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stiv31 [10]
3 years ago
7

What is the quotient of 5s ad 8

Mathematics
1 answer:
butalik [34]3 years ago
4 0

Answer:

0.625 s

Step-by-step explanation:

5s/8

0.625 s

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Three teachers have provided you with an equation that models their salaries in million of dollars each year. teacher
Yuliya22 [10]
Teacher c makes the most money at the end of their career since the model is entirely linear. 
4 0
3 years ago
In relation t a functions? Is the inverse of relation t function
frozen [14]

Answer:

Relation t is a function. The inverse of relation t is not a function ⇒ 3rd

Step-by-step explanation:

* Lets explain how to solve the problem

- A relation is a set of inputs and outputs, and a function is a relation

 with one output for each input

- Ex: T = {(1 , 2) , (3 , 5) , (-4 , 0)} is a function because every input has

  only one output

- To find the inverse of a function we switched the input and the

  out put

- The inverse of a function may not always be a function

* Lets solve the problem

∵ The relation t is the set of ordered pairs of x and y

  x = 0     2      4    6

  y = -8    -7    -4    -4

∵ x = 0 has only y = -8

∵ x = 2 has only y = -7

∵ x = 4 has only y = -4

∵ x = 6 has only y = -4

∴ Every value of x has only one value of y

∴ Relation t is a function

* Lets find its inverse by switching x and y

∵ The inverse function of t is :

   x = -8     -7     -4    -4

   y =  0      2      4     6

∵ x = -8 has only y = 0

∵ x = -7 has only y = 2

∵ x = -4 has y = 4 and y = 6

∴ Not every value of x has only one value of y

∴ The inverse of relation t is not a function

* Relation t is a function. The inverse of relation t is not a function

4 0
3 years ago
Identify the slope in -4x+3y=26​
Strike441 [17]
-4 is the slope in the equation
3 0
3 years ago
Read 2 more answers
For what value of constant c is the function k(x) continuous at x = 0 if k =
nlexa [21]

The value of constant c for which the function k(x) is continuous is zero.

<h3>What is the limit of a function?</h3>

The limit of a function at a point k in its field is the value that the function approaches as its parameter approaches k.

To determine the value of constant c for which the function of k(x)  is continuous, we take the limit of the parameter as follows:

\mathbf{ \lim_{x \to 0^-} k(x) =  \lim_{x \to 0^+} k(x) =  0 }

\mathbf{\implies  \lim_{x \to 0 } \ \  \dfrac{sec \ x - 1}{x}= c }

Provided that:

\mathbf{\implies  \lim_{x \to 0 } \ \  \dfrac{sec \ x - 1}{x}= \dfrac{0}{0} \ (form) }

Using l'Hospital's rule:

\mathbf{\implies  \lim_{x \to 0} \ \  \dfrac{\dfrac{d}{dx}(sec \ x - 1)}{\dfrac{d}{dx}(x)}=  \lim_{x \to 0}   sec \ x  \ tan \ x = 0}

Therefore:

\mathbf{\implies  \lim_{x \to 0 } \ \  \dfrac{sec \ x - 1}{x}=0 }

Hence; c = 0

Learn more about the limit of a function x here:

brainly.com/question/8131777

#SPJ1

5 0
2 years ago
According to cell theory, all cells come from:
kondor19780726 [428]

Answer:

pre-existing cells

Step-by-step explanation:

5 0
4 years ago
Read 2 more answers
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