If g is the midpoint of FH find Fg

Which product has 4 zeroes after the degit 3

valentina_108 [34]

0.3 X 10 to the power of 5!

7 0

2 years ago

Find the value of h(-67) for the function below.

pshichka [43]

Answer:

B. 3158

Step-by-step explanation:

Given function:

h(x) = -49x − 125

Finding h(-67)

h(-67) = -49(-67) - 125 = 3283 - 125 = 3158

Correct option is B.

5 0

2 years ago

Read 2 more answers

Which graph represents y= <img src="https://tex.z-dn.net/?f=3%20%5Csqrt%7B%20%5Ctimes%20%20%2B%206%20-%203%7D%20" id="TexForm

Gnom [1K]

y = ∛ (x + 6 - 3) = ∛ (x +3) -----------> graph this

(see attached)

Check: from formula above, when x = -3, y = 0

Check that this point exists on the graph... it does (check OK!)

edit reason: typo

4 0

3 years ago

Read 2 more answers

Describe the transformation required to obtain the graph of the given function from the basic trigonometric graph.

Shtirlitz [24]

Answer:

Option D is correct .i.e., Vertical translation down 9 units

Step-by-step explanation:

Given Function is y = cosec x - 9

Here basic function or parent function is y = cosec x

1. When constant ' a ' is added to to parent function or basic function then the function is translated vertically upward by a units.

2. When constant ' a ' is subtracted from parent function or basic function then the function is translated vertically downward by a units.

Therefore, Option D is correct .i.e., Vertical translation down 9 units

5 0

3 years ago

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