If WV is a midsegment of STU, find SU. A 4 B 5 C 8 D 10

Use the given information to find the exact value of the trigonometric function

eimsori [14]

$\begin{gathered} \csc \theta=-\frac{6}{5} \\ \tan \theta>0 \\ \cos \frac{\theta}{2}=\text{?} \end{gathered}$

Half Angle Formula

$\cos \frac{\theta}{2}=\pm\sqrt[\square]{\frac{1+\cos\theta}{2}}$ $\tan \theta>0\text{ and csc}\theta\text{ is negative in the third quadrant}$ $\begin{gathered} \csc \theta=-\frac{6}{5}=\frac{r}{y} \\ x^2+y^2=r^2 \\ x=\pm\sqrt[\square]{r^2-y^2} \\ x=\pm\sqrt[\square]{6^2-(-5)^2} \\ x=\pm\sqrt[\square]{36-25} \\ x=\pm\sqrt[\square]{11} \\ \text{x is negative since the angle is on the 3rd quadrant} \end{gathered}$ $\begin{gathered} \cos \theta=\frac{x}{r}=\frac{-\sqrt[\square]{11}}{6} \\ \cos \frac{\theta}{2}=\pm\sqrt[\square]{\frac{1+\cos\theta}{2}} \\ \cos \frac{\theta}{2}is\text{ also negative in the 3rd quadrant} \\ \cos \frac{\theta}{2}=-\sqrt[\square]{\frac{1+\frac{-\sqrt[\square]{11}}{6}}{2}} \\ \cos \frac{\theta}{2}=-\sqrt[\square]{\frac{\frac{6-\sqrt[\square]{11}}{6}}{2}} \\ \cos \frac{\theta}{2}=-\sqrt[\square]{\frac{6-\sqrt[\square]{11}}{12}} \\ \\ \end{gathered}$

Answer:

$\cos \frac{\theta}{2}=-\sqrt[\square]{\frac{6-\sqrt[\square]{11}}{12}}$

Checking:

$\begin{gathered} \frac{\theta}{2}=\cos ^{-1}(-\sqrt[\square]{\frac{6-\sqrt[\square]{11}}{12}}) \\ \frac{\theta}{2}=118.22^{\circ} \\ \theta=236.44^{\circ}\text{ (3rd quadrant)} \end{gathered}$

Also,

$\csc \theta=\frac{1}{\sin\theta}=\frac{1}{\sin (236.44)}=-\frac{6}{5}\text{ QED}$

The answer is none of the choices

7 0

1 year ago

A cubic box is completely filled with 2,179 g of water. What is the length of one side of the box, in meters? m Explain your rea

Mazyrski [523]

Answer:

$l\approx 1.296\,cm$

Step-by-step explanation:

The water density is approximately 1 gram per cubic centimeter, which means that 1 gram of water occupies a volume of a cubic centimeter. Hence, 2.179 grams occupies a volume of 2.179 cubic centimeters.

Let assume volume occupied by the water has a cube-like shape, whose side has the following measure:

$l = \sqrt[3]{2.179\,m^{3}}$

$l\approx 1.296\,cm$

5 0

3 years ago

Form a polynomial f(x) with real coefficients having the given degree and zeros.

dimaraw [331]

There are many polynomials that fit the bill,
f(x)=a(x-r1)(x-r2)(x-r3)(x-r4) where a is any real number not equal to zero.
A simple one is when a=1.
where r1,r2,r3,r4 are the roots of the 4th degree polynomial.
Also note that for a polynomial with *real* coefficients, complex roots *always* come in conjugages, i.e. in the form a&pm;bi [&pm;=+/-]

So a polynomial would be:
f(x)=(x-(-4-5i))(x-(-4+5i))(x--2)(x--2)
or, simplifying
f(x)=(x+4+5i)(x+4-5i)(x+2)^2
=x^4+12x^3+77x^2+196x+164 [if you decide to expand]

6 0

3 years ago

What do i do here and plz help me answer this

stepladder [879]

Answer:

for number 1, you plug -3 in for x in the equation. and for number 2, you plug in 2 for y in the equation. you do this to find the missing values of the variables.

1. y= -10

2. x= -1

Step-by-step explanation:

i hope this helps :)

5 0

3 years ago

Read 2 more answers

9. What is the area of the largest square? A=? A = 36 units A = 64 units.

lisov135 [29]

The answer is A=64 but I’m not pretty sure

4 0

3 years ago

Read 2 more answers