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mr Goodwill [35]
2 years ago
9

Find three consecutive even integers if three times the first integer is two times the third.​

Mathematics
2 answers:
drek231 [11]2 years ago
6 0

Using a system of equations, it is found that the three consecutive even integers are -8, -6 and -4.

<h3>What is a system of equations?</h3>

A system of equations is when two or more variables are related, and equations are built to find the values of each variable.

In this problem, the three consecutive even integers are represented by:

n, n + 2, n + 4

The first integer is two times the third, hence:

n = 2(n + 4)

2n + 8 = n

n = -8

Thus:

n = -8

n + 2 = -8 + 2 = -6

n + 4 = -8 + 4 = -4

The integers are -8, -6 and -4.

To learn more about system of equations, you can take a look at brainly.com/question/26650649

iren [92.7K]2 years ago
5 0

Answer:

8,10,12

Step-by-step explanation:

Step-by-step explanation:

Smallest Integer = 2x - 2

Middle integer = 2x

Largest integer = 2x + 2

All three are every

3(2x - 2) = 2(2x + 2)

6x - 6 = 4x + 4                Add 6 to both sides

6x = 4x + 6 + 4               Combine

6x = 4x + 10                    Subtract 4x from both sides

6x-4x = 10                      Combine

2x = 10                           Divide by 2

x = 10/2

x = 5

2x - 2 = 2*5 - 2 = 8

2x =                     10

2x+2                    12

3*8 = 2*12

24 = 24

So the answers are correct.

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The question is incomplete. The complete question is :

The volume of a right circular cone with radius r and height h is V = pir^2h/3. a. Approximate the change in the volume of the cone when the radius changes from r = 5.9 to r = 6.8 and the height changes from h = 4.00 to h = 3.96.

b. Approximate the change in the volume of the cone when the radius changes from r = 6.47 to r = 6.45 and the height changes from h = 10.0 to h = 9.92.

a. The approximate change in volume is dV = _______. (Type an integer or decimal rounded to two decimal places as needed.)

b. The approximate change in volume is dV = ___________ (Type an integer or decimal rounded to two decimal places as needed.)

Solution :

Given :

The volume of the right circular cone with a radius r and height h is

$V=\frac{1}{3} \pi r^2 h$

$dV = d\left(\frac{1}{3} \pi r^2 h\right)$

$dV = \frac{1}{3} \pi h \times d(r^2)+\frac{1}{3} \pi r^2 dh$

$dV = \frac{2}{3} \pi r h (dr)+\frac{1}{3} \pi r^2 dh$

a). The radius is changed from r = 5.9 to r = 6.8 and the height is changed from h = 4 to h = 3.96

So, r = 5.9  and dr = 6.8 - 5.9 = 0.9

     h = 4  and dh = 3.96 - 4 = -0.04

Now, $dV = \frac{2}{3} \pi r h (dr)+\frac{1}{3} \pi r^2 dh$

$dV = \frac{2}{3} \pi (5.9)(4)(0.9)+\frac{1}{3} \pi (5.9)^2 (-0.04)$

$dV=44.484951 - 1.458117$

$dV=43.03$

Therefore, the approximate change in volume is dV = 43.03 cubic units.

b).  The radius is changed from r = 6.47 to r = 6.45 and the height is changed from h = 10 to h = 9.92

So, r = 6.47  and dr = 6.45 - 6.47 = -0.02

     h = 10  and dh = 9.92 - 10 = -0.08

Now, $dV = \frac{2}{3} \pi r h (dr)+\frac{1}{3} \pi r^2 dh$

$dV = \frac{2}{3} \pi (6.47)(10)(-0.02)+\frac{1}{3} \pi (6.47)^2 (-0.08)$

$dV=-2.710147-3.506930$

$dV= -6.22$

Hence, the approximate change in volume is dV = -6.22 cubic units

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