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nasty-shy [4]
3 years ago
9

Jamal stands on a dock 1.5 meters above the surface of the water. A trout swims 4.8 meters below Jamal. What is the position of

the truck compared to the surface of the water?
Mathematics
2 answers:
Ivanshal [37]3 years ago
7 0

Answer:

s+1.5=4.8

s = 4.8-1.5=3.3

uysha [10]3 years ago
6 0

Answer:

Jamal stands on a dock 1.5 meters above the surface of the water.A trout swims 4.8 meters below Jamal.What is the depth of the trout compared to the surface of the water?

The position of the trout is 3.3 meters below the surface of the water.

Step-by-step explanation:

Given:

Jamal's position from the surface of the water = 1.5 meters

Trout's position with reference to Jamal's position =4.8 meters (Below)

Trout's position from the surface of the water is unknown.

Let

Trouts position be 's'.

And the surface of the water as equivalent to the origin (0,0) co-ordinate of the Cartesian.

Below water we will assign negative height as it lies on negative y-axis (below the origin).

We have to find 's',which can be known from adding both the values.

s+1.5=4.8

s = 4.8-1.5=3.3

So the trout is swimming 3.3 meters below the surface of the water.

We can say that it is below the water so it can also be written as negative 3.3 (according to our coordinates).

The position of the trout is 3.3 meters below the surface of the water.

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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
PLEASE HELP: The mass of a ball is 128 g. It has a density of 0.5 g/cm³. What is the volume of the ball? Show your work and plac
tiny-mole [99]

Answer:

Part 1) The volume of the ball is 256 cm³

Part 2) The radius of the ball is 3.94 cm

Step-by-step explanation:

Part 1) we know that

The density is equal to divide the mass by the volume

D=m/V

Solve for the volume

The volume is equal to divide the mass by the density

V=m/D

In this problem we have

m=128 g

D=0.5 g/cm³

substitute

V=128/0.5=256 cm³

Part 2) what is the radius of the ball?

we know that

The volume of the sphere (ball) is equal to

V=\frac{4}{3}\pi r^{3}

we have

V=256\ cm^{3}

assume

\pi =3.14

substitute and solve for r

256=\frac{4}{3}(3.14)r^{3}

768=(12.56)r^{3}

r^{3}=768/(12.56)

r=3.94\ cm

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Answer:

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Step-by-step explanation:

Finite Math Examples. Popular Problems · Finite Math. Find the Distance Between Two Points (7,2) , (-1,4).

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