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1 year ago

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In quadrilateral ABCD, AD is congruent to BC, and AD is parallel to BC. Andre has written a

1 answer:

Dahasolnce [82]1 year ago

5 0

Answer:

AD is Congruent to BC and it's given, (given just means that it was already said or stated, and you don't need to do the work to find it) Angle DAC would be congruent to angle BAC (if that doesn't work, rearrange them to look like angle CAB). AC would be congruent to DB. Triangle ADC would be congruent to triangle BCD by (since i don't know exactly which way the letters are arranged is would either be SAS or SSA) and that's because you know two of the sides are congruent to each other and one angle.

I tried hard to sketch out what the shape looked like based on the information given, and that's because I need a visual of what the shape looks like. Sorry, this took so long to answer.

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B - (- 7) =-1<br> what is b?

irina1246 [14]

B-(-7)= -1

First negative multiply by negative is POSITIVE.

So: B+7= -1

Subtract 7 from both sides

B= -7-1

B= -8.

Answer: -8.

To Check Answer: Plug in the variable of B in the original equation.

So: -8-(-7)= -1

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If the Solution works out the answer is right.

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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

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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.

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