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3 years ago

A sequence starts 1,5,9,13,17 What’s the formula for the nth term of this sequence??

Mathematics

2 answers:

worty [1.4K]3 years ago

6 0

This is an arithmetic sequence, since it's pattern is +4.

n1 = 1

d = 4

n = 1 + (n-1)(4) <-------------- This is the formula for the nth term of the sequence.

N76 [4]3 years ago

6 0

The pattern is going up by 4 so you take away 4
1-4=-3

The nth term = 4n-3

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4 years ago

MN and KL are parallel. If JM = 6, MK = 12, and LN = 18, what is NJ?

Galina-37 [17]

6/12=X/18
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Answer: NJ=9

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3 years ago

Read 2 more answers

Does anyone know the answer?

umka21 [38]

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Step-by-step explanation: I've attached the explanation

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3 years ago

Simplify each expression <br> 4+8(-2n+3)

artcher [175]

Answer:

28 -16n

Step-by-step explanation:

Use the distributive property, then collect terms.

4 +8(-2n +3)

= 4 +8(-2n) +8(3)

= 4 -16n +24

= 28 -16n

_____

This could also be written

-16n +28

but then the leading sign would be negative. Sometimes leading minus signs get lost, so I prefer the first term to be positive.

3 0

3 years ago

An oil refinery is located on the north bank of a straight river that is 2 km wide. A pipeline is to be constructed from the ref

hichkok12 [17]

Answer:

P is exactly 3km east from the oil refinery.

Step-by-step explanation:

Let's d be the distance in km from the oil refinery to point P. So the horizontal distance from P to the storage is 3 - d and the vertical distance is 2. Hence the diagonal distance is:

$\sqrt{(3 - d)^2 + 2^2} = \sqrt{(3 - d)^2 + 4}$

So the cost of laying pipe under water with this distance is

$800000\sqrt{(3 - d)^2 + 4}$

And the cost of laying pipe over land from the refinery to point P is 400000d. Hence the total cost:

$800000\sqrt{(3 - d)^2 + 4} + 400000d$

We can find the minimum value of this by taking the 1st derivative and set it to 0

$800000\frac{2*0.5*(3-d)(-1)}{\sqrt{(3 - d)^2 + 4}} + 400000 = 0$

We can move the first term over to the right hand side and divide both sides by 400000

$1 = 2\frac{3 - d}{\sqrt{(3 - d)^2 + 4}}$

$\sqrt{(3 - d)^2 + 4} = 6 - 2d$

From here we can square up both sides

$(3 - d)^2 + 4 = (6 - 2d)^2$

$9 - 6d + d^2 + 4 = 36 - 24d + 4d^2$

$3d^2-18d+27 = 0$

$d^2 - 6d + 9 = 0$

$(d - 3)^2 = 0$

$d -3 = 0$

d = 3

So the cost of pipeline is minimum when P is exactly 3km east from the oil refinery.

3 0

3 years ago