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2 years ago

2. √2x-5-√x+6=0(solving the following radical equations and check for extraneous solutions)

Mathematics

1 answer:

Norma-Jean [14]2 years ago

5 0

I assume the equation is

√(2x - 5) - √(x + 6) = 0

Note the domains for the root expressions:

• √(2x - 5) : 2x - 5 ≥ 0 ⇒ x ≥ 5/2

• √(x + 6) : x + 6 ≥ 0 ⇒ x ≥ -6

So any valid solution we find must be at least 5/2.

Move one term to the other side.

√(2x - 5) = √(x + 6)

Take squares.

(√(2x - 5))² = (√(x + 6))²

2x - 5 = x + 6

Solve for x :

x = 11

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Simplify the following expression : (2x + 7y) (x – 3y) – (9x + 4) (3x - 2)

Dmitriy789 [7]

-25x^2+xy-21y^2+6x+8

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3 years ago

You are measuring the height of a statue. You stand 17 feet from the base of the statue. You measure the angle of elevation from

olga_2 [115]

Answer:

height ≈ 31 ft

Step-by-step explanation:

The illustration below will form a right angle triangle. He is standing 17 ft away from the base of the statue. The angle of elevation from the ground to the top of the statue is 61°. The height of the statue can be computed below.

adjacent side of the triangle = 17 ft

opposite side = a

Using SOHCAHTOA principle

tan 61° = opposite/adjacent

tan 61° = a/17

cross multiply

a = 17 tan 61°

a = 17 × 1.80404775527

a = 30.6688118396

a ≈ 31 ft

6 0

3 years ago

A park is 4 times as it is wide. If the distance around the park is 12.5 kilometers what is the area of the park

MaRussiya [10]

Answer:

6.25 km

Step-by-step explanation:

Here is the correct question: A park is 4 times as long as it is wide. If the distance around the park is 12.5 kilometers, what is the area of the park?

Given: Perimeter (Distance) of the park= 12.5 km

Considering park is in rectangular shape.

Let the width of park be x

∴ as given length will be 4x.

Formula for perimeter of rectangle = $2\times (length + width)$

Perimeter is given 12.5 km

⇒ $12.5= 2\times (4x+x)$

⇒ $12.5= 10x$

∴ x= 1.25 km, which means width is 1.25 km and length is 5 km.

Now, finding the area of park

Formula; Area of rectangle= $length\times width$

∴ Area of rectangle= $5\times 1.25 = 6.25 km$

∴Area of park will be 6.25 km.

6 0

3 years ago

According to this diagram, what is cos 28°? 62° 17 8 28° 90° 15 O A. 17 15 O B. 8 17 O c. 15 17 O D. O E. 8 15 15 OF.

VikaD [51]

Answer:

C. $Cos(28) = \frac{15}{17}$

Step-by-step explanation:

Recall: SOHCAHTOA

Thus,

$Cos(\theta) = \frac{Adjacent}{Hypotenuse}$

Reference angle = 28°

Adjacent = 15

Hypotenuse = 17

Plug in the values

$Cos(28) = \frac{15}{17}$

4 0

3 years ago

A research engineer for a tire manufacturer is investigating tire life for a new rubber compound and has built 16 tires and test

frutty [35]

Answer:

The 95% confidence interval for mean tire life is between 52,368.4 kilometers and 67,911 kilometers.

Step-by-step explanation:

We are in posession of the sample's standard deviation, so we use the students t-distribution to solve this question.

The first step to solve this problem is finding how many degrees of freedom, we have. This is the sample size subtracted by 1. So

df = 16 - 1 = 15

95% confidence interval

Now, we have to find a value of T, which is found looking at the t table, with 15 degrees of freedom(y-axis) and a confidence level of [tex]1 - \frac{1 - 0.95}{2} = 0.975. So we have T = 2.68

The margin of error is:

M = T*s = 2.1315*3645.94 = 7771.3

In which s is the standard deviation of the sample.

The lower end of the interval is the sample mean subtracted by M. So it is 60139.7 - 7771.3 = 52,368.4.

The upper end of the interval is the sample mean added to M. So it is 60139.7 + 7771.3 = 67,911.

The 95% confidence interval for mean tire life is between 52,368.4 kilometers and 67,911 kilometers.

3 0

3 years ago

Read 2 more answers