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Tcecarenko [31]

2 years ago

15

The manager of a store increases the price of a popular product by 5%. Let y be the original price of the product. The new price

is t + 0.05t. Find an expression equivalent to t + 0.05t. If the original price was $22.99, what is the new price?

1 answer:

Sphinxa [80]2 years ago

5 0

Answer:

you will multiply the new price plus the original price

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Can anybody tell me the answers to number 1, 2 & 3 ???!!!!

Novay_Z [31]

Problem 1

Domain = {-1, -3, 2, 1}

Range = {5, 0, 2}

The domain is the set of possible inputs and the range is the set of possible outputs. This is a function because each input goes to exactly one output.

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Problem 2

This is a function as well. We do not have any input going to multiple outputs.

Domain = {-2, -3, 5}

Range = {6, 7, 8}

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Problem 3

This is not a function. The input -4 goes to more than one output (outputs 3 and -1 at the same time)

Domain = {-4, -2, 0}

Range = {3, -1, -2, 4}

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3 years ago

1 slope y-intercept equation o 6 5 4 3 2 2 3 5 j-intercept

mars1129 [50]

Answer:

Step-by-step explanation:

(-1,1) and (0,-2) are points on the line

slope of line = (-2-1)/(0-(-1)) = -3

y-intercept = -2

y = -3x - 2

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3 years ago

HURRY PLEASE BEST ANSWER GETS BRAINLIEST!

nekit [7.7K]

Answer:

The value of the second expression is 12, so the expressions are not equivalent.

Step-by-step explanation:

3 0

3 years ago

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HAVE A NICE DAY <br> WILL GIVE BRAINLST :)

muminat

Answer:

1) 2 and 6 ; 4 and 8 ; 1 and 5 ; 3 and 7

2) 3 and 6 ; 4 and 5

3) 1 and 8 ; 2 and 7

7 0

3 years ago

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mina [271]

a) Since both limits are <em>distinct</em> and do not exist, we conclude that x = - 1 is not part of the domain of the <em>rational</em> function.

b) The function $f(x) = \frac{x}{x^{2}+ x}$ is equivalent to the function $g(x) = \frac{1}{x + 1}$ .

<h3>How to determine whether a limit exists or not</h3>

According to theory of limits, a function f(x) exists for x = a if and only if $\lim_{x\to a^{-}} f(x) = \lim_{x \to a^{+}} f(x)$ . This criterion is commonly used to prove continuity of functions.

<em>Rational</em> functions are not continuous for all value of x, as there are x-values that make denominator equal to 0. Based on the figure given below, we have the following <em>lateral</em> limits:

$\lim_{x \to -1^{-}} \frac{x}{x^{2}+x} = - \infty$

$\lim_{x \to -1^{+}} \frac{x}{x^{2}+x} = + \infty$

Since both limits are <em>distinct</em> and do not exist, we conclude that x = - 1 is not part of the domain of the <em>rational</em> function.

In addition, we can simplify the function by <em>algebra</em> properties:

$\frac{x}{x^{2}+ x} = \frac{x}{x\cdot (x + 1)} = \frac{1}{x + 1}$

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

The function $f(x) = \frac{x}{x^{2}+ x}$ is equivalent to the function $g(x) = \frac{1}{x + 1}$ .

To learn more on lateral limits: brainly.com/question/21783151

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7 0

2 years ago