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3 years ago

Please. Help I don't know

Mathematics

2 answers:

olga_2 [115]3 years ago

4 0

I’m pretty sure it’s the last option.

ziro4ka [17]3 years ago

4 0

I’m pretty sure it’s D

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If f(x)=3x^2 and g(x)=4x^3, what is the degree of (f*g)(x)?

Westkost [7]

3x^2 * 4x^3 = 12x^5

Answer is degree 5

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How do you round to the nearest square mile

vladimir2022 [97]

Just like you round to the rearst whole number 4 and below gose down 5 and up gose up

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lance bought a square postage stamp to a mail a card to his cousin. if the stamp has 4 square centimeters.what was the dimension

Lubov Fominskaja [6]

The dimension of one side of the stamp was 2 centimeters

Step-by-step explanation:

The formula of the area of any square is A = l², where l is the

length of the side of the square

lance bought a square postage stamp to a mail a card to his cousin

∵ The stamp has 4 square centimeters

∴ The area of the stamp = 4 cm²

∵ The stamp shaped a square

∵ The formula of the area of the square is A = l²

- Equate the formula of the area of the square by the area of the stamp

∴ l² = 4

- Take √ for both sides to find l

∴ $l=\sqrt{4}$

∴ l = 2 cm

The dimension of one side of the stamp was 2 centimeters

Learn more:

You can learn more about the area of the shapes in brainly.com/question/10677255

#LearnwithBrainly

5 0

3 years ago

Help Me Please I will give brainliest if correct!

il63 [147K]

Answer:

9 $c^{7}$

Step-by-step explanation:

$c^{-4}$ is equal to $\frac{1}{c^{4} }$ . $\frac{9c^{3} }{\frac{1}{c^{4} } }$ is equal to $9c^{7}$ .

3 0

2 years ago

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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

3 years ago