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2 years ago

Shawn found that a solution of x = 3 was extraneous for his function. What would be a possible denominator in Shawn's function?

Mathematics

2 answers:

Alex_Xolod [135]2 years ago

6 0

Your answer is B. x2 -2x -3

Hope this helps!!

laiz [17]2 years ago

3 0

Answer:

Option B is correct

Step-by-step explanation:

Given: A solution x =3 was extraneous for his function.

Extraneous solution defined as a root of a transformed equation that is not a root of the original equation because it was excluded from the domain of the original equation.

The only function from the given options

f(x) = $x^2-2x-3$ has the solution x =3

therefore,

3 is excluded from the domain of the $\frac{x^2-2x-3}{x-3}$ because it would make the denominator of one of the fractions zero and division by zero is not allowed.

Therefore, it cannot be a root of the equation $\frac{x^2-2x-3}{x-3}$

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Which is bigger, 9/17 or 5/6?

d1i1m1o1n [39]

5/6 is bigger than 9/17

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2 years ago

The length of a rectangle is 3 inches more than its width. the area of the rectangle is 40 square inches. what is the length, in

IceJOKER [234]

Since width is the smaller value, set width=x. That wold mean that length=x+3. Since Area=width*length, 40=x(3+x). Distribute the x to give you 40=3x+x^2. Set the equation equal to 0 giving you x^2+3x-40=0. Factor the equation to give you (x+8)(x-5)=0. Set each factor equal to 0. x+8=0 gives you x=-8 & x=-5 gives you x=5. Since we're working with lengths of objects, it can't be negative. Therefore, your width is 5. To find length, substitute width into the original expression (x+3). Therefore, length is 5+3=8.

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2 years ago

A certain square is to be drawn on a coordinate plane. One of the vertices must be on the origin, and the square is to have an a

Scrat [10]

Answer:

The answer is (C) 8

Step-by-step explanation:

First, let's calculate the length of the side of the square.

$A_{square}=a^2$ , where $a$ is the length of the side. Now, let's try to build the square. First we need to find a point which distance from (0, 0) is 10. For this, we can use the distance formula in the plane:

$d = \sqrt{(x_2-x_1)^2 + (y_2-y_1)^2}$ which for $x_1=0$ and $y_1 = 0$ transforms as $d=\sqrt{(x_2)^2 + (y_2)^2}$ . The first point we are looking for is connected to the origin and therefore, its components will form a right triangle in which, the Pythagoras theorem holds, see the first attached figure. Then, $x_2$ , $y_2$ and 10 are a Pythagorean triple. From this, $x_2= 6$ or $x_2=8$ while $y_2= 6$ or $y_2=8$ . This leads us with the set of coordinates:

$(\pm 6, \pm 8)$ and $(\pm 8, \pm 6)$ . (A)

The next step is to find the coordinates of points that lie on lines which are perpendicular to the lines that joins the origin of the coordinate system with the set of points given in (A):

Let's do this for the point (6, 8).

The equation of the line that join the point (6, 8) with the origin (0, 0) has the equation $y = mx +n$ , however, we only need to find its slope in order to find a perpendicular line to it. Thus,

$m = \frac{y_2-y_1}{x_2-x_1} \\m = \frac{8-0}{6-0} \\m = 8/6$

Then, a perpendicular line has an slope $m_{\bot} = -\frac{1}{m} = -\frac{6}{8}$ (perpendicularity condition of two lines). With the equation of the slope of the perpendicular line and the given point (6, 8), together with the equation of the distance we can form a system of equations to find the coordinates of two points that lie on this perpendicular line.

$m_{\bot}=\frac{6}{8} = \frac{8-y}{6-x}\\ 6(6-x)+8(8-y)=0$ (1)

$d^2 = \sqrt{(y_o-y)^2+(x_o-x)^2} \\(10)^2=\sqrt{(8-y)^2+(6-x)^2}\\100 = \sqrt{(8-y)^2+(6-x)^2}$ (2)

This system has solutions in the coordinates (-2, 14) and (14, 2). Until here, we have three vertices of the square. Let's now find the fourth one in the same way we found the third one using the point (14,2). A line perpendicular to the line that joins the point (6, 8) and (14, 2) has an slope $m = 8/6$ based on the perpendicularity condition. Thus, we can form the system:

$\frac{8}{6} =\frac{2-y}{14-x} \\8(14-x) - 6(2-y) = 0$ (1)

$100 = \sqrt{(14-x)^2+(2-y)^2}$ (2)

with solution the coordinates (8, -6) and (20, 10). If you draw a line joining the coordinates (0, 0), (6, 8), (14, 2) and (8, -6) you will get one of the squares that fulfill the conditions of the problem. By repeating this process with the coordinates in (A), the following squares are found:

(0, 0), (6, 8), (14, 2), (8, -6)

(0, 0), (8, 6), (14, -2), (6, -8)
(0, 0), (-6, 8), (-14, 2), (-8, -6)
(0, 0), (-8, 6), (-14, -2), (-6, -8)

Now, notice that the equation of distance between the two points separated a distance of 10 has the trivial solution $(\pm10, 0)$ and $(0, \pm10)$ . By combining this points we get the following squares:

(0, 0), (10, 0), (10, 10), (0, 10)
(0, 0), (0, 10), (-10, 10), (-10, 0)
(0, 0), (-10, 0), (-10, -10), (0, -10)
(0, 0), (0, -10), (-10, -10), (10, 0)

See the attached second attached figure. Therefore, 8 squares can be drawn

8 0

2 years ago

Brad has two lengths of copper pipe to fit together one has a length of 2 5/12 feet and the other has a length of 3 7 12 ft how

leva [86]

To find the total length of the two copper pipes, we need to add them together.

2 5/12 + 3 7/12 = 5 12/12

Since 12/12 = 1 or a whole, we can simplify this further.

5 12/12 = 6

Hope this helps!

4 0

2 years ago

Which of the following statements is not true

Aloiza [94]

Answer:

The slope of AB is different from the slope of BC is NOT TRUE.

Step-by-step explanation:

AB and BC are on the same straight line, so their slopes are the same.

6 0

1 year ago