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Anna [14]
2 years ago
8

Series of math questions

Mathematics
1 answer:
nordsb [41]2 years ago
8 0

\huge\fbox{Answer ☘}

<h2>1 ) </h2>

when x = 3 ; y = 2

\bold\blue{✯\:\:\:2x + y}

substitue x = 3 and y = 2 in the given equation !

\bold{2(3) + 2 = 6 + 2 } \\  \bold{ =  > 8}

<h2>—————————————</h2>

<h2>2 ) </h2>

We've to divide 328 in the ratio of 1 : 3

let the respective distribution be x and 3x.

Then ,

\bold{x + 3x = 328} \\\bold{ 4x = 328 }\\ \bold{x =  \cancel\frac{328}{4} } \\  \\ \bold{x = 82}

the respective shares are :

x = $82

3x = $246

<h2>—————————————</h2>

<h2>3 ) </h2>

6 pens cost $90

therefore ,

cost of 1 pen =

\cancel \frac{90}{6}  = 15 \\

further ,

cost of 9 such pens = $15 × 9 = $135

<h2>—————————————</h2>

<h2>4 )</h2>

0.5 of a pound = 113.398 grams

<h2>—————————————</h2>

hope helpful~

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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):

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Which of the following are solutions to the question below? Check all that apply.
AveGali [126]

Answer:

D

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Given

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Answer:

Step-by-step explanation:

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terms:

6 + -3x = 8 + 5x + -10x

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6 + -3x = 8 + -5x

Solving

6 + -3x = 8 + -5x

Move all terms containing x to the left, all other terms to the right.  (Remember)

Add '5x' to each side of the equation.

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Then '-6' to each side of the equation.

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