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3 years ago

Yuto and Hila attempted to solve the same inequality. Their work is shown below.

Mathematics

2 answers:

Zina [86]3 years ago

8 0

Answer:

The answer is A

Step-by-step explanation:

In other words, Yuto is correct because he isolated the variable correctly and reversed the inequality symbol.

Charra [1.4K]3 years ago

5 0

Answer:

Yuto is correct because he isolated the variable correctly and reversed the inequality symbol.

Step-by-step explanation: hi

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Solve :1/1-x^a-b +1/1-x^b-a

Olegator [25]

Answer:

Step-by-step explanation:

$\frac{1}{1-x^{a-b}}+\frac{1}{1-x^{b-a}}\\\\\frac{1}{1-x^{b-a}}=\frac{1}{1-x^{-(a-b)}}=\frac{1}{1-\frac{1}{x^{a-b}}}\\=\frac{1}{\frac{x^(a-b)-1}{x^(a-b)}}=\frac{x^{a-b}}{x^{a-b}-1}\\\\=\frac{x^{a-b}}{-(1-x^{a-b})}=-\frac{x^{a-b}}{1-x^{a-b}}\\\\\frac{1}{1-x^{a-b}}+\frac{1}{1-x^{b-a}}=\frac{1}{1-x^{a-b}}-\frac{x^{a-b}}{1-x^{a-b}}\\\\=\frac{1-x^{a-b}}{1-x^{a-b}}=1$

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3 years ago

Expand the expression.<br> 0.2(y + 2)

Stolb23 [73]

Answer:

x=0.2(y+2)

Step-by-step explanation:

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3 years ago

an airplane 30000 feet above the ground begins descending at a rate of 2000 feet per minute. write an equation to model the situ

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A = 30,000 - 2000m where A=altitude and m = number of minutes

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3 years ago

Given the function f(x) x^2-2x/x^3 +9x^2-10x Find any holes, vertical asymptotes, and horizontal asymptotes there may be.

yulyashka [42]

Answer:

See below

Step-by-step explanation:

I assume you mean $f(x)=\frac{x^2-2x}{x^3+9x^2-10x}$ :

Holes: Since $f(x)=\frac{x^2-2x}{x^3+9x^2-10x}$ reduces to $f(x)=\frac{x(x-2)}{x(x^2+9x-10)}$ , then there is a hole at $x=0$ as $x$ exists in both the numerator and denominator (however, its limit as x approaches 0 is 1/5).

Vertical Asymptotes: If we further reduce $f(x)=\frac{x(x-2)}{x(x^2+9x-10)}$ to $f(x)=\frac{x-2}{(x+10)(x-1)}$ , then we see that there are vertical asymptotes at $x=-10$ and $x=1$

Horizontal Asymptotes: As the degree of the numerator is less than the degree of the numerator ( $2$ ), then there is a horizontal asymptote at $y=0$

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2 years ago