Factor completely.

What are the solutions to the equation 2(x-3)^2=54 ?

erastova [34]

Answer:

x =(6-√108)/2=3-3√ 3 = -2.196

x =(6+√108)/2=3+3√ 3 = 8.196

Step-by-step explanation:

Step 1 :

Equation at the end of step 1 :

2 • (x - 3)2 - 54 = 0

Step 2 :

2.1 Evaluate : (x-3)2 = x2-6x+9

Step 3 :

Pulling out like terms :

3.1 Pull out like factors :

2x2 - 12x - 36 = 2 • (x2 - 6x - 18)

Adding 9 has completed the left hand side into a perfect square :

x2-6x+9 =

(x-3) • (x-3) =

(x-3)2 (x-3)1 =

x-3

Now, applying the Square Root Principle to Eq. #4.3.1 we get:

x-3 = √ 27

Add 3 to both sides to obtain:

x = 3 + √ 27

Since a square root has two values, one positive and the other negative

x2 - 6x - 18 = 0

has two solutions:

x = 3 + √ 27

or

x = 3 - √ 27

Solve Quadratic Equation using the Quadratic Formula

4.4 Solving x2-6x-18 = 0 by the Quadratic Formula .

According to the Quadratic Formula, x , the solution for Ax2+Bx+C = 0 , where A, B and C are numbers, often called coefficients, is given by :

- B ± √ B2-4AC

x = ————————

2A

In our case, A = 1

B = -6

C = -18

Accordingly, B2 - 4AC =

36 - (-72) =

108

Applying the quadratic formula :

6 ± √ 108

x = —————

2

Can √ 108 be simplified ?

Yes! The prime factorization of 108 is

2•2•3•3•3

To be able to remove something from under the radical, there have to be 2 instances of it (because we are taking a square i.e. second root).

√ 108 = √ 2•2•3•3•3 =2•3•√ 3 =

± 6 • √ 3

√ 3 , rounded to 4 decimal digits, is 1.7321

So now we are looking at:

x = ( 6 ± 6 • 1.732 ) / 2

Two real solutions:

x =(6+√108)/2=3+3√ 3 = 8.196

or:

x =(6-√108)/2=3-3√ 3 = -2.196

7 0

2 years ago

In the given figure L1and L2 are two parallel sides . if the area of the rectangle PQRS is 60cm^2 then what is the area of the p

kiruha [24]

Answer:

Step-by-step explanation:

Believe it or not, the two areas are the same.

The base of the rectangle is PQ

The height of the rectangle is PS

Now look at the parallelogram.

The base is PQ

The height is PS

The area has to be the same in both cases. There is no other way to interpret what is happening.

4 0

3 years ago

Help please! Step by steps

marin [14]

Area of the square bottom = L x W = 5 x 5 = 25 square cm

Area of triangle = 1/2 x base x height = 1/2 x 5 x 6 = 15 square cm

Total area = 25 + 15 = 40 square cm.

7 0

3 years ago

Jason went to a carnival where he could goon all rides for a flat fee of $30 but he had to pay $2 for each arcade game he played

enot [183]

Answer:

He played 7 arcade games.

Step-by-step explanation:

The amount paid in relation to the number of games played can be modeled by a linear function in the following format:

$A(n) = F + gn$

In which F is the flat rate and g is the price of each game.

Flat fee of $30 but he had to pay $2 for each arcade game he played.

This means that $F = 30, g = 2$

So

$A(n) = 30 + 2n$

Jason spent $44. How many arcade games did he play?

This is n for which A(n) = 44. So

$A(n) = 30 + 2n$

$44 = 30 + 2n$

$2n = 14$

$n = \frac{14}{2}$

$n = 7$

He played 7 arcade games.

4 0

2 years ago

A consumer products company is formulating a new shampoo and is interested in foam height (in mm). Foam height is approximately

Genrish500 [490]

Answer:

a) 0.057

b) 0.5234

c) 0.4766

Step-by-step explanation:

a)

To find the p-value if the sample average is 185, we first compute the z-score associated to this value, we use the formula

$z=\frac{\bar x-\mu}{\sigma/\sqrt N}$

where

$\bar x=mean\; of\;the \;sample$

$\mu=mean\; established\; in\; H_0$

$\sigma=standard \; deviation$

N = size of the sample.

So,

$z=\frac{185-175}{20/\sqrt {10}}=1.5811$

$\boxed {z=1.5811}$

As the sample suggests that the real mean could be greater than the established in the null hypothesis, then we are interested in the area under the normal curve to the right of 1.5811 and this would be your p-value.

We compute the area of the normal curve for values to the right of 1.5811 either with a table or with a computer and find that this area is equal to 0.0569 = 0.057 rounded to 3 decimals.

So the p-value is

$\boxed {p=0.057}$

b)

Since the z-score associated to an α value of 0.05 is 1.64 and the z-score of the alternative hypothesis is 1.5811 which is less than 1.64 (z critical), we cannot reject the null, so we are making a Type II error since 175 is not the true mean.

We can compute the probability of such an error following the next steps:

Step 1

Compute $\bar x_{critical}$