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pishuonlain [190]
3 years ago
9

Marsha has some math exercises to do for homework. She did one half during study period and two thirds of those remaining while

waiting for a friend after school. He had three to finish at home that evening. How many exercises did she have for homework?
Mathematics
2 answers:
Paha777 [63]3 years ago
8 0

Answer: She has 1 5/6 exercises for homework.

Step-by-step explanation:

1/2 + 2/3 = 1 1/6

2 * 3 = 6 3 * 2 = 6

1 * 3 = 3 2 * 2 = 4

4 + 3/ 3 + 3

7/6 = 1 1/6

So, she finished 1 1/6 of her problems.

3 - 1 1/6 = 1 5/6

3 - 1 = 2

2 - 1/6 = 1 5/6

pshichka [43]3 years ago
3 0

Answer:  18

Step-by-step explanation:

Let x represents the number of exercises she had for homework.

Then, the expression for the number of exercises she did during study period =\dfrac{x}{2}

Remaining exercise = x-\dfrac{x}{2}=\dfrac{x}{2}

The number of exercises she did while waiting for a friend after school=\dfrac{2}{3}\times\dfrac{x}{2}=\dfrac{x}{3}

Also, she had three to finish at home that evening.

Then, we have

x=\dfrac{x}{2}+\dfrac{x}{3}+3\\\\\Rightarrow\ x-\dfrac{x}{2}-\dfrac{x}{3}=3\\\\\Rightarrow\dfrac{6x-3x-2x}{6}=3\\\\\Rightarrow\dfrac{x}{6}=3\\\\\Rightarrow\ x=6\times3=18

Hence, she had 18 exercises for homework.

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3 years ago
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Alenkinab [10]
<h2>Answer:</h2>

y=\pm (\frac{x}{6})^{\frac{1}{4}} \ is \ not \ a \ function

<h2>Step by step solution:</h2>

A function f from a set A to a set B is a relation that assigns to each element x in the set A exactly one element y in the set B. The set A is the domain (also called the set of inputs) of the function and the set B contains the range (also called the set of outputs). On the other hand, a function has an inverse function if and only if passes the Horizontal Line Test for Inverse Functions. This test tells us that a function f has an inverse function if and only if there is no any horizontal line that intersects the graph of f at more than one point. So the function is called one-to-one. The graph of f is shown below. As you can see, this function does not pass the Horizontal Line Test, therefore the inverse is not a function. However, let's find f^-{1}(x):

f(x)=6x^4 \\ \\ Substitute \ f(x) \ by \ y \\ \\ y=6x^4 \\ \\ Interchange \ x \ and \ y: \\ \\ x=6y^4 \\ \\ Solve \ for \ y: \\ \\ y^4=\frac{x}{6} \\ \\ Solving \\ \\y=\pm \sqrt[4]{\frac{x}{6}} \\ \\ \boxed{y=\pm \left(\frac{x}{6}\right)^{\frac{1}{4}}}

and this is not a function because there are elements in the set of inputs that match with two elements in the set of outputs.

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3 years ago
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