) ft

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Read 2 more answers

I will give you 50 point please help and get. it right

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3 years ago

Part B

alisha [4.7K]

Question:

Consider ΔABC, whose vertices are A (2, 1), B (3, 3), and C (1, 6); let the line segment AC represent the base of the triangle.

(a) Find the equation of the line passing through B and perpendicular to the line AC

(b) Let the point of intersection of line AC with the line you found in part A be point D. Find the coordinates of point D.

Answer:

$y = \frac{1}{5}x + \frac{12}{5}$

$D = (\frac{43}{26},\frac{71}{26})$

Step-by-step explanation:

Given

$\triangle ABC$

$A = (2,1)$

$B = (3,3)$

$C = (1,6)$

Solving (a): Line that passes through B, perpendicular to AC.

First, calculate the slope of AC

$m = \frac{y_2 - y_1}{x_2 - x_1}$

Where:

$A = (2,1)$ --- $(x_1,y_1)$

$C = (1,6)$ --- $(x_2,y_2)$

The slope is:

$m = \frac{6- 1}{1 - 2}$

$m = \frac{5}{-1}$

$m = -5$

The slope of the line that passes through B is calculated as:

$m_2 = -\frac{1}{m}$ --- because it is perpendicular to AC.

So, we have:

$m_2 = -\frac{1}{-5}$

$m_2 = \frac{1}{5}$

The equation of the line is the calculated using:

$m_2 = \frac{y_2 - y_1}{x_2 - x_1}$

Where:

$m_2 = \frac{1}{5}$

$B = (3,3)$ --- $(x_1,y_1)$

$(x_2,y_2) = (x,y)$

So, we have:

$\frac{1}{5} = \frac{y - 3}{x - 3}$

Cross multiply

$5(y-3) = 1(x - 3)$

$5y - 15 = x - 3$

$5y = x - 3 + 15$

$5y = x +12$

Make y the subject

$y = \frac{1}{5}x + \frac{12}{5}$

Solving (b): Point of intersection between AC and $y = \frac{1}{5}x + \frac{12}{5}$

First, calculate the equation of AC using:

$y = m(x - x_1) + y_1$

Where:

$A = (2,1)$ --- $(x_1,y_1)$

$m = -5$

So:

$y=-5(x - 2) + 1$

$y=-5x + 10 + 1$

$y=-5x + 11$

So, we have:

$y=-5x + 11$ and $y = \frac{1}{5}x + \frac{12}{5}$

Equate both to solve for x

i.e.

$y = y$

$-5x + 11 = \frac{1}{5}x + \frac{12}{5}$

Collect like terms

$-5x -\frac{1}{5}x = \frac{12}{5} - 11$

Multiply through by 5

$-25x-x = 12 - 55$

Collect like terms

$-26x = -43$

Solve for x

$x = \frac{-43}{-26}$

$x = \frac{43}{26}$

Substitute $x = \frac{43}{26}$ in $y=-5x + 11$

$y = -5 * \frac{43}{26} + 11$

$y = \frac{-5 *43}{26} + 11$

Take LCM

$y = \frac{-5 *43+11 * 26}{26}$

$y = \frac{71}{26}$

Hence, the coordinates of D is:

$D = (\frac{43}{26},\frac{71}{26})$

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2 years ago