What is the area of 45 30 36 54

Can someone do 15?!?!?

andrew-mc [135]

Given:

The given triangle is a right triangle.

The length of the hypotenuse is 31 units.

The length of the leg is 23 units.

One of the angle is x.

We need to determine the value of x.

Value of x:

The value of x can be determined using the trigonometric ratio.

Thus, we have;

$sin \ \theta=\frac{opp}{hyp}$

Substituting $\theta=x$ , the side opposite to the angle x measures 23 units and the hypotenuse is 31.

Thus, we have;

$sin \ x=\frac{23}{31}$

Dividing, we get;

$sin \ x=0.74$

Taking $sin ^{-1}$ on both sides of the equation, we get;

$x=sin^{-1}(0.74)$

$x=47.7$

Rounding off to the nearest whole integer, we get;

$x=48$

Thus, the value of x is 48°

Hence, Option c is the correct answer.

5 0

3 years ago

Plz help anwser these questions

Anna35 [415]

Answer:

idk

Step-by-step explanation:

ummmmm idkidkidkidk

3 0

3 years ago

Read 2 more answers

HELP PLEASE

Fantom [35]

Answer:

Option C

Step-by-step explanation:

You forgot to attach the expression that models the cost of the camping trip during the three days. However, by analyzing the units, the answer can be reached.

The total cost has to be in units of $.

There are two types of costs in the problem:

Those that depend on the number of days ($/day )

Those that depend on the number of students and the number of days ($/(student * day))

If there are 3 days of camping and b students, then you have to multiply the costs that depend on the days by the number of days (3), and the costs that depend on the number of students you have to multiply them by 'b'

So, if the costs that must be multiplied by 'b' are only those that depend on the number of students, the coefficient of b must be:

3 days (Cost of training + Cost of food Miscellaneous expenses :).

Therefore the correct answer is option C:

C. It is the total cost of 3 days per student of Mr. Brown, with training, food and miscellaneous expenses.

The expression that represents the total expense should have a formula similar to this:

$y = (3 days) *([\frac{20.dollars}{(day * student)} + \frac{30.dollars}{(student * day)} + \frac{50.dollars}{(student * day)}] b + \frac{200}{day}) + 1050.dollars$