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4 years ago

Multiply and Simplify cos(theta) sin(theta)(sec (theta) + csc (theta))

Mathematics

1 answer:

fredd [130]4 years ago

3 0

Before I put out my answer, I just want make note that sec() = 1 / cos() and csc () = 1 / sin ()

1) Once you distribute cos()sin() to sec(), the cos() will cancel out with the sec() to make it just sin(), and once cos()sin() is distributed to csc(), the sin() will cancel out with the cos()

2) After doing that you should just be left with sin() + cos()

3) And since no more simplification can be done, that is your answer :D

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Please help!

Mandarinka [93]

Answer:0.11111111111

Step-by-step explanation:0.1 with a repeating decimal over the 1 because when you multiply 1/3 by 1/3 thats what you get.

8 0

3 years ago

Lydia drives from city a to city b to transport goods. her return speed is 3 times her departure speed and she takes 40 minutes

gizmo_the_mogwai [7]

Answer:

1 hour

Step-by-step explanation:

Hello, let's say that her departure trip takes t in minutes, as her return speed is 3 times her departure speed, she took t/3 for the return and we know that this 40 minutes less, so we can write.

t/3=t-40

We can multiply by 3

t = 3t -40*3 = 3t - 120

This is equivalent to

3t -120 = t

We subtract t

2t-120 = 0

2t = 120

We divide by 2

t = 120/2 = 60

So this is 60 minutes = 1 hour.

Thank you.

4 0

3 years ago

Segment AB is congruent to segment AB.<br> This statement shows the<br> property.

lions [1.4K]

It shows the reflective property

4 0

3 years ago

Read 2 more answers

Colton has 8 88 comedy movies, 6 66 action movies, 5 55 fantasy movies, and 1 11 drama movie on his list of favorite movies. For

Ostrovityanka [42]

Answer:

(Choice A) A comedy

Step-by-step explanation:

The total number of movies is 8 + 6 + 5 + 1 = 20.

2 movies out of every 5 movies corresponds to 2/5.

For 20 movies, this 2/5 × 20 = 8 movies.

This corresponds to comedy movies.

6 0

3 years ago

If A, B, and C are integers between 1 and 10 (inclusive), how many different combinations of A, B, and C exist such that A

Marianna [84]

$\fontsize{18}{10}{\textup{\textbf{The number of different combinations is 120.}}}$

Step-by-step explanation:

A, B and C are integers between 1 and 10 such that A<B<C.

The value of A can be minimum 1 and maximum 8.

If A = 1, B = 2, then C can be one of 3, 4, 5, 6, 7, 8, 9, 10 (8 options).

If A = 1, B = 3, then C has 7 options (4, 5, 6, 7, 8, 9, 10).

If A = 1, B = 4, then C has 6 options (5, 6, 7, 8, 9, 10).

If A = 1, B = 5, then C has 5 options (6, 7, 8, 10).

If A = 1, B = 6, then C has 4 options (7, 8, 9, 10).

If A = 1, B = 7, then C has 3 options (8, 9, 10).

If A = 1, B = 8, then C has 2 options (9, 10).

If A = 1, B = 9, then C has 1 option (10).

So, if A = 1, then the number of combinations is

$n_1=1+2+3+4+5+6+7+8=\dfrac{8(8+1)}{2}=36.$

Similarly, if A = 2, then the number of combinations is

$n_2=1+2+3+4+5+6+7=\dfrac{7(7+1)}{2}=28.$

If A = 3, then the number of combinations is

$n_3=1+2+3+4+5+6=\dfrac{6(6+1)}{2}=21.$

If A = 4, then the number of combinations is

$n_4=1+2+3+4+5=\dfrac{5(5+1)}{2}=15.$

If A = 5, then the number of combinations is

$n_5=1+2+3+4=\dfrac{4(4+1)}{2}=10.$

If A = 6, then the number of combinations is

$n_6=1+2+3=\dfrac{3(3+1)}{2}=6.$

If A = 7, then the number of combinations is

$n_7=1+2=\dfrac{2(2+1)}{2}=3.$

If A = 3, then the number of combinations is

$n_8=1.$

Therefore, the total number of combinations is

$n\\\\=n_1+n_2+n_3+n_4+n_5+n_6+n_7+n_8\\\\=36+28+21+15+10+6+3+1\\\\=120.$

Thus, the required number of different combinations is 120.

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Question : ow many types of zygotic combinations are possible between a cross AaBBCcDd × AAbbCcDD?

Link : https://brainly.in/question/4909567.

8 0

3 years ago