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3 years ago

Help me plz i have 8 mins!!

Mathematics

2 answers:

romanna [79]3 years ago

3 0

Remember the PEMDAS method. The operation inside the parenthesis should be done first. The sum of 2 and 7 is 9, so this leaves us with 5(9), which is 45. If you look at the last option and apply a similar method, this makes 10 + 35, which is also 45. Your answer is the last option.

tiny-mole [99]3 years ago

3 0

D. they both equal 45.

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Solve 2x+3y=6 for y Answers listed: y=-2/3x+2 y=2/3x-2 y=2x-3 y=2x+6

iren2701 [21]

2x + 3y = 6 <=>

3y = -2x + 6 <=>

y = -2/3x + 6/3 <=>

y = -2/3x + 2

The answer is the first one.

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2 years ago

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Sara bought a computer that was originally priced at $896. The computer was for 12% off. When she bought the computer, a 7% tax

lozanna [386]

Answer:

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Step-by-step explanation:

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2 years ago

2x-5y=17 ..solve for y

AlladinOne [14]

$2x-5y=17\ \ \ \ | subtract\ 2x\\\\
-5y=17-2x\ \ \ | divide\ by\ -5\\\\
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7 0

2 years ago

If y(x) = 3x – 2 and g(x) = 2x+1, find (f- g)(x).

r-ruslan [8.4K]

Answer:

(f- g)(x) = x - 3

Step-by-step explanation:

(f- g)(x) = f(x) - g(x)

This means that you have to subtract the two equation.

If f(x) = 3x - 2 and g(x) = 2x + 1

Then

f(x) - g(x) = [3x - 2] - [2x+1]

= 3x - 2- 2x - 1

= x - 3

∴ (f- g)(x) = x - 3

6 0

2 years ago

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).

Roots of f(x) means the value of x where f(x) = 0

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

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

2 years ago