What is the initial value of the function represented by this table?

Use a half-angle identity to find the exact value

Tatiana [17]

Given:

$\cos 15^{\circ}$

To find:

The exact value of cos 15°.

Solution:

$$\cos 15^{\circ}=\cos\frac{ 30^{\circ}}{2}$

Using half-angle identity:

$$\cos \left(\frac{x}{2}\right)=\sqrt{\frac{1+\cos (x)}{2}}$

$$\cos \frac{30^{\circ}}{2}=\sqrt{\frac{1+\cos \left(30^{\circ}\right)}{2}}$

Using the trigonometric identity: $\cos \left(30^{\circ}\right)=\frac{\sqrt{3}}{2}$

$$=\sqrt{\frac{1+\frac{\sqrt{3}}{2}}{2}}$

Let us first solve the fraction in the numerator.

$$=\sqrt{\frac{\frac{2+\sqrt{3}}{2}}{2}}$

Using fraction rule: $\frac{\frac{a}{b} }{c}=\frac{a}{b \cdot c}$

$$=\sqrt{\frac {2+\sqrt{3}}{4}}$

Apply radical rule: $\sqrt[n]{\frac{a}{b}}=\frac{\sqrt[n]{a}}{\sqrt[n]{b}}$

$$=\frac{\sqrt{2+\sqrt{3}}}{\sqrt{4}}$

Using $\sqrt{4} =2$ :

$$=\frac{\sqrt{2+\sqrt{3}}}{2}$

$$\cos 15^\circ=\frac{\sqrt{2+\sqrt{3}}}{2}$

5 0

3 years ago

Audrey, an astronomer is searching for extra-solar planets using the technique of relativistic lensing. Though there are believe

Stolb23 [73]

Answer:

Step-by-step explanation:

The model $N (t)$ , the number of planets found up to time $t$ , as a Poisson process. So, the $N (t)$ has distribution of Poison distribution with parameter $(\lambda t)$

a)

The mean for a month is, $\lambda = \frac{1}{3}$ per month

$E[N(t)]= \lambda t\\\\=\frac{1}{3}(24)\\\\=8$

(Here. t = 24)

For Poisson process mean and variance are same,

$Var[N (t)]= Var[N(24)]\\= E [N (24)]\\=8$

(Poisson distribution mean and variance equal)

The standard deviation of the number of planets is,

$\sigma( 24 )] =\sqrt{Var[ N(24)]}=\sqrt{8}= 2.828$

b)

For the Poisson process the intervals between events(finding a new planet) have independent exponential distribution with parameter $\lambda$ . The sum of $K$ of these independent exponential has distribution Gamma $(K, \lambda)$ .

From the given information, $k = 6$ and $\lambda =\frac{1}{3}$

Calculate the expected value.

$E(x)=\frac{\alpha}{\beta}\\\\=\frac{K}{\lambda}\\\\=\frac{6}{\frac{1}{3}}\\\\=18$

(Here, $\alpha =k$ and $\beta=\lambda$ )

C)

Calculate the probability that she will become eligible for the prize within one year.

Here, 1 year is equal to 12 months.

P(X ≤ 12) = (1/Г (k)λ^k)(x)^(k-1).(e)^(-x/λ)

$=\frac{1}{Г (6)(\frac{1}{3})^6}(12)^{6-1}e^{-36}\\\\=0.2148696\\=0.2419\\=21.49%$

Hence, the required probability is 0.2149 or 21.49%

5 0

2 years ago

Someone help! I’m not good at these problems!

Evgen [1.6K]

Answer:

I beliive it is 16, but haven't done this in a bit haha

Step-by-step explanation:

So bisect means to "split in half" aka segment BD is directly in the middle of triangle ABC. This means we can look at the half of the triangle we know (ABD). We see a 16, so therefore x should be 16.

(again, hopefully I am correct)

5 0

2 years ago

Read 2 more answers

I need help for 7 and 8. i also need the solution on how you get the answer. thanks in advance!

ycow [4]

7) Rectangular pyramid:
Length = 8m ; width = 4.6m ; Volume = 88m³

Volume of a rectangular pyramid = (Length * Width * Height)/3
88m³ = (8m * 4.6m * height)/3
88m³ * 3 = 36.8m² * height
264m³ = 36.8m² * height
264m³ / 36.8m² = height
7.2 m = height

8) Cone:
r = 5 in ; volume = 487 in³

Volume of a cone = π r² h/3
487 in³ = 3.14 * (5in)² * h/3
487 in³ * 3 = 3.14 * 25in² * h
1,461 in³ = 78.5 in² * h
1,461 in³ / 78.5 in² = height
18.6 in = height

8 0

3 years ago

What is the area of a triangle with side lengths 13-14-15?

Novosadov [1.4K]

84. Area of a triangle is (base x height) / 2

3 0

3 years ago