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3 years ago

1, 4, 13, 40, 121 what is the next number in the pattern

1 answer:

7 0

Add powers of 3.
1 + 3 = 4
4 + 9 = 13 = 4 + 3^2
13 + 27 = 40 = 13 + 3^3
40 + 81 = 121 = 40 + 3^4
Now you need to add 3^5 to 121.

121 + 3^5 = 121 + 243 = 364

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Okay Ju|ian Foley simplifies this polynomial and then write it in standard form if Ju|ian wrote it the last term

Fudgin [204]

Answer:

the second choice

Step-by-step explanation:

8 0

3 years ago

Prove that: (b²-c²/a)CosA+(c²-a²/b)CosB+(a²-b²/c)CosC = 0

IRISSAK [1]

Prove that:

$\:\:\sf\:\:\left(\dfrac{b^2-c^2}{a}\right)\cos A+\left(\dfrac{c^2-a^2}{b}\right)\cos B +\left(\dfrac{a^2-b^2}{c}\right)\cos C=0$

Proof: 

We know that, by Law of Cosines,

$\sf \cos A=\dfrac{b^2+c^2-a^2}{2bc}$
$\sf \cos B=\dfrac{c^2+a^2-b^2}{2ca}$
$\sf \cos C=\dfrac{a^2+b^2-c^2}{2ab}$

Taking LHS

$\left(\dfrac{b^2-c^2}{a}\right)\cos A+\left(\dfrac{c^2-a^2}{b}\right)\cos B +\left(\dfrac{a^2-b^2}{c}\right)\cos C$

Substituting the value of cos A, cos B and cos C,

$\longmapsto\left(\dfrac{b^2-c^2}{a}\right)\left(\dfrac{b^2+c^2-a^2}{2bc}\right)+\left(\dfrac{c^2-a^2}{b}\right)\left(\dfrac{c^2+a^2-b^2}{2ca}\right)+\left(\dfrac{a^2-b^2}{c}\right)\left(\dfrac{a^2+b^2-c^2}{2ab}\right)$

$\longmapsto\left(\dfrac{(b^2-c^2)(b^2+c^2-a^2)}{2abc}\right)+\left(\dfrac{(c^2-a^2)(c^2+a^2-b^2)}{2abc}\right)+\left(\dfrac{(a^2-b^2)(a^2+b^2-c^2)}{2abc}\right)$

$\longmapsto\left(\dfrac{(b^2-c^2)(b^2+c^2)-(b^2-c^2)(a^2)}{2abc}\right)+\left(\dfrac{(c^2-a^2)(c^2+a^2)-(c^2-a^2)(b^2)}{2abc}\right)+\left(\dfrac{(a^2-b^2)(a^2+b^2)-(a^2-b^2)(c^2)}{2abc}\right)$

$\longmapsto\left(\dfrac{(b^4-c^4)-(a^2b^2-a^2c^2)}{2abc}\right)+\left(\dfrac{(c^4-a^4)-(b^2c^2-a^2b^2)}{2abc}\right)+\left(\dfrac{(a^4-b^4)-(a^2c^2-b^2c^2)}{2abc}\right)$

$\longmapsto\dfrac{b^4-c^4-a^2b^2+a^2c^2}{2abc}+\dfrac{c^4-a^4-b^2c^2+a^2b^2}{2abc}+\dfrac{a^4-b^4-a^2c^2+b^2c^2}{2abc}$

On combining the fractions,

$\longmapsto\dfrac{(b^4-c^4-a^2b^2+a^2c^2)+(c^4-a^4-b^2c^2+a^2b^2)+(a^4-b^4-a^2c^2+b^2c^2)}{2abc}$

$\longmapsto\dfrac{b^4-c^4-a^2b^2+a^2c^2+c^4-a^4-b^2c^2+a^2b^2+a^4-b^4-a^2c^2+b^2c^2}{2abc}$

Regrouping the terms,

$\longmapsto\dfrac{(a^4-a^4)+(b^4-b^4)+(c^4-c^4)+(a^2b^2-a^2b^2)+(b^2c^2-b^2c^2)+(a^2c^2-a^2c^2)}{2abc}$

$\longmapsto\dfrac{(0)+(0)+(0)+(0)+(0)+(0)}{2abc}$

$\longmapsto\dfrac{0}{2abc}$

$\longmapsto\bf 0=RHS$

LHS = RHS proved.

7 0

2 years ago

As a contestant on a televised game show, Daniel gets to spin the big prize wheel, which has a diameter of 4 meters. What is the

shusha [124]

Area of a circle = pir^2
The radius is half the diameter so is 2 metres
So r^2 is 4
So the answer is 4pi or 12.57 metres squared
Let me know if you have any questions

8 0

3 years ago

The angles in a triangle are in the ratio 1:2:3 a) Show that the triangle is a right -angled triangle b) The hypotenuse of the t

Hitman42 [59]

Answer:

8.5 cm

Step-by-step explanation:

Angles of the triangle:

x+2x+3x= 180
6x= 180
x= 30

30°, 60° and 90°

Right triangle as it has 90° angle

Sides are a, b and 19 cm

sin 30°= a/19 ⇒ a= 19 sin 30°= 19*0.5= 8.5 cm = shortest side

sin 60°= b/19 ⇒ b= 19 sin 60°= 19*0.866= 16.45 cm

5 0

3 years ago

A) –2n + 9m – 17p

finlep [7]

-2n + 9mn - 17np

All you do is simplify the equation by distributing and multiplying the 1/3n to each of the numbers.

Hope this helps!

6 0

3 years ago

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