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3 years ago

A rectangles width is one-fourth of its length . its area is 9 square units

Mathematics

1 answer:

Paha777 [63]3 years ago

4 0

Answer:

Length = 6

Width = 3/2

Step-by-step explanation:

You can choose to go about this question in lots of different ways, but this is how i did it. I drew a diagram of the rectangle and labeled it: the height/length as x and the width as x/4 (they told us the width is a quarter of the length). They also told us that the area is 9 square units.

We know that length x width = area

So: length x width = 9

I'm gonna bring in the variables i used in the diagram, so:

$x$ x $\frac{x}{4}$ = 9

Keep solving:

$\frac{x^{2} }{4}$ = 9

Times both sides by 4:

$x^{2}$ = 36

Square root both sides for x:

* remember that when you square root, its possible to have a positive or negative version of the value

x = ±6

Since its impossible to have a negative length, we can say that x = 6.

X was just the variable we gave to the length of the rectangle, so now we know the length is 6. If you needed to find width as well, you can do 6/4, which simplifies to 3/2.

To double-check

- 6 x 3/2 should equal 9, since length x width = area

which it is, so we're correct!

Hope that helped : )

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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

A concert hall has three sections: the main floor, the balcony, and the gallery. For a recent concert, 1020 tickets were sold. T

Angelina_Jolie [31]

Answer:

Let the number of balcony seats be b

Let the number of gallery tickets be g

Let the number of main floor tickets be m

b + g + m = 1020

b = 2g

m = b + 225

The number of balcony tickets = 318

The number of gallery tickets = 159

The number of main floor tickets = 543

Step-by-step explanation:

Let the number of balcony seats be b

Let the number of gallery tickets be g

Let the number of main floor tickets be m

i) Therefore we can write b + g + m = 1020

ii) there were twice as many balcony tickets sold as gallery tickets

therefore b = 2g $\Rightarrow$ g = 0.5b

iii) there were 225 more main floor tickets than balcony tickets

therefore m = b + 225

iv) using equation ii) and equation iii) in equation i) we can write

b + g + m = b + 0.5b + b + 225 = 1020 $\Rightarrow$ 2.5b = 795 $\Rightarrow$ b = 318

v) g = 0.5 b = 0.5 $\times$ 318 = 159

vi) m = b + 225 = 318 + 225 = 543

5 0

3 years ago

Read 2 more answers