The sum of five consecutive even integers is 1310. find the integers

5 consecutive  even integers: n, n+2,n+4,n+6,n+8

n+(n+2)+(n+4)+(n+6)+(n+8)=1310

combine like terms

5n+20 = 1310

subtract 20 from each side

5n = 1290

divide by 5

n=258

n+2 = 260

n+4=262

n+6=264

n+8=266

258,260,262,264,266

Step-by-step explanation:

Let the first of these even numbers = n

The second one is n + 2

The third one is n + 4

The fourth one is n + 6

The fifth one is n + 8

Their sum is 1310

Equation

n + n + 2 + n + 4 + n + 6 + n + 8 = 1310

Solution

There are 5 ns.

The total of the numbers is 2 + 4 + 6 + 8 = 20

Simplify the given equation

5n +20 = 1310                        Subtract  20 from both sides.

5n + 20 - 20 = 1310 - 20       Combine

5n = 1290                               Divide by 5

5n/5 = 1290/5                         Divide

Answers

n1 = 258

n2 = 260

n3 = 262

n4 = 264

n5 = 266

Check

The total is 258 + 260 + 262 + 264 + 266 = 1310

It checks.

the probability is 0

answer: (-8,4) are the coordinates of a

step-by-step explanation:

let the side of square abcd is x .

thus the area of square abcd is .

now, by the diagram area of triangle ead= 1/2 × ea × ad = 1/2 × ea × x

( because ad is the side of square)

and, according to the question,

the areas of square abcd and right ead are equal.

⇒

⇒ea=2x

thus, ea: ab=2: 1

therefore, point a divides line segment eb into the ratio 2: 1.

⇒ by sectional formula coordinates of a are,

=

=

=(-8, 4)

thus the coordinates of a are (-8,4)