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Zolol [24]
3 years ago
11

Evaluate the expression (x - y)^2 – x when x=3 and y=8 HELP PLEASE‼️‼️

Mathematics
2 answers:
vovikov84 [41]3 years ago
3 0

Answer:

22

Step-by-step explanation:

Step 1: fill in the equation

(3-8)^2-3

Step 2: simplify

= -5^2-3

= 25-3

Step 3: solve

= 22

Doss [256]3 years ago
3 0

Answer:

This would change into (9 - 64) - 3 which would change to, -55 - 3 = -58

Step-by-step explanation:

Hope this helped!

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Answer:

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Step-by-step explanation:

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2 years ago
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
Suppose z varies directly with x and inversely with the
bija089 [108]

Answer:

K = 27

z = 1.89

Step-by-step explanation:

Check attachment below

5 0
4 years ago
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3 years ago
One leg of a 45°-45°-90° triangle is 3.<br> Find the length of the hypotenuse.
alexira [117]
The answer should be 3 square root 2
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3 years ago
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