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2 years ago

13

Select the equation of the line parallel to the equation y = -2x - 7 that passes through the point (3, 1). y = -2x + 7 y = -2x +

6 y = -2x - 2 y = 3x - 8

1 answer:

Vilka [71]2 years ago

7 0

Y - y₁ = m(x - x₁)
y - 1 = -2(x - 3)
y - 1 = -2(x) + 2(3)
y - 1 = -2x + 6
+ 1 + 1
y = -2x + 7

The answer is A.

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State whether f is a function

Alenkinab [10]

<h2>Answer:</h2>

$y=\pm (\frac{x}{6})^{\frac{1}{4}} \ is \ not \ a \ function$

<h2>Step by step solution:</h2>

A function $f$ from a set $A$ to a set $B$ is a relation that assigns to each element $x$ in the set $A$ exactly one element $y$ in the set $B$ . The set $A$ is the domain (also called the set of inputs) of the function and the set $B$ contains the range (also called the set of outputs). On the other hand, a function has an inverse function if and only if passes the Horizontal Line Test for Inverse Functions. This test tells us that a function $f$ has an inverse function if and only if there is no any horizontal line that intersects the graph of $f$ at more than one point. So the function is called one-to-one. The graph of $f$ is shown below. As you can see, this function does not pass the Horizontal Line Test, therefore the inverse is not a function. However, let's find $f^-{1}(x)$ :

$f(x)=6x^4 \\ \\ Substitute \ f(x) \ by \ y \\ \\ y=6x^4 \\ \\ Interchange \ x \ and \ y: \\ \\ x=6y^4 \\ \\ Solve \ for \ y: \\ \\ y^4=\frac{x}{6} \\ \\ Solving \\ \\y=\pm \sqrt[4]{\frac{x}{6}} \\ \\ \boxed{y=\pm \left(\frac{x}{6}\right)^{\frac{1}{4}}}$

and this is not a function because there are elements in the set of inputs that match with two elements in the set of outputs.

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What I've done is, first draw a triangle with a 90º angle and write down the height (324m) and the tourist's distance (100m).

So I calculated the tangent of the angle of the tourist.

The formula is tan=Oposite side/adjacent

$tan=\frac{opposite side}{adjacent}$

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And now with your calculator you press "shift" and "tan". Then introduce 3,24 and it will give you the result (80,94º).

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