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Sergio039 [100]
3 years ago
11

Consider the equation |(x-3)^2-4|=b. Determine the value of b so that the equation has no solution.

Mathematics
2 answers:
oee [108]3 years ago
3 0
As an absolute value makes any value positive, any negative value of b will cause the equation to be unsolvable. 
saul85 [17]3 years ago
3 0
<span>|(x-3)^2-4|=b  since this is an absolute value equation

b can have any value less than zero so that the above is never true, in interval notation:

b=(-oo,0)</span>
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(x⁴)(x³)

x⁷

(12x⁵)/(36x)

x⁴/3

(x⁴y³)/(x⁴y)

y²

(-2x³)(-4x²)

8x⁵

(-3x²)³

-27x⁶

(x³y⁵)/(xy²)

x²y³

2x³(10x)³

2000x⁶

(x⁷y²)/(x⁴y²)

x³

x⁻⁴

1/x⁴

(-21x⁵y²)/(7x⁴y⁵)

-(3x)/y³

3x⁻³

3/x³

(32x³y²z⁵)/(-8xyz²)

-4x²yz³

...

x⁴y⁴

(4x⁷/7y²)²

(16x¹⁴)/(49y⁴)

...

8x³

4⁻⁴

1/(4⁴) or 1/256

...

a⁵

8⁻²

1/(8²) or 1/64

...

2c⁷

x⁻²

1/x²

...

9x²

x⁻³

1/x³

...

a⁴

x⁻⁴

1/x⁴

...

4c⁵

x²y⁻³

(x²)/(y³)

When multiplying monomials with the same base, ___________ the exponents

add

x³y⁻²

(x³)/(y²)

When dividing monomials with the same base, ______________ the exponents

subtract

x²y³z⁻⁴

(x²y³)/(z⁴)

When raising a power to a power, ________________ the exponents

multiply

x²y⁻³z⁴

(x²z⁴)/(y³)

x⁻²y⁻³

1/(x²y³)

A base raised to a zero exponent equals________________

one

(2m²)(2m³)

4m⁵

(x/y)⁻¹

y/x

4r⁻³(2r²)

8÷r

(x²/y²)⁻¹

y²/x²

2x³y⁻³(2x⁻¹y³)

4x²

(x³/y³)⁻¹

y³/x³

2y²(3x)

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(x/y)⁻²

y²/x²

4a³b²(3a⁻⁴b⁻³)

12÷(ab)

(x²/y²)⁻³

y⁶/x⁶

4r⁰

4

(2/3)⁻²

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(4xy)⁻¹

1÷(4xy)

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(3/4)⁻²

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(3/4)⁻¹

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(x³/x⁻⁶)

x⁹

(x²/x⁻⁵)

x⁷

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1

y⁰

1

x²y⁰

x²

x²y³z⁰

x²y³

y⁰

1

100⁰

1

(-99)⁰

1

x⁰(x⁴)(x⁻⁶)

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(x⁻³)/(x⁴)

1/x⁷

(x⁻⁴)/(x⁵)

1/x⁹

(4x³/2x⁵)⁰

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2 years ago
What is the scale factor of the two triangles below?
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The correct answer is:  [A]:  " \frac{3}{4} " .
______________________________________________________

<u>Note</u>:  "3/4"  =  "6/8"  =  "15/20" .
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