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3 years ago

9 folders cost $2.50 9 notebooks cost $4 write equivalent expressions and then find the total cost

Mathematics

1 answer:

Leokris [45]3 years ago

8 0

$2.50(f) + $4(n) = Total Cost

$2.50(9) + $4(9) = Total Cost

$22.50 + $36 = Total Cost

$58.50 = Total Cost

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Number of the day 3.59

sleet_krkn [62]

Please be a little specific.

7 0

3 years ago

Which statement best reflects the solution(s) of the equation?

Inessa [10]

x=2 is only solution while x=1 is extraneous solution

Option C is correct.

Step-by-step explanation:

We need to solve the equation $\frac{1}{x-1}+\frac{2}{x}=\frac{x}{x-1}$ and find values of x.

Solving:

Find the LCM of denominators x-1,x and x-1. The LCM is x(x-1)

Multiply the entire equation with x(x-1)

$\frac{1}{x-1}+\frac{2}{x}=\frac{x}{x-1}\\\frac{1}{x-1}*x(x-1)+\frac{2}{x}*x(x-1)=\frac{x}{x-1}*x(x-1)\\Cancelling\,\,out\,\,the\,\,same\,\,terms:\\x+2(x-1)=x^2\\x+2x-2=x^2\\3x-2=x^2\\x^2-3x+2=0$

Now, factoring the term:

$x^2-2x-x+2=0\\x(x-2)-1(x-2)=0\\(x-1)(x-2)=0\\x-1=0\,\,and\,\, x-2=0\\x=1\,\,and\,\, x=2$

The values of x are x=1 and x=2

Checking for extraneous roots:

Extraneous roots: The root that is the solution of the equation but when we put it in the equation the answer turns out not to be right.

If we put x=1 in the equation, $\frac{1}{x-1}+\frac{2}{x}=\frac{x}{x-1}$ the denominator becomes zero i.e

$\frac{1}{1-1}+\frac{2}{1}=\frac{1}{1-1}\\\frac{1}{0}+2=\frac{1}{0}$

which is not correct as in fraction anything divided by zero is undefined. So, x=1 is an extraneous solution.

If we put x=2 in the equation,

$\frac{1}{x-1}+\frac{2}{x}=\frac{x}{x-1}$

$\frac{1}{2-1}+\frac{2}{2}=\frac{2}{2-1}\\\frac{1}{1}+1=\frac{2}{1}\\1+1=2\\2=2$

So, x=2 is only solution while x=1 is extraneous solution

Option C is correct.

Keywords: Solving Equations and checking extraneous solution

Learn more about Solving Equations and checking extraneous solution at:

#learnwithBrainly

5 0

3 years ago

PLEASE HELP

balu736 [363]

Answer:

slope = 1

Step-by-step explanation:

-3 | -3

5 | 5

_____

8 | 8

8/8= 1

7 0

2 years ago

If two of these shapes are randomly chosen, one after the other without replacement, what is the probability that the first will

aleksandrvk [35]

Answer:

$P(T\ n\ S) = \frac{1}{14}$

Step-by-step explanation:

*Missing Part of the Question*

4 stars

5 triangles

3 circles

3 squares

Required

Determine the probability of triangle being first then square being second

$Total = 4 + 5 + 3 + 3$

$Total = 15$

Represent the triangle with T and square with S

So, we're solving for P(T n S)

$P(T\ n\ S) = P(T) * P(S)$

Solving for P(T)

$P(T) = \frac{n(T)}{Total}$

$P(T) = \frac{5}{15}$

Solving for P(S)

The question implies a probability without replacement;

Hence Total has now been reduced by 1

Total = 14

$P(S) = \frac{n(S)}{Total}$

$P(S) = \frac{3}{14}$

Recall that

$P(T\ n\ S) = P(T) * P(S)$

$P(T\ n\ S) = \frac{5}{15} * \frac{3}{14}$

$P(T\ n\ S) = \frac{15}{15 * 14}$

$P(T\ n\ S) = \frac{1}{14}$

Hence, the required probability is

$P(T\ n\ S) = \frac{1}{14}$

3 0

2 years ago

Why are different intervals used when restricting the domains of the sine and cosine functions in the process of defining their

umka21 [38]

Answer:

See below

Step-by-step explanation:

The interval must be chosen so that the function is one-to-one, and the sine and cosine functions are not one-to-one on the same intervals.

3 0

2 years ago