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3 years ago

Which equation can be used to solve for c?

Mathematics

2 answers:

AlladinOne [14]3 years ago

5 0

From the given triangle in the photo, the lengths of the sides are 3m and a, the hypotenuse is c. One angle, opposite of 3m is given to be 50 degrees C. The right equation that would solve c is,
sin (50 degrees C) = 3m / c
Thus, the answer is the first choice.

Liula [17]3 years ago

5 0

Answer:

Option (a) is correct.

Step-by-step explanation:

Given a right angled triangle with given dimensions.

We have to find an equation that can be used to solve for side c,

Consider the given triangle ABC,

We know for ∠B = 50° , we have perpendicular = 3 m and hypotenuse is c,

Trigonometric ratio that defines the relation between perpendicular and hypotenuse is Sine.

Using trigonometric ratio,

$\sin\theta=\dfrac{Perpendicular}{ hypotenuse}$

$\sin 50^{\circ}=\frac{AC}{AB}$

$\sin 50^{\circ}=\frac{3}{c}$

Thus, Option (a) is correct.

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3 years ago

The height after t seconds of an object projected upward with an initial velocity of 48 feet per second from a 210-foot tower ca

svetlana [45]

Given:

The height h of an object after t seconds is

$h=-16t^2+48t+210$

The height of a neighboring 50-foot tall building is modeled by the equation h=50.

The time (t) when the object will be at the same height as the building is found to be t = –2 and t = 5.

To find:

The statement which describes the validity of these solutions.

Solution:

We have,

$h=-16t^2+48t+210$

Here, t is the time in seconds.

For t=-2,

$h=-16(2)^2+48(-2)+210$

$h=-64-96+210$

$h=50$

For t=5,

$h=-16(5)^2+48(5)+210$

$h=-400+240+210$

$h=50$

So, the value of h is 50 at t=-2 and t=5.

We know that time is always positive so it cannot be negative value. It means t=-2 is not possible.

The solution t = 5 is the only valid solution to this system since time cannot be negative.

Therefore, the correct option is C.

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3 years ago

Find the two small as possible solutions

solong [7]

Answer: $\bold{b)\quad \dfrac{5}{6},\ \dfrac{25}{6}}$

<u>Step-by-step explanation:</u>

$12cos\bigg(\dfrac{2\pi}{5}x\bigg)+10=16\\\\\\12cos\bigg(\dfrac{2\pi}{5}x\bigg)=6\\\\\\cos\bigg(\dfrac{2\pi}{5}x\bigg)=\dfrac{1}{2}\\\\\\cos^{-1}\bigg[cos\bigg(\dfrac{2\pi}{5}x\bigg)\bigg]=cos^{-1}\bigg(\dfrac{1}{2}\bigg)$

$\text{Use the Unit Circle to determine when}\ cos=\dfrac{1}{2}\quad \rightarrow \quad\text{at}\ \dfrac{\pi}{3}\ and \ \dfrac{5\pi}{3}\\\\\\\dfrac{2\pi}{5}x=\dfrac{\pi}{3}\qquad and\qquad \dfrac{2\pi}{5}x=\dfrac{5\pi}{3}\\\\\\\bigg(\dfrac{5}{2\pi}\bigg)\dfrac{2\pi}{5}x=\dfrac{\pi}{3}\bigg(\dfrac{5}{2\pi}\bigg)\quad and\quad \bigg(\dfrac{5}{2\pi}\bigg)\dfrac{2\pi}{5}x=\dfrac{5\pi}{3}\bigg(\dfrac{5}{2\pi}\bigg)\\\\\\.\qquad \qquad \large\boxed{x=\dfrac{5}{6}\qquad and \qquad x=\dfrac{25}{6}}$

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