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3 years ago

What are the answers to the last two questions?

Mathematics

1 answer:

ivanzaharov [21]3 years ago

8 0

The answer to 15 is addition and multiplication

the answer to 16 is 0

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Solve for x<br> <img src="https://tex.z-dn.net/?f=%5Cfrac%7B1%7D%7Bx-9%7D" id="TexFormula1" title="\frac{1}{x-9}" alt="\frac{1}{

Anna [14]

r = 1/15

a branliest from the answer will be appreciated

8 0

3 years ago

Read 2 more answers

To decorate the gym for a dance, Zoe cut crepe paper into 10-foot strips. She attached one end of each strip 8 feet up on the wa

ohaa [14]

Answer:

there is no question

8 0

3 years ago

PLEASE HELPP!!! I HAVE 3 MINUTES

Rzqust [24]

Answer:

340 - 25 = 315

315/45 = 7

7 days

4 0

3 years ago

Consider the functions f and g defined by \[f(x) = \sqrt{\dfrac{x+1}{x-1}}\qquad\qquad\text{and}\qquad\qquad g(x) = \dfrac{\sqrt

tino4ka555 [31]

Answer:

The given functions are not same because the domain of both functions are different.

Step-by-step explanation:

The given functions are

$f(x)= \sqrt{\dfrac{x+1}{x-1}}$

$g(x) = \dfrac{\sqrt{x+1}}{\sqrt{x-1}}$

First find the domain of both functions. Radicand can not be negative.

Domain of f(x):

$\dfrac{x+1}{x-1}>0$

This is possible if both numerator or denominator are either positive or negative.

Case 1: Both numerator or denominator are positive.

$x+1\geq 0\Rightarrow x\geq -1$

$x-1\geq 0\Rightarrow x\geq 1$

So, the function is defined for x≥1.

Case 2: Both numerator or denominator are negative.

$x+1\leq 0\Rightarrow x\leq -1$

$x-1\leq 0\Rightarrow x\leq 1$

So, the function is defined for x≤-1.

From case 1 and 2 the domain of the function f(x) is (-∞,-1]∪[1,∞).

Domain of g(x):

$x+1\geq 0\Rightarrow x\geq -1$

$x-1\geq 0\Rightarrow x\geq 1$

So, the function is defined for x≥1.

So, domain of g(x) is [1,∞).

Therefore, the given functions are not same because the domain of both functions are different.

4 0

4 years ago

9.84615384615 in a fraction

choli [55]

Answer:

Step-by-step explanation:

128/13

3 0

3 years ago