All categories

3 years ago

How would the transition rule below move the figure (x,y)

Mathematics

1 answer:

exis [7]3 years ago

5 0

D. 7 units down because the y axis deals with movement up and down and since it's negative it moves down

You might be interested in

Find the area of the shape shown below.

erastova [34]

Answer:10 is the area

Step-by-step explanation:

6 0

3 years ago

Read 2 more answers

Find the fourth term of a geometric sequence if a=6 and r=2. Use the formula an=ar^n-1

Vikentia [17]

Answer:

Step-by-step explanation:

here, we are using an = ar^n-1

so, we have to find a4= ar^4-1 = ar^3

now, putting the given values in the equation,

a4= (6)(2)^3 = 6(8) = 48

therefore, the 4th term is 48.

I have done this question previously on Gauthmath app, you can check that app out, really cool app.

All the best with your exam. 

8 0

3 years ago

Use the given pair of functions to find and simplify expressions for the following functions and state the domain of each using

Tamiku [17]

Answer:

Given functions,

$f(x) = 3 - x^2$

$g(x) = \sqrt{x}+1$

Since, by the compositions of functions,

1. (g◦f)(x) = g(f(x))

$=g(3-x^2)$

$=\sqrt{3-x^2}+1$

Since, (g◦f) is defined,

If 3 - x² ≥ 0

⇒ 3 ≥ x²

⇒ -√3 ≤ x ≤ √3

Thus, Domain = [-√3, √3]

2. (f◦g)(x) = f(g(x))

$=f(\sqrt{x}+1)$

$=3-(\sqrt{x}+1)^2$

Since, (g◦f) is defined,

If x ≥ 0

Thus, Domain = [0, ∞)

3. (f◦f)(x) = f(f(x))

$=f(3-x^2)$

$=3-(3-x^2)^2$

$=3-9-x^4+6x^2$

$=-6+6x^2-x^4$

Since, (f◦f) is a polynomial,

We know that,

A polynomial is defined for all real value of x,

Thus, Domain = (-∞, ∞)

8 0

3 years ago

Find the six trigonometric function values for angle ∅ where its adjacent side is -9 and its hypotenuse is 41. (Theta is located

arsen [322]

Check the picture below.

$\bf \textit{using the pythagorean theorem}
\\\\
c^2=a^2+b^2\implies \sqrt{c^2-a^2}=b
\qquad 
\begin{cases}
c=hypotenuse\\
a=adjacent\\
b=opposite\\
\end{cases}
\\\\\\
\sqrt{41^2-(-9)^2}=b\implies \sqrt{1681-81}=b\\\\\\ \sqrt{1600}=b\implies 40=b\\\\
-------------------------------$

$\bf sin(\theta )=\cfrac{\stackrel{opposite}{40}}{\stackrel{hypotenuse}{41}}\qquad~~ cos(\theta )=\cfrac{\stackrel{adjacent}{-9}}{\stackrel{hypotenuse}{41}}\qquad~~ tan(\theta )=\cfrac{\stackrel{opposite}{40}}{\stackrel{adjacent}{-9}}
\\\\\\
csc(\theta )=\cfrac{\stackrel{hypotenuse}{41}}{\stackrel{opposite}{40}}\qquad ~~sec(\theta )=\cfrac{\stackrel{hypotenuse}{41}}{\stackrel{adjacent}{-9}}\qquad ~~cot(\theta )=\cfrac{\stackrel{adjacent}{-9}}{\stackrel{opposite}{40}}$

3 0

3 years ago

Help me pleaseeeeeeeeee

crimeas [40]

Answer:

x=10 and x=-9

Step-by-step explanation:

Factor:

(x-10)(x+9)=0

x-10=0

x=10

and

x+9=0

x=-9

3 0

3 years ago