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Andrews [41]
3 years ago
10

Someone give me the answers to this

Mathematics
2 answers:
strojnjashka [21]3 years ago
6 0
Range: the set of all y-values of a graph or equation.

Y-intercept: the y-value where the graph intersects the y-axis.

Domain: the set of all x-values of a graph or equation.

X-intercept: the x-values where the graph intersects the x-axis.

Hope this helps! :)
kolezko [41]3 years ago
5 0

Answer:

The set of all y-values of a graph or equation.

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HELPP <br> Complete the ratio table.<br> 5 1<br> 30 6<br> 60 _<br> 75 _<br> 80 _
goldfiish [28.3K]

Answer:

12 27 32 it could be wrong though

5 0
2 years ago
1. Determine if the two expressions are equivalent and explain your reasoning.
Elena-2011 [213]
1. Yes because of the constant and coefficient if you add it. Like the answer is 5m +4.

6 0
3 years ago
Find the inverse of the function f f (x )<br><br> equals 9 x + 7
MrMuchimi

<em>Note: Your question seems a little bit ambiguous. So, I am assuming the given function f(x)=9x+7.</em>

<em>Thus, I am solving based on it. It would still clear your concept. </em>

Answer:

The inverse of f(x)=9x+7

  • \frac{x-7}{9}

Step-by-step explanation:

Given the function

f(x)=9 x + 7

A function g is the inverse of function f if for y=f(x), x=g(y)

Replace x with y

x=9y+7

solve for y

9y=x-7

y=\:\frac{x-7}{9}

Therefore,

The inverse of f(x)=9x+7 is:

  • \frac{x-7}{9}

i.e.

\mathrm{Inverse\:of}\:9x+7:\quad \frac{x-7}{9}

6 0
3 years ago
Consider the figure below. Abraham cut regular pentagon ABCDE into quadrilateral BEGC. Which of the following statements is true
Genrish500 [490]

Let 'n' be the number of side of a regular polygon each angle are equal and sum of  each angle equal to (n-2)×180. \\∴For a regular pentagon have five equal side, n=5\\ ∴ sum of each angle = (n-2)×180=3×180 = 540. and ecah angle = 540÷5=108°. Therefore ∠BCG = 108° and ∠ABC =108° \\ Here BE is one of the diagonal of regular pentagon ABCDE. BE divide ∠ABC in 1:2 ratio \\ Therefore ∠ABE : ∠CBE = 36: 72\\∴ ∠GCB + ∠CBE = 108° +72°=180°

5 0
3 years ago
Read 2 more answers
Use De Moivre’s Theorem to compute the following: <img src="https://tex.z-dn.net/?f=%5B3%28cos27%2Bisin27%29%7D%5E5" id="TexForm
zhannawk [14.2K]
Hello: 
<span>Use De Moivre’s Theorem : 
</span>(3(cos27 +isin27)^5 = 3^5( cos(27 × 5) +isin(27 × 5))
                                = 3^5 ( cos(135)+i sin(135))
                                = 3^5(-√2/2+i √2/2) 
because :  cos(135) = -√2/2    and   sin(135) = √2/2
(3(cos27 +isin27)^5 = (- 3^5√2/2)+ i ( 3^5√2/2)  ...(form : a+ib  when
a= (- 3^5√2/2)  and  b = ( 3^5√2/2)
4 0
3 years ago
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