How many itegers from 15 to 85, inclusive are multibles of 8

9

Step-by-step explanation:

First multiple of 8 in that range is 8(2)=16.

The last multiple of 8 in that range is 8(10)=80.

So we just need to find how many numbers there are between 2 and 10. inclusive.

10-2+1=9

It's also not that long to write out and count.

1-2

2-3

3-4

4-5

5-6

6-7

7-8

8-9

9-10

9 numbers there are

Multiply those 9 numbers by i you will have all multiples of 8 btw 15 and 85.

8(2)=16

8(3)=24

8(4)=32

8(5)=40

8(6)=48

8(7)=56

8(8)=64

8(9)=72

8(10)=80

answer is : muchas/tipicas

have a nice day!

~recklesschicken

x = 2

step-by-step explanation:

if sr parallel to ba then the sides bc and ac are in proportion, that is

= ← substitute values

= ( cross- multiply )

27x = 6(4x + 1)

27x = 24x + 6 ( subtract 24x from both sides )

3x = 6 ( divide both sides by 3 )

x = 2