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3 years ago

For positive acute angles A and B, it is know that tan A= 8/15 and sin B= 11/61. Find the value of cos (a+b) in simplest form

Mathematics

1 answer:

iragen [17]3 years ago

7 0

Answer:

812/1037

Step-by-step explanation:

To solve this, we have to use trigonometric identities.

Cos (A + B) is given as Cos A Cos B - Sin A Sin B. And from the question, we have that

Tan A = 8/15.

We know that in a triangle, the Tan angle is represented Opp/Adj and thus the Opp is 8, and the Adj is 15. Using Pythagoras, we have

hyp² = opp² + adj²

hyp² = 8² + 15²

hyp² = 64 + 225

hyp² = 289

hyp = √289 = 17

The identity of Cos is Adj/Hyp and that of Sin is Opp/Hyp.

Cos A = 15/17

Sin A = 8/17

Repeating the same process for B, we have

Sin B = 11/61

adj² = hyp² - opp²

adj² = 61² - 11²

adj² = 3721 - 121

adj² = 3600

adj = √3600 = 60

Cos B = 60/61

Now, using the earlier stated trigonometric identity, we have

cos (a + b) = CosA CosB - SinA SinB

cos (a + b) = 15/17 * 60/61 - 8/17 * 11/61

cos (a + b) = 900/1037 - 88/1037

cos (a + b) = 812/1037

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can anyone help me with this ??

Mars2501 [29]

Answer:

A. 336

B. Arranging 3 people in 8 chairs.

C. 42

B. Arranging 2 people in 7 chairs.

Step-by-step explanation:

$P(n, r) = P(8, 3)$

Formula for Permutation is given as:

$P(n, r) = \dfrac{n!}{(n-r)!}$

Putting the value of $n$ = 8 and $r =3$ , we get:

$P(8, 3) = \dfrac{8!}{(8-3)!} \\\Rightarrow \dfrac{8\times 7 \times 6 \times 5\times 4 \times 3 \times 2\times 1}{5\times 4 \times 3\times 2\times 1}\\\Rightarrow 8\times 7 \times 6 \\\Rightarrow 336$

B. Real world situation for part A:

Arranging 3 people on 8 chairs.

First person has 8 options.

Second person has 7 options.

Third person has 6 options.

$P(n, r) = P(7, 2)$

Formula for Permutation is given as:

$P(n, r) = \dfrac{n!}{(n-r)!}$

Putting the value of $n$ = 7 and $r$ = 2 , we get:

$P(7, 2) = \dfrac{7!}{(7-2)!} \\\Rightarrow \dfrac{7 \times 6 \times 5\times 4 \times 3 \times 2\times 1}{5\times 4 \times 3\times 2\times 1}\\\Rightarrow 7 \times 6 \\\Rightarrow 42$

D. Real world situation for part AC:

Arranging 2 people on 7 chairs.

First person has 7 options.

Second person has 6 options.

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3 years ago

Calculate two iterations of Newton's Method for the function using the given initial guess. (Round your answers to four decimal

pickupchik [31]

Answer:

The first and second iteration of Newton's Method are 3 and $\frac{11}{6}$ .

Step-by-step explanation:

The Newton's Method is a multi-step numerical method for continuous diffentiable function of the form $f(x) = 0$ based on the following formula:

$x_{i+1} = x_{i} -\frac{f(x_{i})}{f'(x_{i})}$

Where:

$x_{i}$ - i-th Approximation, dimensionless.

$x_{i+1}$ - (i+1)-th Approximation, dimensionless.

$f(x_{i})$ - Function evaluated at i-th Approximation, dimensionless.

$f'(x_{i})$ - First derivative evaluated at (i+1)-th Approximation, dimensionless.

Let be $f(x) = x^{2}-8$ and $f'(x) = 2\cdot x$ , the resultant expression is:

$x_{i+1} = x_{i} -\frac{x_{i}^{2}-8}{2\cdot x_{i}}$

First iteration: ( $x_{1} = 2$ )

$x_{2} = 2-\frac{2^{2}-8}{2\cdot (2)}$

$x_{2} = 2 + \frac{4}{4}$

$x_{2} = 3$

Second iteration: ( $x_{2} = 3$ )

$x_{3} = 3-\frac{3^{2}-8}{2\cdot (3)}$

$x_{3} = 2 - \frac{1}{6}$

$x_{3} = \frac{11}{6}$

7 0

3 years ago

Whats the answer to this???

Leno4ka [110]

Measure angle EFG = 15 degrees
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3 years ago

If a test of H subscript 0 colon space mu subscript D equals 0 space v s. space H subscript a colon space space mu subscript D g

lara [203]

Answer:

$p_v =P(t_{(n-1)}>t_{calculated}) =0.0601$

The p value on this case is given by the problem.

If we compare the p value with a significance level assumed $\alpha=0.05$ , we see that $p_v > \alpha$ and we can conclude that we FAIL to reject the null hypothesis that the difference mean between after and before is less or equal than 0.

Step-by-step explanation:

A paired t-test is used to compare two population means where you have two samples in which observations in one sample can be paired with observations in the other sample. For example if we have Before-and-after observations (This problem) we can use it.

Let put some notation

x=test value before , y = test value after

The system of hypothesis for this case are:

Null hypothesis: $\mu_y- \mu_x \leq 0$