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3 years ago

Please help me with these two questions

Mathematics

1 answer:

KonstantinChe [14]3 years ago

4 0

Answer:

#1 is 60°, #2 is 55°

Step-by-step explanation:

all angles in a triangle add up to 180° so u can just subtract from that for the missing angles. :)

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1. - x + 5y =-23 4x – y=16

Tpy6a [65]

Answer:

Step-by-step explanation:

Solve for the first variable in one of the equations, then substitute the result into the other equation.

Point Form:(4,0)

Equation form:x=4 , y=0

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3 years ago

How many square feet of outdoor carpet will we need for this hole

Leokris [45]

Can you describe it a little more- I have no clue how I’m suppose to solve that.

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3 years ago

Jacob was out at a restaurant for dinner when the bill came. His dinner came to $18. After adding in a tip, before tax, he paid

yuradex [85]

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Step-by-step explanation:

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3 years ago

Determine whether the relation is a function. {(5, 0), (8, 1), (1, 3), (5, 2), (3, 8)}

mariarad [96]

It is a function because all of the inputs have exactly one output.

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4 years ago

Determine the domain and range of (g ○ f)(x) if f of x is equal to 4 over the quantity x squared minus 4 end quantity and g(x) =

lara [203]

The domain and range of a function are the possible x and y values of the function.

The domain and the range of the function is: (a) $\mathbf{D:\{x \in R|x \ne -2,2\}}$ and $\mathbf{R:\{(-\infty,1) \cup (2,\infty)\}}$

The functions are given as:

$\mathbf{f(x) = \frac{4}{x^2 - 4}}$

$\mathbf{g(x) = x + 2}$

(g o f)(x) is calculated as:

$\mathbf{(g\ o\ f)(x) = g(f(x))}$

So, we have:

$\mathbf{(g\ o\ f)(x) = \frac{4}{x^2 - 4} + 2}$

Take LCM

$\mathbf{(g\ o\ f)(x) = \frac{4 + 2x^2 - 8}{x^2 - 4} }$

$\mathbf{(g\ o\ f)(x) = \frac{2x^2 - 4}{x^2 - 4} }$

Represent the denominator as follows, to calculate the domain

$\mathbf{x^2 - 4 \ne 0 }$

Add 4 to both sides

$\mathbf{x^2 \ne 4 }$

Take square roots of bot sides

$\mathbf{x \ne \±2 }$

Hence, the domain of the function is:

$\mathbf{D:\{x \in R|x \ne -2,2\}}$

On the graph of $\mathbf{(g\ o\ f)(x) = \frac{2x^2 - 4}{x^2 - 4} }$ (see attachment), the function does not have a value from y = 1 to 2.

Hence, the range is:

$\mathbf{R:\{(-\infty,1) \cup (2,\infty)\}}$

Read more about domain and range at:

brainly.com/question/1632425

5 0

2 years ago

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