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2 years ago

9% of 750............................

Mathematics

2 answers:

babunello [35]2 years ago

8 0

67.5 is your answer.

KatRina [158]2 years ago

7 0

67.5

thats the answer

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A family paid $30600 as a down payment for a home. If this represents %17 of the price of the home, find the price of the hom

snow_lady [41]

Answer:

180000

Step-by-step explanation:

30600/17=1800

1800•100=180000

7 0

3 years ago

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La raíz cuadrada de 196

Verizon [17]

14 i hope this is what you were looking for

4 0

3 years ago

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Identify any extraneous solution. See above pic

Sav [38]

Answer:

There are no extraneous solutions

Reasoning:

An extraneous solution is a solution that isn't valid, it might be imaginary like the square root of a negative number.

first we want to isolate z:

1+sqrt(z)=sqrt(z+5)

^2 all ^2 all

(1+sqrt(z))(1+sqrt(z))=z+5

expand

1+2sqrt(z)+z=z+5

-1 -z -z -1

2sqrt(z)=4

/2 /2

sqrt(z)=2

^2 all ^2 all

z=4

Since there is one solution and it is a real number, there are no extraneous solutions.

4 0

3 years ago

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1/2x-2 <br> One half multiply negative 2

sattari [20]

Answer:

Step-by-step explanation:

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2 years ago

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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

3 years ago