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4 years ago

6

find 3 positive consecutive integers such that the product of the first and the third integer is 17 more than three times the se

cond integer

1 answer:

MariettaO [177]4 years ago

4 0

The required three consecutive numbers are 5,6 and 7.

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My name is Ann [436]

2^2= 4
4x4= 16
answer is 16

4 0

3 years ago

Read 2 more answers

Complete the ratio table below to show ratios equivalent to 4:18.

liubo4ka [24]

The blanks are weird. I don't understand them!

3 0

3 years ago

Read 2 more answers

I can't quite figure out what ( 7- 4n ) x 6 is with distributive property.

Tema [17]

Answer:

(42-24n)

Step-by-step explanation:

(7-4n)×6

take the 6 and multiply it to the integers in the paraenthesis.

so you take the ouside number which is 6 and multiply it with the first number which is 7 in the paraenthesis.

6×7=42

(42-4n)

then take the next number 4n and multiply by 6

4n × 6 = 24n

finally put it all together

(42-24n)

sorry if that didnt help

7 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

What is the difference between the temperature - 7 Celsius and - 12 Celsius on a scatter diagram

xeze [42]

Answer:

I have to go get my car from the doctor office today

3 0

3 years ago