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3 years ago

How do you take away 3 7/8 from 15 1/2

2 answers:

8 0

Well you would make the dynamiter the same like this: 3 7/8 and then change the 1/2 to a 15 4/8 and then subtract it. was that easier?

mr_godi [17]3 years ago

3 0

Chang the denominators

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kolbaska11 [484]

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3 years ago

What is the area of the kitchen floor in this floor plan?

Illusion [34]

The answer is A=lw
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4 years ago

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Which box and whiskers plot represents the data set. 10, 5, 8, 14, 21, 7, 13, 17, 17.

navik [9.2K]

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3 years ago

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Find the vertical and horizontal asymptotes, domain, range, and roots of f (x) = -1 / x-3 +2.

gladu [14]

Answer:

Vertical asymptote: $x=3$

Horizontal asymptote: $f(x) =2$

Domain of f(x) is all real numbers except 3.

Range of f(x) is all real numbers except 2.

Step-by-step explanation:

Given:

Function:

$f (x) = -\dfrac{1 }{ x-3} +2$

One root, $x = 3.5$

To find:

Vertical and horizontal asymptote, domain, range and roots of f(x).

Solution:

First of all, let us find the roots of f(x).

<em>Roots of f(x) means the value of x where f(x) = 0</em>

$0= -\dfrac{1 }{ x-3} +2\\\Rightarrow 2= \dfrac{1 }{ x-3}\\\Rightarrow 2x-2 \times 3=1\\\Rightarrow 2x=7\\\Rightarrow x = 3.5$

One root, $x = 3.5$

Domain of f(x) i.e. the values that we give as input to the function and there is a value of f(x) defined for it.

For x = 3, the value of f(x) $\rightarrow \infty$

For all, other values of $x$ , $f(x)$ is defined.

Hence, Domain of f(x) is all real numbers except 3.

Range of f(x) i.e. the values that are possible output of the function.

f(x) = 2 is not possible in this case because something is subtracted from 2. That something is $\frac{1}{x-3}$ .

Hence, Range of f(x) is all real numbers except 2.

Vertical Asymptote is the value of x, where value of f(x) $\rightarrow \infty$ .

$-\dfrac{1 }{ x-3} +2 \rightarrow \infty$

It is possible only when

$x-3=0\\\Rightarrow x=3$

$\therefore$ vertical asymptote: $x=3$

Horizontal Asymptote is the value of f(x) , where value of x $\rightarrow \infty$ .

$x\rightarrow \infty \Rightarrow \dfrac{1 }{ x-3} \rightarrow 0\\\therefore f(x) =-0+2 \\\Rightarrow f(x) =2$

$\therefore$ Horizontal asymptote: $f(x) =2$

Please refer to the graph of given function as shown in the attached image.

5 0

3 years ago

Whoever answers my question will get a brainliest.

Maurinko [17]

Answer:

see explanation

Step-by-step explanation:

Calculate the distance d using the distance formula

d = $\sqrt{(x_{2}-x_{1})^2+(y_{2}-y_{1})^2 }$

with (x₁, y₁ ) = ((- 2, - 3 ) and (x₂, y₂ ) = (x, 5x + 9 )

d = $\sqrt{(x-(-2))^2+(5x+9-(-3))^2}$

= $\sqrt{(x+2)^2+(5x+9+3)^2}$

= $\sqrt{(x+2)^2+(5x+12)^2}$

4 0

2 years ago