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andriy [413]
2 years ago
8

sewer lines in a system of equations with no solution have slopes that must be the same or different ?​

Mathematics
1 answer:
Paul [167]2 years ago
3 0

Answer:

Same slopes

Step-by-step explanation:

If the lines don't have the exact same slope then they will intersect at a point.

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Super easy question just don’t know it
Doss [256]

Answer:

Integer

Step-by-step explanation:

6 0
3 years ago
What is the value of the expression i 0 × i 1 × i 2 × i 3 × i 4?
astraxan [27]
ANSWER


The value of the expression is
- 1


EXPLANATION

Method 1: Rewrite as product of
{i}^{2}


The expression given to us is,

{i}^{0}  \times {i}^{1}  \times {i}^{2}  \times {i}^{3}  \times {i}^{4}


We use the fact that
{i}^{2}  =  - 1
to simplify the above expression.



{i}^{0}  \times {i}^{1}  \times {i}^{2}  \times {i}^{3}  \times {i}^{4}  =  {i}^{0}  \times {i}^{1}  \times {i}^{3}   \times {i}^{2}   \times {i}^{4}


This implies,


{i}^{0}  \times {i}^{1}  \times {i}^{2}  \times {i}^{3}  \times {i}^{4}  =  {i}^{0}  \times {i}^{2}  \times {i}^{2}   \times {i}^{2}   \times {i}^{2} \times {i}^{2}


We substitute to obtain,

{i}^{0}  \times {i}^{1}  \times {i}^{2}  \times {i}^{3}  \times {i}^{4}  =  1\times  - 1 \times  - 1  \times  - 1\times  - 1 \times  - 1


{i}^{0}  \times {i}^{1}  \times {i}^{2}  \times {i}^{3}  \times {i}^{4}  =  1\times  1 \times   1  \times  - 1 =  - 1


Method 2: Use indices to solve.



{i}^{0}  \times {i}^{1}  \times {i}^{2}  \times {i}^{3}  \times {i}^{4}  = {i}^{0 + 1 + 2 + 3 + 4}



This implies that,


{i}^{0}  \times {i}^{1}  \times {i}^{2}  \times {i}^{3}  \times {i}^{4}  = {i}^{10}




{i}^{0}  \times {i}^{1}  \times {i}^{2}  \times {i}^{3}  \times {i}^{4}  =  (  {{i}^{2}} )^{5}


{i}^{0}  \times {i}^{1}  \times {i}^{2}  \times {i}^{3}  \times {i}^{4}  =  (   - 1 )^{5}   =  - 1


8 0
3 years ago
Read 2 more answers
Y^2 - 12y + 27 / y^2 - 6y - 27 <br> simply the rational equation.
Ulleksa [173]

The simplified rational expression is (y - 3)/(y + 3). Where y ≠ -3.

<h3>How to simplify a rational expression?</h3>

A rational expression is in the p/q form. Where p and q are polynomial functions.

To simplify this rational equation,

  • Factorize the polynomials in both numerator and denomiantor.
  • Cancel out common factors if any.
  • If the denominator and the numerator have no common factors except 1, then that is said to be the simplest form of the given rational expression.

<h3>Calculation:</h3>

The given rational equation is

\frac{y^2 - 12y + 27 }{y^2 - 6y - 27}

Factorizing the expression in the numerator:

y² - 12y + 27 = y² - 9y - 3y + 27

⇒ y(y - 9) - 3(y - 9)

⇒ (y - 3)(y - 9)

Factorizing the expression in the denominator:

y² - 6y - 27 = y² - 9y + 3y - 27

⇒ y(y - 9) + 3(y - 9)

⇒ (y + 3)(y - 9)

Since they have (y - 9) as the common factor, we can simplify,

\frac{y^2 - 12y + 27 }{y^2 - 6y - 27}=\frac{(y-3)(y-9)}{(y+3)(y-9)}

⇒ (y - 3)/(y + 3) where y ≠ -3(denomiantor)

Here there are no more common factors except 1; this is the simplest form of the given rational expression.

Learn more about simplifying rational expressions here:

brainly.com/question/1928496

#SPJ9

3 0
1 year ago
Solve for h in the literal equation g(h+3/2)=5
joja [24]

Answer:  h= 5/g−3/2

Step-by-step explanation:

h=  5 over g minus   3 over 2

5 0
2 years ago
Read 2 more answers
LAST ATTEMPT! MARKING AS BRAINLIEST!! ( simplifying radicals)
klasskru [66]

\huge \bf༆ Answer ༄

Let's solve ~

\qquad \looparrowright \sf - 2 \sqrt{192 {k}^{4} }

\qquad \looparrowright \sf - 2 \sqrt{2 {}^{6} \times   3{k}^{4} }

\qquad \looparrowright \sf - 2 \times 2 {}^{3}  \times  {k}^{2}   \times \sqrt{3}

\qquad \looparrowright \sf  - 16 {k}^{2}  \sqrt{3}

Hence, the correct choice is D

5 0
2 years ago
Read 2 more answers
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