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3 years ago

What is the solution to the compound inequality in interval notation?

Mathematics

2 answers:

mars1129 [50]3 years ago

6 0

Answer: The correct option is

(D) (−∞, −5] or (0, ∞).

Step-by-step explanation: We are given to select the solution to the following inequality in interval notation :

$2(x+3)>6~~\textup{or}~~2x+3\leq-7~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~(i)$

To find the correct solution, we need to solve both the inequalities in (i) separately.

The solution of (i) is as follows :

$2(x+3)>6\\\\\Rightarrow x+3>\dfrac{6}{2}\\\\\Rightarrow x+3>3\\\\\Rightarrow x>3-3\\\\\Rightarrow x>0$

and

$2x+3\leq -7\\\\\Rightarrow 2x\leq-7-3\\\\\Rightarrow 2x\leq-10\\\\\Rightarrow x\leq-\dfrac{10}{2}\\\\\Rightarrow x\leq -5.$

That is, the solution is given by

$x\epsilon (-\infty,-5]~or~(0,\infty).$

Thus, (D) is the correct option.

Eduardwww [97]3 years ago

5 0

First solve the 2 inequalities for x:-

x > 0 or x <= -5

The last choice is the correct one.

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5p+2= -10 Show your work

Monica [59]

Answer:

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Step-by-step explanation:

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2 years ago

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julsineya [31]

Answer:

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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;

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