All categories

3 years ago

If theta is an acute angle and sin=1/2 , then cos theta =

Mathematics

1 answer:

lilavasa [31]3 years ago

7 0

$\pi /6$ is the only acute angle whose sin = 1/2. Therefore, cos ( $\pi /6$ ) = $\sqrt{3} /2$

You might be interested in

Will mark the BRANLIEST.

True [87]

Answer:

see below

Step-by-step explanation:

Let r = number of rides

total amount of money spent has to be less than or equal to 50

costs are admission and food and rides

rides cost 1.50 each

50≥ admission + food + rides

50 ≥ 15 +10 + 1.50r

Combine like terms

50 ≥25 + 1.50 r

Subtract 25 from each side

50-25 ≥25-25 + 1.50 r

25 ≥ 1.50 r

Divide by 1.5 on each side

25/1.5 ≥ 1.5r/1.5

50/3 ≥ r

Changing this to a mixed number

16 2/3 ≥r

We can only take a whole number of rides

r = 16

Beverly has not spent all of her 50 dollars since there was a fraction for the rides

cost = 15 +10 + 1.50r

15+10 + 1.5*16

25+24

She has 1 dollar left

5 0

3 years ago

Over five different weeks, Irina tracked the hours she spent exercising and the hours she spent playing video games. What is the

madam [21]

Answer:

moderate negative correlation

6 0

3 years ago

Read 2 more answers

There are 3072 books in the school library if 96 students reads the same number of books how many books would each student need

s2008m [1.1K]

Answer:

32 books each

Step-by-step explanation:

5 0

3 years ago

2.2f+0.3f−13−4= Pls hurry!

Aleks [24]

Answer:

-14.5

Step-by-step explanation:

To get the calculator and type 2.2+0.3-13-4

6 0

3 years ago

Help me with these questions

Elena L [17]

Answer: a = 4, b = 163,422

Step-by-step explanation:

$P = P_{0} e^{kt}$ P is the new population, P₀ is the initial population, k is the growth rate, t is the time elapsed.

Part a) What do we know?

P = 7003

P₀ = unknown

k = .21 (converted 21% into a decimal)

t = 36 yrs (2006 - 1970)

$P = P_{0} e^{kt}$ need to solve for P₀

$7003 = P_{0} e^{(.21)(36)}$

$7003 = P_{0} e^{7.56}$

7003 = P₀ (1920)

$\frac{7003}{1920} = P_{0}$

3.6 = P₀

rounded to the nearest whole number = 4

Part b) What do we know?

P = unknown

P₀ = 7003

k = .21 (converted 21% into a decimal)

t = 15 yrs (2021 - 2006)

$P = P_{0} e^{kt}$ need to solve for P₀

$P = 7003 e^{(.21)(15)}$

$P = 7003 e^{3.15}$

P = 7003(23.33)

P = 163,422.46

rounded to the nearest whole number = 163,422

*************************************************************

Answer: 2, -1

Step-by-step explanation:

eˣ² = eˣ * e²

eˣ² = eˣ⁺²

x² = x + 2

x² - x - 2 = 0

(x - 2)(x + 1) = 0

x - 2 = 0 or x + 1 = 0

x = 2 or x = -1

Check both answer for validity:

eˣ² = eˣ⁺²

e⁴ = e⁴ or e¹ = e¹

TRUE TRUE

*****************************************

Answer: 0 = ln 1

Step-by-step explanation:

$e^{-10} = \frac{1}{e^{10}}$

$ln(e^{-10}) = ln(\frac{1}{e^{10}})$

-10 = ln 1 - 10 log rules say division is subtraction

0 = ln 1

5 0

3 years ago