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Flauer [41]
2 years ago
15

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?​
Mathematics
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
8 0
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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Answer:

c true

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1st u read the problem and thensolve the problem

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