Answer:
14, 28, 42, 56, and 70 are the first 5 common multiples of 14
20, 40, 60, 80, and 100 are the first 4 common multiples of 20
Answer:
Step-by-step explanation:
Answer:
23rd term of the arithmetic sequence is 118.
Step-by-step explanation:
In this question we have been given first term a1 = 8 and 9th term a9 = 48
we have to find the 23rd term of this arithmetic sequence.
Since in an arithmetic sequence
here a = first term
n = number of term
d = common difference
since 9th term a9 = 48
48 = 8 + (9-1)d
8d = 48 - 8 = 40
d = 40/8 = 5
Now 
= 8 + (23 -1)5 = 8 + 22×5 = 8 + 110 = 118
Therefore 23rd term of the sequence is 118.
The answer is A because the bigger/ larger the Apple the longer time it takes to eat it
Answer:
32 5/8 (or 32 and 5/8)
Step-by-step explanation:
Okay, so this problem is asking for us to solve this problem with the substitution of a variable: x = -4. Before we fully solve this problem, we need to replace all of those x variables with -4 so that it is easier to solve.
-7 1/4(-4) + 3 5/8.
To make this problem even easier to solve, let's turn these mixed numbers into improper fractions. To do this, multiply the denominator by the whole number. Then add the numerator to this number. The new number that you just got is now your new numerator for this number.
-29/4(-4) + 3 5/8.
Repeat the last step to turn the other mixed number into an improper fraction.
-29/4(-4) + 29/8.
Now let's multiply that -4 by 29/4. Because a whole number technically has a denominator of 1, we can now set up the next part of our problem. (The symbol * means multiplication since we can't use x since it is already being used as a variable in this equation.)
(-29/4 * -4/1) + 29/8.
29 + 29/8.
Now to solve the rest of this problem, let's convert the whole number of 29 so that it has a denominator of 8. This is that these two numbers are addable.
<u>29</u> x <u>8</u> = <u>232</u>
1 x 8 = 8
Now these numbers are addable. So:
232/8 + 29/8 = 261/8 = 32 5/8.
I hope that this helps.
