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tester [92]
2 years ago
5

Simplify: √36a^3b^4 A) 6ab^2√a B) 6ab^2 C) 6ab^2√ab D) 6ab^3

Mathematics
1 answer:
UkoKoshka [18]2 years ago
8 0
Your answer is A !!
Can you mark me BRAINLEST
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Polar form of (x+6)^2 +y^2=36
Sindrei [870]

Answer:

  r = -12cos(θ)

Step-by-step explanation:

The usual translation can be used:

  • x = r·cos(θ)
  • y = r·sin(θ)

Putting these relationships into the formula, we have ...

  (r·cos(θ) +6)² +(r·sin(θ))² = 36

  r²·cos(θ)² +12r·cos(θ) +36 +r²·sin(θ)² = 36

  r² +12r·cos(θ) = 0 . . . . subtract 36, use the trig identity cos²+sin²=1

  r(r +12cos(θ)) = 0

This has two solutions for r:

  r = 0 . . . . . . . . a point at the origin

  r = -12cos(θ) . . . the circle of interest

4 0
3 years ago
A rectangle's width is twice as long as its length. If the perimeter is 36 in., what is the length?
Bumek [7]
Okay here: 

Let L be the length and W the width of the rectangle. 

1. w=2•L
2. 2L+2W=36

Substitute eq. 1 into eq. 2, 

2L+2•(2•L)=36

2L+4L=36

6L=36

L=9

Not done yet, then from eq. 1, 
 
W=2•L
W=2•9
W=18 <------- Your answer. :)
6 0
3 years ago
What is the greatest common multiple of 96,48,84
never [62]

Answer: 12

Step-by-step explanation: 96-84=12

I dont really know how i did it I just did.

5 0
3 years ago
Select all expressions shown that are equivalent to
Murrr4er [49]

Answer:

A.  and C.

if you simplify the expressions, you will see which ones are equal to the expression given

4 0
2 years ago
Drag the tiles to the correct boxes to complete the pairs. Not all tiles will be used. Match each verbal description of a sequen
galben [10]

Answer:

I think the question is wrong so, I will try and explain with some right questions

Step-by-step explanation:

We are give 6 sequences to analyse

1. an = 3 · (4)n - 1

2. an = 4 · (2)n - 1

3. an = 2 · (3)n - 1

4. an = 4 + 2(n - 1)

5. an = 2 + 3(n - 1)

6. an = 3 + 4(n - 1)

1. This is the correct sequence

an=3•(4)^(n-1)

If this is an

Let know an+1, the next term

an+1=3•(4)^(n+1-1)

an+1=3•(4)^n

There fore

Common ratio an+1/an

r= 3•(4)^n/3•(4)^n-1

r= (4)^(n-n+1)

r=4^1

r= 4, then the common ratio is 4

Then

First term is when n=1

an=3•(4)^(n-1)

a1=3•(4)^(1-1)

a1=3•(4)^0=3.4^0

a1=3

The first term is 3 and the common ratio is 4, it is a G.P

2. This is the correct sequence

an=4•(2)^(n-1)

Therefore, let find an+1

an+1=4•(2)^(n+1-1)

an+1= 4•2ⁿ

Common ratio=an+1/an

r=4•2ⁿ/4•(2)^(n-1)

r=2^(n-n+1)

r=2¹=2

Then the common ratio is 2,

The first term is when n =1

an=4•(2)^(n-1)

a1=4•(2)^(1-1)

a1=4•(2)^0

a1=4

It is geometric progression with first term 4 and common ratio 2.

3. This is the correct sequence

an=2•(3)^(n-1)

Therefore, let find an+1

an+1=2•(3)^(n+1-1)

an+1= 2•3ⁿ

Common ratio=an+1/an

r=2•3ⁿ/2•(3)^(n-1)

r=3^(n-n+1)

r=3¹=3

Then the common ratio is 3,

The first term is when n =1

an=2•(3)^(n-1)

a1=2•(3)^(1-1)

a1=2•(3)^0

a1=2

It is geometric progression with first term 2 and common ratio 3.

4. I think this correct sequence so we will use it.

an = 4 + 2(n - 1)

Let find an+1

an+1= 4+2(n+1-1)

an+1= 4+2n

This is not GP

Let find common difference(d) which is an+1 - an

d=an+1-an

d=4+2n-(4+2(n-1))

d=4+2n-4-2(n-1)

d=4+2n-4-2n+2

d=2.

The common difference is 2

Now, the first term is when n=1

an=4+2(n-1)

a1=4+2(1-1)

a1=4+2(0)

a1=4

This is an arithmetic progression of common difference 2 and first term 4.

5. I think this correct sequence so we will use it.

an = 2 + 3(n - 1)

Let find an+1

an+1= 2+3(n+1-1)

an+1= 2+3n

This is not GP

Let find common difference(d) which is an+1 - an

d=an+1-an

d=2+3n-(2+3(n-1))

d=2+3n-2-3(n-1)

d=2+3n-2-3n+3

d=3.

The common difference is 3

Now, the first term is when n=1

an=2+3(n-1)

a1=2+3(1-1)

a1=2+3(0)

a1=2

This is an arithmetic progression of common difference 3 and first term 2.

6. I think this correct sequence so we will use it.

an = 3 + 4(n - 1)

Let find an+1

an+1= 3+4(n+1-1)

an+1= 3+4n

This is not GP

Let find common difference(d) which is an+1 - an

d=an+1-an

d=3+4n-(3+4(n-1))

d=3+4n-3-4(n-1)

d=3+4n-3-4n+4

d=4.

The common difference is 4

Now, the first term is when n=1

an=3+4(n-1)

a1=3+4(1-1)

a1=3+4(0)

a1=3

This is an arithmetic progression of common difference 4 and first term 3.

5 0
3 years ago
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