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stiv31 [10]
3 years ago
13

Write a quadratic equation in standard form that has 5 3 and 7 3 as its roots. A) 9x2 − 36x + 35 = 0 B) 9x2 + 36x − 35 = 0 C) 9x

2 + 36x + 35 = 0 D) 9x2 − 36x − 35 = 0
Mathematics
1 answer:
Serggg [28]3 years ago
5 0

Answer:

Option A)

9x^2 - 36x + 35 = 0

Step-by-step explanation:

We are given the following in the question:

Roots of quadratic equation are:

\alpha = \dfrac{5}{3}, \beta = \dfrac{7}{3}

The sum of the roots and the product of the roots can be calculated as:

\alpha + \beta = \dfrac{5+7}{3} = 4\\\\\alpha\beta = \dfrac{5}{3}\times \dfrac{7}{3} = \dfrac{5}{9}

Standard form of quadratic equation:

x^2-(\alpha + \beta)x+\alpha\beta = 0

Putting values, we get,

x^2 - 4x + \dfrac{35}{9} = 0\\\\9x^2 - 36x + 35 = 0

is the required quadratic equation.

Thus, the correct answer is

Option A)

9x^2 - 36x + 35 = 0

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Answer:

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Step-by-step explanation:

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Now, equation (1) is in the slope-intercept form and the slope of the line is 3.

Let, m is the slope of the required line.

So, 3m = -1

{Since, the product of the slopes of two perpendicular straight lines is -1}

⇒ m = - \frac{1}{3}

Therefore, the equation of the required line in slope intercept form is  

y = - \frac{1}{3} x + c {Where c is a constant}

Now, this above equation passes through the point (-9,2) point.

So, 2 = -  \frac{1}{3} \times (-9) + c

⇒ 2 = 3 + c

⇒ c = - 1  

Therefore, the equation of the required straight line is y = - \frac{1}{3}x - 1 (Answer)

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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
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