What’s the 8th term in 4, 12, 36, 108

Mathematics

2 answers:

Virty [35]3 years ago

8 0

The 8th term is 8,748

DIA [1.3K]3 years ago

3 0

The 8th term is 1024

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Classify the following triangle. check all that apply

Karo-lina-s [1.5K]

Answer:

Options (A), (B)

Step-by-step explanation:

Option A.

Two sides of the given triangle are equal. Therefore, the given triangle is an isosceles triangle.

Option (B).

Given triangle has a right angle. Therefore, it's a right triangle.

Option (C).

Only two sides of the given triangle are equal. Therefore, it's not an equilateral triangle.

Option (D).

Although this triangle has two acute angles (less than 90°) but the third angle is 90°. Therefore, this triangle is not an acute triangle. It's a right triangle.

Option (E).

A scalene triangle is a triangle having all sides different in measure but the given triangle is an isosceles triangle. Therefore, it's not a scalene.

Option (F).

Any angle of this triangle is not more than 90°. Therefore, the given triangle is not an obtuse triangle.

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3 years ago

Factor completely 81x^4-16

mestny [16]

The will be (3x - 2) x (3x + 2) x (9x2 + 4)

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2 years ago

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Sin theta+costheta/sintheta -costheta+sintheta-costheta/sintheta+costheta=2sec2/tan2 theta -1

sleet_krkn [62]

$\dfrac{sin\theta + cos\theta}{sin\theta-cos\theta}+\dfrac{sin\theta-cos\theta}{sin\theta+cos\theta}=\dfrac{2sec^2\theta}{tan^2\theta-1}$

From Left side:

$\dfrac{sin\theta + cos\theta}{sin\theta-cos\theta}\bigg(\dfrac{sin\theta+cos\theta}{sin\theta+cos\theta}\bigg)+\dfrac{sin\theta-cos\theta}{sin\theta+cos\theta}\bigg(\dfrac{sin\theta-cos\theta}{sin\theta-cos\theta}\bigg)$

$\dfrac{sin^2\theta+2cos\thetasin\theta+cos^2\theta}{sin^2\theta-cos^2\theta}+\dfrac{sin^2\theta-2cos\thetasin\theta+cos^2\theta}{sin^2\theta-cos^2\theta}$