Ask question

Ask question

All categories

2 years ago

11

Which angle has a sine of −12 and a cosine of −3√2 ? A 7π6 B 4π3 C 5π6 D 5π3

1 answer:

kobusy [5.1K]2 years ago

6 0

Answer:

A. 7π/6.

Step-by-step explanation:

I think you mean a sine of -1/2 and cosine of -√3/2 ( the values in the question are not possible)

Angle π/6 has a sine of 1/2 and a cosine of √3/2 ( from the 30-60-90 triangle).

This angle must be in the third quadrant because the sin and cos are negative.

So the answer is π +π/6 = 7π/6.

You might be interested in

The factorization of 8x3 -125 is (2x-5)(jx2 +kx+25)

lbvjy [14]

$\bf ~\hspace{10em}\textit{difference and sum of cubes} \\\\ a^3+b^3 = (a+b)(a^2-ab+b^2) ~\hfill a^3-b^3 = (a-b)(a^2+ab+b^2) \\\\[-0.35em] \rule{34em}{0.25pt}\\\\ 8x^3-125~~ \begin{cases} 8=2\cdot 2\cdot 2\\ \qquad 2^3\\ 125=5\cdot 5\cdot 5\\ \qquad 5^3 \end{cases}\implies 2^3x^3-5^3\implies (2x)^3-5^3 \\[2em] [2x-5][(2x)^2+(2x)(5)+5^2]\implies (2x-5)(\stackrel{\stackrel{j}{\downarrow }}{4}x^2+\stackrel{\stackrel{k}{\downarrow }}{10}x+25)$

3 0

2 years ago

Markiya deposited $500 in a savings account. The annual interest rate is 2%, and the interest is compounded monthly. Write a com

motikmotik

I=Prt
I=500*0.02*4
I=$40

So, basically the answer is $40.

7 0

2 years ago

Read 2 more answers

A metallic spherical ball of radius 14 cm is being melted and recast it into spherical balls of radius 2 cm. How many spherical

mojhsa [17]

Volume of sphere = 4/3 x pi x r^3

Volume of large sphere:

4/3 x 3.14 x 14^3 = 11488.21 cubic cm

Volume of smaller sphere:

4/3 x 3.14 x 2^3 = 33.49 cubic cm

Number of small spheres = volume of large sphere/ volume of small sphere

11488.21 / 33.49 = 343.03

Round to 343 small spheres

4 0

2 years ago

What is x divided by 3

Andrews [41]

Answer:

x divided by 3 would be x/3 or 1/3x

5 0

2 years ago

Read 2 more answers

Based on a sample of 100 employees a 95% confidence interval is calculated for the mean age of all employees at a large firm. Th

babymother [125]

Answer:

a

$\= x = 40.85$

b

$E = 5.85$

Ca

$t_c = 2.08$

Cb

$t_c = 1.282$

Explanation:

From the question we are told that

The sample size is n = 100

The upper limit of the 95% confidence interval is b = 47.2 years

The lower limit of the 95% confidence interval is a = 34.5 years

Generally the sample mean is mathematically represented as

$\= x = \frac{a + b }{2}$

=> $\= x = \frac{47.2 + 34.5 }{2}$

=> $\= x = 40.85$

Generally the margin of error is mathematically represented as

$E = \frac{b- a }{ 2}$

=> $E = \frac{47.2- 34.5 }{ 2}$

=> $E = 5.85$

Considering question C a

From the question we are told the confidence level is 90% , hence the level of significance is

$\alpha = (100 - 90 ) \%$

=> $\alpha = 0.10$

The sample size is n = 22

Given that the sample size is not sufficient enough i.e $n < 30$ we will make use of the student t distribution table

Generally the degree of freedom is mathematically represented as

$df = n- 1$

=> $df = 22 - 1$

=> $df = 21$

Generally from the student t distribution table the critical value of $\frac{\alpha }{2}$ at a degree of freedom of 21 is

$t_c =t_{\frac{\alpha }{2} , 21 } = 2.08$

Considering question C b

From the question we are told the confidence level is 80% , hence the level of significance is

$\alpha = (100 - 80 ) \%$

=> $\alpha = 0.20$

Generally from the normal distribution table the critical value of $\frac{\alpha }{2}$ is

$t_c =Z_{\frac{\alpha }{2} } = 1.282$

4 0

2 years ago