Write the facts family numbers 15,5,and 3

If 5^n x 5^3 = 5^6, then n is _____. A) 18 B) 2 C) 3 D) 9

Eddi Din [679]

<h3>Answer: C) 3</h3>

The rule we'll use is a^b*a^c = a^(b+c). So we add the exponents.

That means 5^n*5^3 = 5^(n+3)

So 5^n*5^3 = 5^6 turns into 5^(n+3) = 5^6

The bases are equal to 5, so the exponents be equal to one another.

n+3 = 6

n+3-3 = 6-3

n = 3

So 5^3*5^3 = 5^(3+3) = 5^6.

6 0

2 years ago

Read 2 more answers

I have several other questions

Licemer1 [7]

Answer:

81 m²

Step-by-step explanation:

A = 6*10 + 7*6/2

= 60 + 7*3

= 60 + 21

= 81 m²

6 0

3 years ago

WILL GIVE BRAINLIEST:

vfiekz [6]

Answer:

you roll a 6-sided die 6 times, what is the best prediction possible for the number of times you will roll a five? 5. 1. 2. 6. Submit. Submit. Continue. Correct!

Step-by-step explanation:

6 0

3 years ago

Read 2 more answers

Evaluate when a = 8 and b = 4. ab^2 a−b

eduard

Evaluate when a = 8 and b = 4.

ab^2
a−b

ab² = (8)(4)² = 8*16= 128

a - b = 8 - 4 = 4

3 0

2 years ago

Sin4x.sin5x+sin4x.sin3x-sin2x.sinx=0

andreev551 [17]

Recall the angle sum identity for cosine:

cos(x + y) = cos(x) cos(y) - sin(x) sin(y)

cos(x - y) = cos(x) cos(y) + sin(x) sin(y)

==> sin(x) sin(y) = 1/2 (cos(x - y) - cos(x + y))

Then rewrite the equation as

sin(4x) sin(5x) + sin(4x) sin(3x) - sin(2x) sin(x) = 0

1/2 (cos(-x) - cos(9x)) + 1/2 (cos(x) - cos(7x)) - 1/2 (cos(x) - cos(3x)) = 0

1/2 (cos(9x) - cos(x)) + 1/2 (cos(7x) - cos(3x)) = 0

sin(5x) sin(-4x) + sin(5x) sin(-2x) = 0

-sin(5x) (sin(4x) + sin(2x)) = 0

sin(5x) (sin(4x) + sin(2x)) = 0

Recall the double angle identity for sine:

sin(2x) = 2 sin(x) cos(x)

Rewrite the equation again as

sin(5x) (2 sin(2x) cos(2x) + sin(2x)) = 0

sin(5x) sin(2x) (2 cos(2x) + 1) = 0

sin(5x) = 0 or sin(2x) = 0 or 2 cos(2x) + 1 = 0

sin(5x) = 0 or sin(2x) = 0 or cos(2x) = -1/2

sin(5x) = 0 ==> 5x = arcsin(0) + 2nπ or 5x = arcsin(0) + π + 2nπ

… … … … … ==> 5x = 2nπ or 5x = (2n + 1)π

… … … … … ==> x = 2nπ/5 or x = (2n + 1)π/5

sin(2x) = 0 ==> 2x = arcsin(0) + 2nπ or 2x = arcsin(0) + π + 2nπ

… … … … … ==> 2x = 2nπ or 2x = (2n + 1)π

… … … … … ==> x = nπ or x = (2n + 1)π/2

cos(2x) = -1/2 ==> 2x = arccos(-1/2) + 2nπ or 2x = -arccos(-1/2) + 2nπ

… … … … … … ==> 2x = 2π/3 + 2nπ or 2x = -2π/3 + 2nπ

… … … … … … ==> x = π/3 + nπ or x = -π/3 + nπ

(where n is any integer)

5 0

2 years ago