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3 years ago

10

A 2-inch-wide frame is to be built around theregular decagonal window shown. At what angles a and bshould the corners of each pi

ece be cut?

1 answer:

OLEGan [10]3 years ago

7 0

Right angles is one angle.

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For 0 ≤ θ < 2 π what are the solutions to sin^2(θ) =2sin^2(θ/2)

gregori [183]

Recall the half-angle identity for sine,

sin²(x/2) = (1 - cos(x))/2

Then the given equation is identical to

sin²(θ) = 1 - cos(θ)

Also recall the Pythagorean identity,

sin²(θ) + cos²(θ) = 1

Then we rewrite the equation as

1 - cos²(θ) = 1 - cos(θ)

Factoring the left side, we have

(1 - cos(θ)) (1 + cos(θ)) = 1 - cos(θ)

and so

(1 - cos(θ)) (1 + cos(θ)) - (1 - cos(θ)) = 0

and we factor this further as

(1 - cos(θ)) (1 + cos(θ) - 1) = 0

which gives

cos(θ) (1 - cos(θ)) = 0

Then either

cos(θ) = 0 or 1 - cos(θ) = 0

cos(θ) = 0 or cos(θ) = 1

[θ = arccos(0) + 2nπ or θ = -arccos(0) + 2nπ]

… or [θ = arccos(1) + 2nπ or θ = -arccos(1) + 2nπ]

(where n is any integer)

[θ = π/2 + 2nπ or θ = -π/2 + 2nπ] or [θ = 0 + 2nπ]

In the interval 0 ≤ θ < 2π, we get three solutions:

• first solution set with n = 0 ⇒ θ = π/2

• second solution set with n = 1 ⇒ θ = 3π/2

• third solution set with n = 0 ⇒ θ = 0

So, the first choice is correct.

6 0

2 years ago

What is an equivalent expression for 5(2x+y-3z)

Delicious77 [7]

The equivalent expression is 10x+5y-15z

6 0

2 years ago

Read 2 more answers

suppose that you have 10 coins consisting of quarters and dimes. if you have a total of $1.90, find the number of each type of c

Anna [14]

You would have 6 quarters and 4 dimes.

4 0

3 years ago

In a group of 40 people, 35% have never been abroad. Two people are selected at random without replacement and are asked about t

Alisiya [41]

Answer:

$\frac{7}{60}$

Step-by-step explanation:

Given that in a group of 40 people, 35% have never been abroad. Two people are selected at random without replacement and are asked about their past travel experience.

Since the two are drawn without replacement the first trial affects the second trial. In other words, each trial is not independent of the other.

a) Hence this cannot be binomial experiment. The reason is p=probability for each trial is not constant as first draw affect the probability for Ii draw

b) The probability that in a random sample of 2, no one has been abroad

=Prob of selecting both people from the group where no one has been abroad

We have in the group 35% i.e. 14 people never been to abroad

So required probability = Prob of selecting both from 14 people

= $\frac{14C2}{40C2} \\=\frac{14*13}{40*39} \\=\frac{7}{60}$

3 0

2 years ago

Solve for y: -2(5y-1)-y=-4(y-3)

Talja [164]

Answer:

<h2> $y = - \frac{10}{7}$ </h2>

Step-by-step explanation:

- 2(5y - 1) - y = - 4(y - 3)

Expand the terms in the bracket

That's

-10y + 2 - y = - 4y + 12

Group like terms

Send the constants to the right side of the equation and those with variables to the left side

That's

- 10y - y + 4y = 12 - 2

- 7y = 10

Divide both sides by - 7

We have the final answer as

<h3> $y = - \frac{10}{7}$ </h3>

Hope this helps you

5 0

3 years ago