I really need help on this it’s really urgent. Thank you :)

Mathematics

1 answer:

Flauer [41]2 years ago

6 0

Answer:

Step-by-step explanation:

$-\frac{1}{3}y - \frac{1}{3} x$

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A ship leaves port at 1:00 P.M. and travels S35°E at the rate of 27 mi/hr. Another ship leaves the same port at 1:30 P.M. and tr

Len [333]

To solve this problem you must apply the proccedure shown below:

1. You must apply the Law of Cosines, as you can see in the figure attached. Then:

- The first ship travels at $27 mi/hr$ in for two hours. Therefore, the side $a$ is:

$a=(27 mi/hr)(2 hr)=54mi$

- The second ship travels at $18 mi/hr$ for $1.5 hours$ . Therefore, the side $b$ is:

$b=(18mi/hr)(1.5hr)=27mi$

- Now, you can calculate $c$ :

$c=\sqrt{54^{2}+27^{2}-2(54)(27)Cos(55)}=44 mi$

The answer is: $44 miles$

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2 years ago

A statue is mounted on top of a 16 foot hill. From the base of the hill to where you are standing is 77 feet and the statue subt

DanielleElmas [232]

Consider right triangle with vertices B - base of the hill, S - top of the statue and Y - you. In this triangle angle B is right and angle Y is 13.2°. If h is a height of the statue, then the legs YB and BS have lengths 77 ft and 16+h ft.

You have lengths of two legs and measure of one acute angle, then you can use tangent to find h:

$\tan 13.2^{\circ}=\dfrac{\text{opposite leg}}{\text{adjacent leg}} =\dfrac{16+h}{77}, \\ \\ 0.2345=\dfrac{16+h}{77},\\ \\ 16+h=0.2345\cdot 77=18.0565,\\ h=18.0565-16=2.0565$ ft.