You didn’t attach anything?..
Answer:
hi there ☺️
Here we will use algebra to find three consecutive integers whose sum is 345. We start by assigning X to the first integer. Since they are consecutive, it means that the 2nd number will be X + 1 and the 3rd number will be X + 2 and they should all add up to 345. Therefore, you can write the equation as follows:
(X) + (X + 1) + (X + 2) = 345
To solve for X, you first add the integers together and the X variables together. Then you subtract three from each side, followed by dividing by 3 on each side. Here is the work to show our math:
X + X + 1 + X + 2 = 345
3X + 3 = 345
3X + 3 - 3 = 345 - 3
3X = 342
3X/3 = 342/3
X = 114
Which means that the first number is 114, the second number is 114 + 1 and the third number is 114 + 2. Therefore, three consecutive integers that add up to 345 are 114, 115, and 116.
114 + 115 + 116 = 345
We know our answer is correct because 114 + 115 + 116 equals 345 as displayed above.
Step-by-step explanation:
pls rate me the brainiest
10 / (1/3); she has 30 pounds of seeds in her supply.
There are 10,000 total four-digit numbers (1000 through 9999).
Multiples of 2 end in 0, 2, 4, 6, and 8. There are 9*10*10*5 = 4500 four-digit multiples of 2.
Multiples of 5 end in 0 or 5. There are 9*10*10*2 = 1800 four-digit multiples of 5.
There is redundancy between the two sets of numbers, namely those that end in 0, which are both multiples of 2 and 5. There are 9*10*10*1 = 900 four-digit multiples of both 2 and 5.
Then there are 4500 + 1800 - 900 = 5400 total four-digit numbers that are either multiples of 2 or 5, which means the remaining 4600 numbers are neither multiples of 2 nor 5.
Isn’t it h?
extra words so i can actually post this reply
