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2 years ago

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Type the correct answer in each box. Use numerals instead of words. The function f(x)=x^1/2 is transformed tp get function w. w(

x)=-(3x)^1/2-4 What are the domain and the range of function w?

1 answer:

ludmilkaskok [199]2 years ago

4 0

Answer:

domain(-1, infinite)

range(2, infinite)

Step-by-step explanation:

i took da test

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Neko [114]

Expansions and compressions are transformations that change the length or width of the graph of a function.

To graph y = a*f (x)
if a> 1, the graph of y = f (x) is expanded vertically by a factor a.
We have then:
f (x) = 2/5 x ^ 2-2
The function g (x) is a vertical stretch of f (x) by a factor of 2:
g (x) = 2f (x)
g (x) = 2 (2/5 x ^ 2-2)
g (x) = 4/5 x ^ 2-4

Answer:
The equation of g (x) is:
g (x) = 4/5 x ^ 2-4

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2 years ago

Given: ABC is a right triangle with right angle C. AC=15 centimeters and m∠A=40∘ . What is BC ? Enter your answer, rounded to th

konstantin123 [22]

In order to answer this question, the figure in the first picture will be helpful to understand what a right triangle is. Here, a right angle refers to $90\°$ .

However, if we want to solve the problem we have to know certain things before:

In the second figure is shown a general right triangle with its three sides and another given angle, we will name it $\alpha$ :

The side <u>opposite to the right angle</u> is called The Hypotenuse (h)

The side <u>opposite to the angle $\alpha$ </u> is called the Opposite (O)

The side <u>next to the angle $\alpha$ </u> is called the Adjacent (A)

So, going back to the triangle of our question (first figure):

The Hypotenuse is AB

The Opposite is BC

The Adjacent is AC

Now, if we want to find the length of each side of a right triangle, we have to use the <u>Pythagorean Theorem</u> and T<u>rigonometric Functions:</u>

Pythagorean Theorem

$h^{2}=A^{2} +O^{2}$ (1)

Trigonometric Functions (here are shown three of them):

Sine: $sin(\alpha)=\frac{O}{h}$ (2)

Cosine: $cos(\alpha)=\frac{A}{h}$ (3)

Tangent: $tan(\alpha)=\frac{O}{A}$ (4)

In this case the function that works for this problem is cosine (3), let’s apply it here:

$cos(40\°)=\frac{AC}{h}$

$cos(40\°)=\frac{15}{h}$ (5)

And we will use the Pythagorean Theorem to find the hypotenuse, as well:

$h^{2}=AC^{2}+BC^{2}$

$h^{2}=15^{2}+BC^{2}$ (6)

$h=\sqrt{225+BC^2}$ (7)

Substitute (7) in (5):

$cos(40\°)=\frac{15}{\sqrt{225+BC^2}}$

Then clear BC, which is the side we want:

${\sqrt{225+BC^2}}=\frac{15}{cos(40\°)}$

${{\sqrt{225+BC^2}}^2={(\frac{15}{cos(40\°)})}^2$

$225+BC^{2}=\frac{225}{{(cos(40\°))}^2}$

$BC^2=\frac{225}{{(cos(40\°))}^2}-225$

$BC=\sqrt{158,41}$

$BC=12.58$

Finally $BC$ is approximately 13 cm

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