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natulia [17]
2 years ago
15

The table of ordered pairs (x,y) gives an exponential function.

Mathematics
1 answer:
Andreyy892 years ago
8 0

The equation for the exponential function containing the points (0, 8) and (1, 32) is y = 8(4)ˣ

What is an equation?

An equation is an expression that shows the relationship between two or more numbers and variables.

An independent variable is a variable that does not depend on any other variable for its value while a dependent variable is a variable that depends on other variable.

An exponential function is in the form:

y = abˣ

Where a is the initial value and b is the multiplication factor.

At point (0, 8):

8 = ab⁰

a = 8

At point (1, 32):

32 = 8b

b = 4

The equation for the exponential function containing the points (0, 8) and (1, 32) is y = 8(4)ˣ

Find out more on equation at: brainly.com/question/2972832

#SPJ1

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Find the exact value of csc theta if tan theta = sqrt3 and the terminal side of theta is in Quadrant III.
WARRIOR [948]

Answer:

3rd option

Step-by-step explanation:

Using the identities

cot x = \frac{1}{tanx}

csc² x = 1 + cot² x

Given

tanθ = \sqrt{3} , then cotθ = \frac{1}{\sqrt{3} }

csc²θ = 1 + (\frac{1}{\sqrt{3} } )² = 1 + \frac{1}{3} = \frac{4}{3}

cscθ = ± \sqrt{\frac{4}{3} } = ± \frac{2}{\sqrt{3} }

Since θ is in 3rd quadrant, then cscθ < 0

cscθ = - \frac{2}{\sqrt{3} } × \frac{\sqrt{3} }{\sqrt{3} } = - \frac{2\sqrt{3} }{3}

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3 years ago
The diameter of a circle is two times its radius. Which
atroni [7]
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3 0
3 years ago
MATH HELP PLEASE!!! PLEASE HELP!!!
grandymaker [24]
The answer is:  [C]:  " f(c) = \frac{9}{5} c  + 32 " .
________________________________________________________

Explanation:

________________________________________________________
Given the original function:  

" c(y) = (5/9) (x <span>− 32) " ; in which "x = f" ; and "y = c(f) " ;
________________________________________________________
</span>→  <span>Write the original function as:  " y = </span>(5/9) (x − 32) " ; 

Now, change the "y" to an "x" ; and the "x" to a "y"; and rewrite; as follows:
________________________________________________________
    x = (5/9) (y − 32) ; 

Now, rewrite THIS equation; by solving for "y" ; in terms of "x" ; 
_____________________________________________________
→ That is, solve this equation for "y" ; with "c" as an "isolated variable" on the
 "left-hand side" of the equation:

We have:

→  x  =  " (  \frac{5}{9}  ) * (y − 32) " ;

Let us simplify the "right-hand side" of the equation:
_____________________________________________________

Note the "distributive property" of multiplication:
__________________________________________
a(b + c) = ab + ac ;  <u><em>AND</em></u>:

a(b – c) = ab – ac
.
__________________________________________

As such:
__________________________________________

" (\frac{5}{9}) * (y − 32) " ; 

=  [ (\frac{5}{9}) * y ]   −  [ (\frac{5}{9}) * (32) ] ; 


=  [ (\frac{5}{9}) y ]  − [ (\frac{5}{9}) * (\frac{32}{1})" ;

=  [ (\frac{5}{9}) y ]  − [ (\frac{(5*32)}{(9*1)} ] ; 

=  [ (\frac{5}{9}) y ]  −  [ (\frac{(160)}{(9)} ] ; 

= [ (\frac{5y}{9}) ]  −  [ (\frac{(160)}{(9)} ] ; 

= [ \frac{(5y-160)}{9} ] ;  
_______________________________________________
And rewrite as:  

→  " x  =  \frac{(5y-160)}{9} "  ;

We want to rewrite this; solving for "y";  with "y" isolated as a "single variable" on the "left-hand side" of the equation ;

We have:

→  " x  =  \frac{(5y-160)}{9} "  ; 

↔  " \frac{(5y-160)}{9} = x ; 

Multiply both sides of the equation by "9" ; 

 9 * \frac{(5y-160)}{9}  =  x * 9 ; 

to get:

→  5y − 160 = 9x ; 

Now, add "160" to each side of the equation; as follows:
_______________________________________________________

→  5y − 160 + 160 = 9x + 160 ; 

to get:

→  5y  =  9x + 160 ; 

Now,  divided Each side of the equation by "5" ; 
      to isolate "y" on one side of the equation; & to solve for "y" ; 

→  5y / 5  = (9y + 160) / 5 ; 

to get: 
 
→  y = (9/5)x + (160/5) ; 

→  y =  (9/5)x + 32 ; 

 →  Now, remember we had substituted:  "y" for "c(f)" ; 

Now that we have the "equation for the inverse" ;
     →  which is:  " (9/5)x  + 32" ; 

Remember that for the original ("non-inverse" equation);  "y" was used in place of "c(f)" .  We have the "inverse equation";  so we can denote this "inverse function" ; that is, the "inverse" of "c(f)" as:  "f(c)" .

Note that "x = c" ; 
_____________________________________________________
So, the inverse function is: "  f(c) = (9/5) c  + 32 " .
_____________________________________________________

 The answer is:  " f(c) = \frac{9}{5} c  + 32 " ;
_____________________________________________________
 →  which is:  

→  Answer choice:  [C]:  " f(c) = \frac{9}{5} c  + 32 " .
_____________________________________________________
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where <em>a</em> is some unknown constant. By definition of the derivative,

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