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Arturiano [62]
2 years ago
15

Calculate the area of this triangle. 17 cm 8 cm 15 cm- Area - Triangles Overview

Mathematics
1 answer:
Bezzdna [24]2 years ago
4 0

Answer:

There are 3 possible answers because you didn't state which value was base or height.

Going off the assumption that the base is 15 cm and the height is 8 cm, the area is 60 cm. (same answer if base is 8 cm and height is 15 cm)

A = 1/2(b*h)

A = 1/2(15*8)

A = 1/2(120)

A = 60

Going off the assumption that the base is 17 cm and the height is 8 cm, the area is 68 cm. (same answer if base is 8 cm and height is 17 cm)

A = 1/2(b*h)

A = 1/2(17*8)

A = 1/2(136)

A = 68

Going off the assumption that the base is 15 cm and the height is 17 cm, the area is 127.5 cm. (same answer if base is 17 cm and the height is 15 cm)

A = 1/2(b*h)

A = 1/2(15*17)

A = 1/2(255)

A = 127.5

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Find an explicit formula for the sequence 2, 4, 8, 16, 32
MA_775_DIABLO [31]

Answer:

a

n

=

4

⋅

2

n

−

1

if it is a geometric sequence.

or could be:  

a

n

=

1

3

(

2

n

3

−

6

n

2

+

16

n

)

if not.

Explanation:

There is a common ratio between successive pairs of terms:

8

4

=

2

16

8

=

2

32

16

=

2

So this looks like a geometric sequence with initial term  

a

=

4

and common ratio  

r

=

2

.

If so, the formula for the  

n

th term is:

a

n

=

a

r

n

−

1

=

4

⋅

2

n

−

1

This is probably the answer expected by the questioner.

However, note that any finite sequence of terms does not determine an infinite sequence - unless you are told what kind of sequence it is - e.g. arithmetic, geometric, harmonic.

For example, we can match these first  

4

terms with a cubic formula as follows:

Write down the sequence as a list:

4

,

8

,

16

,

32

Write down the sequence of differences between each pair of terms:

4

,

8

,

16

Write down the sequence of differences of this sequence:

4

,

8

Write down the sequence of differences of this sequence:

4

Having reached a constant sequence (albeit consisting of only one term), we can use the initial term of each of the sequences we have found as coefficients of a formula for the  

n

th term:

a

n

=

4

0

!

+

4

1

!

(

n

−

1

)

+

4

2

!

(

n

−

1

)

(

n

−

2

)

+

4

3

!

(

n

−

1

)

(

n

−

2

)

(

n

−

3

)

=

4

+

(

4

n

−

4

)

+

(

2

n

2

−

6

n

+

4

)

+

(

2

3

n

3

−

4

n

2

+

22

3

n

−

4

)

=

2

3

n

3

−

2

n

2

+

16

3

n

=

1

3

(

2

n

3

−

6

n

2

+

16

n

Step-by-step explanation:

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