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Alenkinab [10]
3 years ago
15

Use the quadratic formula to find both solutions to the quadratic equation given below x^2+6x=27

Mathematics
2 answers:
Dmitriy789 [7]3 years ago
8 0

Answer:

option (d) and (f) is correct.

The solution of given quadratic equation is 3 and -9.

Step-by-step explanation:

Given quadratic equation x^2+6x=27

We have to solve the given quadratic equation using quadratic formula.

Consider x^2+6x=27 , we can rewrite it as x^2+6x-27=0

For the general quadratic equation ax^2+bx+c=0 the quadratic formula is given as x_{1,\:2}=\frac{-b\pm \sqrt{b^2-4ac}}{2a}

Here a = 1 , b= 6 and c = -27

Substitute, we get,

x_{1,\:2}=\frac{-6\pm \sqrt{6^2-4\cdot \:1\left(-27\right)}}{2\cdot \:1}

Solving further , we get,

x_{1,2}=\frac{-6\pm\sqrt{6^2+4\cdot \:1\cdot \:27}}{2\cdot \:1}

x_{1,2}=\frac{-6\pm\sqrt{144}}{2\cdot \:1}

We know \sqrt{144}=12, we get,

x_{1,2}=\frac{-6\pm 12}{2\cdot \:1}

x_{1}=\frac{-6+12}{2} and x_{2}=\frac{-6-12}{2}

Solving we get,

x_{1}=\frac{6}{2} and x_{2}=\frac{-18}{2}

x_{1}=3 and x_{2}=-9

Thus, the solution of given quadratic equation is 3 and -9.

Thus, option (d) and (f) is correct.

 

docker41 [41]3 years ago
7 0

Answer:

x=3 or x=−9

Step-by-step explanation:

Step 1: Subtract 27 from both sides.

x2+6x−27=27−27

x2+6x−27=0

Step 2: Factor left side of equation.

(x−3)(x+9)=0

Step 3: Set factors equal to 0.

x−3=0 or x+9=0

x=3 or x=−9

Answer:

x=3 or x=−9

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Step-by-step explanation:

Given that :

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H_o: \mu_1 = \mu_2

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This is because of the ≠ sign in the alternative hypothesis which signifies that the rejection region in the alternative hypothesis are at the both sides of the hypothesized mean difference .

Decision Rule: at the level of significance ∝ = 0 . 10

The decision rule is to reject the null hypothesis if z < - 1 . 64 and z > 1 . 64

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3 years ago
If tanA=a <br>then find sin4A-2sin2A/ sin4A+2sin2A​
anygoal [31]

Answer:

The value of the given expression is

\frac{sin4A-2sin2A}{sin4A+2sin2A}=-a^2

Step by step Explanation:

Given that tanA=a

To find the value of \frac{sin4A-2sin2A}{sin4A+2sin2A}

Let us find the value of the expression :

\frac{sin4A-2sin2A}{sin4A+2sin2A}=\frac{2cos2Asin2A-2sin2A}{2cos2Asin2A+2sin2A} ( by using the formula sin2A=2cosAsinA here A=2A)  

=\frac{2sin2A(cos2A-1)}{2sin2A(cos2A+1)}

=\frac{(cos2A-1)}{(cos2A+1)}

=\frac{(-(1-cos2A))}{(1+cos2A)}(using  sin^2A+cos^2A=1  here A=2A)

=\frac{-(sin^2A+cos^2A-(cos^2A-sin^2A))}{sin^2A+cos^2A+(cos^2A-sin^2A)}(using cos2A=cos^2A-sin^2A here A=2A)

=\frac{-(sin^2A+cos^2A-cos^2A+sin^2A)}{sin^2A+cos^2A+(cos^2A-sin^2A)}

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Therefore \frac{sin4A-2sin2A}{sin4A+2sin2A}=-a^2

6 0
3 years ago
There are 5 green marbles, 4 red marbles, 3 blue marbles and 3 yellow marbles in a bag. Bryan select's a marble at random. Deter
Taya2010 [7]

Answer:

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We can tell if this is a factor of the polynomial shown by using the remainder theorem.

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x + 1 = 0

Now, we solve for x.

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3 years ago
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