Find four consecutive integers such that 3 times the sum of the first and the fourth was 114 less than the product of -5 and the sum of the first two consecutive integers. what are the integers?

The four  such consecutive integers are 8, 9 , 10 and 11.

Step-by-step explanation:

Let first integer = k

Then (k+1), ((k+1)+1), and (((k+1)+1)+1) are the next consecutive integers are.

Or, k , (k+1), (k+2) and (k+3) are four consecutive integers.

Now according to the question,

3(k +(k+3)) = -5(k + (k+1))-114

or simplifying  this , we get

3(2k+3)  = -5(2k+1)  - 114

or, 6k + 9 = -10k - 5 - 114

⇒ 6k + 10k = -119 - 9

or, 16k = -128

or, k = 128/16 =8

Hence, the first integer = k = 8

The next consecutive integers are (k+1), (k+2) and (k+3) = 9, 10 and 11

Hence, the four such consecutive integers are 8, 9 , 10 and 11.

hi! i think the answer is 1/2 : )

the answer is b: 18

step-by-step explanation:

use the pythagorean theorem to .

a^2+b^2=c^2

plug 18 for "r"

the 2 legs are 18 and 24

18^2+24^2=c^2

324+576=c^2

900=c^2

c=30

30-12=18

18 is the radius

: )