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3 years ago

Which of the following is a solution to 2cos^2x– 2 = 0 ?

Mathematics

2 answers:

Sphinxa [80]3 years ago

8 0

Answer:

A and C

Step-by-step explanation:

Given

2cos²x - 2 = 0 ( add 2 to both sides )

2cos²x = 2 ( divide both sides by 2 )

cos²x = 1 ( take the square root of both sides )

cos x = ± $\sqrt{1}$ = ± 1

cos x = 1 ⇒ x = $cos^{-1}$ (1) = 0°

cos x = - 1 ⇒ x = $cos^{-1}$ (- 1) = 180°

gogolik [260]3 years ago

5 0

Answer:

2 possible results

0° and 180°

Step-by-step explanation:

2cos² x– 2 = 0 (add 2 to both sides)

2cos² x = 2 (divide both sides by 2)

cos² x = 2/2

cos² x = 1

cos x = ±√1

cos x = ±1

because the range of x is not stated, there are 2 possible results for x that will cause cos x to be either 1 or -1

0° and 180°

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3 years ago

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Quality control studies for Speedy Jet Computer Printers show the lifetime of the printer follows a normal distribution with mea

Anika [276]

Answer:

About 3 years Speedy Jet printers be guaranteed.

Step-by-step explanation:

Given : Quality control studies for Speedy Jet Computer Printers show the lifetime of the printer follows a normal distribution with

mean $\mu = 4$ and standard deviation $\sigma = 0.78$

If the company wishes to replace no more than 10% of the printers.

To find : How long should Speedy Jet printers be guaranteed ?

Solution :

Using Z- table,

Z value corresponding to 10% or 0.10 = -1.28.

Applying z-score formula,

$z=\frac{x-\mu}{\sigma}$

The value of x is given by,

$x=(z\times \sigma)+\mu$

Substitute the values,

$x=(-1.28\times 0.78)+4$

$x=-0.9984+4$

$x=3.0016$

So, approx 3 years.

Therefore, About 3 years Speedy Jet printers be guaranteed.

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3 years ago

What is the height ? Please help, I need it! Urgent !

Vika [28.1K]

Answer:

207.24 inch^2

Step-by-step explanation:

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3 years ago

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3 years ago

A ship leaves port at 10 miles per hour, with a heading of N 35° W. There is a warning buoy located 5 miles directly north of th

Leno4ka [110]

The value of the angle subtended by the distance of the buoy from the

port is given by sine and cosine rule.

The bearing of the buoy from the is approximately 307.35°

Reasons:

Location from which the ship sails = Port

The speed of the ship = 10 mph

Direction of the ship = N35°W

Location of the warning buoy = 5 miles north of the port

Required: The bearing of the warning buoy from the ship after 7.5 hours.

Solution:

The distance travelled by the ship = 7.5 hours × 10 mph = 75 miles

By cosine rule, we have;

a² = b² + c² - 2·b·c·cos(A)

Where;

a = The distance between the ship and the buoy

b = The distance between the ship and the port = 75 miles

c = The distance between the buoy and the port = 5 miles

Angle ∠A = The angle between the ship and the buoy = The bearing of the ship = 35°

Which gives;

a = √(75² + 5² - 2 × 75 × 5 × cos(35°))

By sine rule, we have;

$\displaystyle \frac{a}{sin(A)} = \mathbf{ \frac{b}{sin(B)}}$

Therefore;

$\displaystyle sin(B)= \frac{b \cdot sin(A)}{a}$

Which gives;

$\displaystyle sin(B) = \mathbf{\frac{75 \cdot sin(35^{\circ})}{\sqrt{75^2 + 5^2 - 2 \times 75\times5\times cos(35^{\circ}) } }}$

$\displaystyle B = arcsin\left( \frac{75 \cdot sin(35^{\circ})}{\sqrt{75^2 + 5^2 - 2 \times 75\times5\times cos(35^{\circ}) } }\right) \approx 37.32^{\circ}$

Similarly, we can get;

$\displaystyle B = arcsin\left( \frac{75 \cdot sin(35^{\circ})}{\sqrt{75^2 + 5^2 - 2 \times 75\times5\times cos(35^{\circ}) } }\right) \approx \mathbf{ 142.68^{\circ}}$

The angle subtended by the distance of the buoy from the port, C is therefore;

C ≈ 180° - 142.68° - 35° ≈ 2.32°

By alternate interior angles, we have;

The bearing of the warning buoy as seen from the ship is therefore;

Bearing of buoy ≈ 270° + 35° + 2.32° ≈ 307.35°

Learn more about bearing in mathematics here:

brainly.com/question/23427938

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2 years ago