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3 years ago

13

PLEASE HELP a right triangle is removed from a rectangle to create the shaded region shown below. Find the area of the shaded re

gion.

1 answer:

UNO [17]3 years ago

3 0

Answer:

56m²

Step-by-step explanation:

First, find the area of the entire rectangle:

10 × 8 = 80

Then, find the area of the triangle:

1/2(6)(8) = 24

Finally, subtract the two areas to find the shaded area:

80 - 24 = 56

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Llana [10]

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3 0

2 years ago

12) Write the equation of a rational function

aev [14]

Answer:

$\frac{4(x - 5)(x + 7)(x - 12)}{(x + 1)(x)(x - 12)}$

Step-by-step explanation:

A rational function is

$\frac{p(x)}{q(x)}$

where q(x) doesn't equal zero.

If p is a asymptote, or hole at that value, then we will use

$(x - p)$

Step 1: We have asymptote as 0 and -1 so our denomiator will include

$(x - 0)(x - ( - 1)$

Which is

$(x)(x + 1)$

So our denomator so far is

$\frac{p(x)}{x(x + 1)}$

Step 2: Find Holes.

Since 12 is the value of the hole,

$(x - 12)$

is a the binomial.

This will be both on the numerator and denomator so qe have

$\frac{(x - 12)}{x(x + 1)(x - 12)}$

Step 3: Put the x intercepts in the numerator.

Since 5 and -7 is the intercepts,

$\frac{(x - 12)(x - 5)(x + 7)}{x(x + 1)(x - 12)}$

Step 4: Horinzontal Asymptotes,

Multiply the numerator and denomiator out fully,

$\frac{ {x}^{3} - 10 {x}^{2} - 59x + 420 }{ {x}^{3} - 12 {x}^{2} + x - 12}$

Take a L

look at the coefficients,

Notice they have the same degree,3, this means if we divide the leading coefficents, we will get our horinzonral asymptote.

Multiply the numerator by 4.

$\frac{4(x - 12)(x - 5)(x - 7)}{x(x + 1)(x - 12)}$

Above is the function,

5 0

2 years ago

Solve the quadratic equation by completing the square.

Step2247 [10]

Answer:

x1, x2 = 4.74 , -2.74

Step-by-step explanation:

To find the roots of a quadratic function we have to use the bhaskara formula

ax^2 + bx + c

x^2 - 2x - 13

a = 1 b = -2 c = -13

x1 = (-b + √ b^2 - 4ac)/2a

x2 =(-b - √ b^2 - 4ac)/2a

x1 = (2 + √ (2^2 - 4 * 1 * (-13)))/2 * 1

x1 = (2 + √ (4 + 52)) / 2

x1 = (2 + √ 56 ) / 2

x1 = (2 + 7.48) / 2

x1 = 9.48 / 2

x1 = 4.74

x2 = (2 - √ (2^2 - 4 * 1 * (-13)))/2 * 1

x2 = (2 - √ (4 + 52)) / 2

x2 = (2 - √ 56 ) / 2

x2 = (2 - 7.48) / 2

x2 = -5.48 / 2

x2 = -2.74

6 0

3 years ago

Sara is working on a Geometry problem in her Algebra class. The problem requires Sara to use the two quadrilaterals below to ans

zloy xaker [14]

Part A:

Given a square with sides 6 and x + 4. Also, given a rectangle with sides 2 and 3x + 4

The perimeter of the square is given by 4(x + 4) = 4x + 16

The area of the rectangle is given by 2(2) + 2(3x + 4) = 4 + 6x + 8 = 6x + 12

For the perimeters to be the same

4x + 16 = 6x + 12
4x - 6x = 12 - 16
-2x = -4
x = -4 / -2 = 2

The value of x that makes the <span>perimeters of the quadrilaterals the same is 2.

Part B:

The area of the square is given by

$Area=(x+4)^2=x^2+8x+16$

The area of the rectangle is given by 2(3x + 4) = 6x + 8

For the areas to be the same

$x^2+8x+16=6x+8 \\ \\ \Rightarrow x^2+8x-6x+16-8=0 \\ \\ \Rightarrow x^2+2x+8=0 \\ \\ \Rightarrow x= \frac{-2\pm\sqrt{2^2-4(8)}}{2} \\ \\ = \frac{-2\pm\sqrt{4-32}}{2} = \frac{-2\pm\sqrt{-28}}{2} \\ \\ = \frac{-2\pm2i\sqrt{7}}{2} =-1\pm i\sqrt{7}$

Thus, there is no real value of x for which the area of the quadrilaterals will be the same.
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7 0

3 years ago

Solve the following equation for x: 6(4x+5)= 3(x+8)+3. Round to the nearest hundredth.

gavmur [86]

6(4x + 5) = 3(x + 8) + 3
24x + 30 = 3x + 24 + 3
24x + 30 = 3x + 27
24x - 3x = 27 - 30
21x = - 3
x = -3/21
x = - 0.143 <==

8 0

2 years ago

Read 2 more answers