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Nina [5.8K]
3 years ago
13

Cos(a) 63/65 to find sin(a) and tan(a)

Mathematics
2 answers:
wlad13 [49]3 years ago
4 0

Answer:

Sin(a) = 16/65

Tan(a) = 16/63

Step-by-step explanation:

Cos(a) = adj/hyp

opp² = 65² - 63²

Opp = 16

Sin(a) = 16/65

Tan(a) = 16/63

Assuming a is acute.

If angle is in the fourth quadrant, sin and tan will be negative

Semenov [28]3 years ago
3 0

Answer:

\huge\boxed{\sin\alpha=-\dfrac{16}{65},\ \tan\alpha=-\dfrac{16}{63}}\\\\or\\\\\huge\boxed{\sin\alpha=\dfrac{16}{65},\ \tan\alpha=\dfrac{16}{63}}

Step-by-step explanation:

\cos\alpha=\dfrac{63}{65}\\\\\text{use}\ \sin^2\alpha+\cos^2\alpha=1\\\\\sin^2\alpha+\left(\dfrac{63}{65}\right)^2=1\\\\\sin^2\alpha+\dfrac{3969}{4225}=1\qquad\text{subtract}\ \dfrac{3969}{4225}\ \text{from both sides}\\\\\sin^2\alpha=\dfrac{4225}{4225}-\dfrac{3969}{4225}\\\\\sin^2\alpha=\dfrac{256}{4225}\to\sin\alpha=\pm\sqrt{\dfrac{256}{4225}}\\\\\sin\alpha=\pm\dfrac{\sqrt{256}}{\sqrt{4225}}\\\\\sin\alpha=\pm\dfrac{16}{65}

\text{use}\ \tan\alpha=\dfrac{\sin\alpha}{\cos\alpha}\\\\\text{substitute:}\\\\\tan\alpha=\dfrac{\pm\frac{16}{65}}{\frac{63}{65}}=\pm\dfrac{16}{65}\cdot\dfrac{65}{63}=\pm\dfrac{16}{63}

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LCM of 4/6 and 4/8????
Svet_ta [14]

Answer:

24 is the LCM of 4,6,8. I think thats what your looking for.

Hope this helps!

Step-by-step explanation:

5 0
2 years ago
Use the standard norml distribution or the t-distribution to construct a 99% confidence interval for the population mean. Justif
dalvyx [7]

Answer:

E) we will use t- distribution because is un-known,n<30

the confidence interval is (0.0338,0.0392)

Step-by-step explanation:

<u>Step:-1</u>

Given sample size is n = 23<30 mortgage institutions

The mean interest rate 'x' = 0.0365

The standard deviation 'S' = 0.0046

the degree of freedom = n-1 = 23-1=22

99% of confidence intervals t_{0.01} =2.82  (from tabulated value).

The mean value = 0.0365

x±t_{0.01} \frac{S}{\sqrt{n-1} }

0.0365±2.82 \frac{0.0046}{\sqrt{23-1} }

0.0365±2.82 \frac{0.0046}{\sqrt{22} }

0.0365±2.82 \frac{0.0046}{4.690 }

using calculator

0.0365±0.00276

Confidence interval is

(0.0365-0.00276,0.0365+0.00276)

(0.0338,0.0392)

the mean value is lies between in this confidence interval

(0.0338,0.0392).

<u>Answer:-</u>

<u>using t- distribution because is unknown,n<30,and the interest rates are not normally distributed.</u>

4 0
3 years ago
Confusion...help please
kifflom [539]

Given:

In triangle KLM, KL = 123 cm and measure of angle K is 35 degrees.

To find:

The length of the side KM to the nearest tenth of a centimeter.

Solution:

In a right angle triangle,

\cos \theta =\dfrac{Base}{Hypotenuse}

In the given right triangle KLM,

\cos K=\dfrac{KM}{KL}

\cos (35^\circ)=\dfrac{KM}{123}

0.819152=\dfrac{KM}{123}

Multiply both sides by 123.

0.819152\times 123=KM

100.755696=KM

KM\approx 100.8

The measure of side KM is 100.8 cm.

Therefore, the correct option is (2).

8 0
2 years ago
SOMEONE PLEASE HELP!!!
SVETLANKA909090 [29]
ST = 1/2 (RV) = VQ = 6

answer

ST = 6

hope it helps
8 0
3 years ago
hi, i dont undertand number 20 because i was absent in class today and i rerally need help, i will appraciate with the help, and
Mariulka [41]

Given:

The equation is,

2\log _3x-\log _3(x-2)=2

Explanation:

Simplify the equation by using logarthimic property.

\begin{gathered} 2\log _3x-\log _3(x-2)=2 \\ \log _3x^2-\log _3(x-2)=2_{}\text{      \lbrack{}log(a)-log(b) = log(a/b)\rbrack} \\ \log _3\lbrack\frac{x^2}{x-2}\rbrack=2 \end{gathered}

Simplify further.

\begin{gathered} \log _3\lbrack\frac{x^2}{x-2}\rbrack=2 \\ \frac{x^2}{x-2}=3^2 \\ x^2=9(x-2) \\ x^2-9x+18=0 \end{gathered}

Solve the quadratic equation for x.

\begin{gathered} x^2-6x-3x+18=0 \\ x(x-6)-3(x-6)=0 \\ (x-6)(x-3)=0 \end{gathered}

From the above equation (x - 6) = 0 or (x - 3) = 0.

For (x - 6) = 0,

\begin{gathered} x-6=0 \\ x=6 \end{gathered}

For (x - 3) = 0,

\begin{gathered} x-3=0 \\ x=3 \end{gathered}

The values of x from solving the equations are x = 3 and x = 6.

Substitute the values of x in the equation to check answers are valid or not.

For x = 3,

\begin{gathered} 2\log _3(3^{})-\log _3(3-2)=2 \\ 2\log _33-\log _31=2 \\ 2\cdot1-0=2 \\ 2=2 \end{gathered}

Equation satisfy for x = 3. So x = 3 is valid value of x.

For x = 6,

\begin{gathered} 2\log _36-\log _3(6-2)=2 \\ 2\log _36-\log _34=2 \\ \log _3(6^2)-\log _34=2 \\ \log _3(\frac{36}{4})=2 \\ \log _39=2 \\ \log _3(3^2)=2 \\ 2\log _33=2 \\ 2=2 \end{gathered}

Equation satifies for x = 6.

Thus values of x for equation are x = 3 and x = 6.

6 0
1 year ago
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