All categories

3 years ago

Law of Sines, due by tonight

Mathematics

1 answer:

ivann1987 [24]3 years ago

5 0

Answer:

$21^{\circ}$

Step-by-step explanation:

The Law of Sines is given by $\frac{\sin A}{a}=\frac{\sin B}{b}=\frac{\sin C}{c}$ and works for every triangle.

Substituting given values, we have the following equation:

$\frac{\sin 68^{\circ}}{18}=\frac{\sin C}{7}, \\\\\sin C=\frac{7\sin 68^{\circ}}{18},\\\\C=\sin^{-1}(0.36057149899),\\\\C\approx \boxed{21^{\circ}}$

*Note that because $\sin \theta=\sin(180-\theta)$ , when solving for an angle with the Law of Sines, there may be two answers. However, since the problem designates angle C as an acute angle, the other angle is negligible.

You might be interested in

PLS SOMEONE HELP ME ASAP

babymother [125]

Answer:

A. (2i)(8) = d. 16i

B. 16i³ = b. -16i

C. (2i)⁴ = a. 16

D. (2i)(8i) = c. -16

Step-by-step explanation:

A. Multiply 2i by 8 to get 16i, which corresponds to d.

B. The exponent is 3 more than a mulitple of 4 in 16i³, so the answer is negative. -16i corresponds to b.

C. (2i)⁴ has an exponent that is a multiple of 4, so the i isn't needed. 16 corresponds to a.

D. (2i)(8i) simplifies to 2(8) * i². The exponent is 2 more than a multiple of 4, so the answer is negative, without an i. -16 corresponds to c.

8 0

2 years ago

Which image shows a plane slicing a double-napped cone to produce a hyperbola? Answer is D

Butoxors [25]

The second image in the diagram is a hyperbola. As can be seen, the plane cutting the cone can be at any angle but never equal to the slant angle of the cone. This has a very important implication. The plane cuts both cones of the double-napped cone. The third double-napped cone of Figure 3 shows two hyperbolas.

4 0

2 years ago

F(t)=t^2+19t+60 What are the zeros of the function and what is the vertex of the parabola?

SVEN [57.7K]

Answer: $-4,-15;\ \left(-\dfrac{19}{2},-\dfrac{121}{4}\right)$

Step-by-step explanation:

Given

Function is $F(t)=t^2+19+60$

Zeroes of the function are

$t^2+19t+60=0\\\\\Rightarrow t=\dfrac{-19\pm\sqrt{19^2-4\times 1\times 60}}{2\times 1}\\\\\Rightarrow t=\dfrac{-19\pm \sqrt{121}}{2}\\\\\Rightarrow t=\dfrac{-19\pm 11}{2}\\\\\Rightarrow t=-4,-15$

Using completing the square method

$y=t^2+2\times \dfrac{19}{2}t+\dfrac{19^2}{2^2}-\dfrac{19^2}{2^2}+60\\\\y=\left(t+\dfrac{19}{2}\right)^2+60-\dfrac{361}{4}\\\\y=\left(t+\dfrac{19}{2}\right)^2-\dfrac{121}{4}\\\\y+\dfrac{121}{4}=\left(t+\dfrac{19}{2}\right)^2\\\\\left(y-(-\dfrac{121}{4}\right)=\left(t-(-\dfrac{19}{2})\right)^2\\\\\text{The vertex is }\left(-\dfrac{19}{2},-\dfrac{121}{4}\right)$

4 0

2 years ago

Plz help urgent !!!! will give the brainliest !!

Mkey [24]

Answer:

$X = \begin{bmatrix}1&3\\ 2&4\end{bmatrix}$

Step-by-step explanation:

The question we have at hand is, in other words,

$\begin{bmatrix}4&2\\ \:1&1\end{bmatrix}\left(X\right)=\begin{bmatrix}8&20\\ \:3&7\end{bmatrix}$ - where we have to solve for the value of $X$

If we have to isolate $X$ here, then we would have to take the inverse of the following matrix ...

$\begin{bmatrix}4&2\\ \:1&1\end{bmatrix}$ ... so that it should be as follows ... $\begin{bmatrix}4&2\\ \:1&1\end{bmatrix}^{-1}$

Therefore, we can conclude that the equation as to solve for " $X$ " will be the following,

$X=\begin{bmatrix}4&2\\ 1&1\end{bmatrix}^{-1}\begin{bmatrix}8&20\\ 3&7\end{bmatrix}$ - First find the 2 x 2 matrix inverse of the first portion,

$\begin{bmatrix}4&2\\ 1&1\end{bmatrix}^{-1}$ = $\frac{1}{\det \begin{pmatrix}4&2\\ 1&1\end{pmatrix}}\begin{pmatrix}1&-2\\ -1&4\end{pmatrix}$ = $\frac{1}{2}\begin{bmatrix}1&-2\\ -1&4\end{bmatrix}$ = $\begin{bmatrix}\frac{1}{2}&-1\\ -\frac{1}{2}&2\end{bmatrix}$

At this point we have to multiply the rows of the first matrix by the rows of the second matrix,

$X = \begin{bmatrix}\frac{1}{2}&-1\\ -\frac{1}{2}&2\end{bmatrix}\begin{bmatrix}8&20\\ 3&7\end{bmatrix}$ ,

$X = \begin{pmatrix}\frac{1}{2}\cdot \:8+\left(-1\right)\cdot \:3&\frac{1}{2}\cdot \:20+\left(-1\right)\cdot \:7\\ \left(-\frac{1}{2}\right)\cdot \:8+2\cdot \:3&\left(-\frac{1}{2}\right)\cdot \:20+2\cdot \:7\end{pmatrix}$ - Simplifying this, we should get ...

$\begin{bmatrix}1&3\\ 2&4\end{bmatrix}$ ... which is our solution.

7 0

3 years ago

PLEASE HELP ME!!!!!! I NEED HELP REALLY FAST!! I WILL BE MARKING BRAINLIEST!!!!!!

viva [34]

Common difference is +6 and I think nth term is 5 (pretty sure it's always the first term) and the second one im not so sure about I'm sorry.

4 0

3 years ago

Read 2 more answers