Factors of 84: 1, 2<span>, </span>3<span>, 4, 6, </span>7<span>, 12, </span>14<span>, </span>21<span>, </span>28<span>, </span>42<span>, 84. Prime factorization: 84 = </span>2<span> x </span>2<span> x </span>3<span>x </span>7<span> which can also be written (</span>2^2<span>) x </span>3<span> x </span>7<span>.</span>
Answer:
30
Step-by-step explanation:
15cm for 5
5 x 2=10
15 x 2 = 30
Answer:
The solution is
.
Step-by-step explanation:
Given:
The inequality given is:

In order to simplify for 'x', we first isolate 'x' on one side.
Adding -4 on both sides, we get:

Now,
is an absolute value function which is defined as:

Therefore, the given inequality can be rewritten as:
and 
Therefore, the solution is
.
The 24th term is 152
Step-by-step explanation:
The formula of the nth term of an arithmetic sequence is:
, where
- a is the first term
- d is the common difference between consecutive terms
The third term means n = 3
∵ 
∴ 
∵
= 5
- Equate the right hand sides of the third term
∴ a + 2d = 5 ⇒ (1)
The fifth term means n = 5
∵ 
∴ 
∵
= 19
- Equate the right hand sides of the fifth term
∴ a + 4d = 19 ⇒ (2)
Now we have a system of equations to solve it
Subtract equation (1) from equation (2) to eliminate a
∴ 2d = 14
- Divide both sides by 2
∴ d = 7
- Substitute the value of d in equation (1) to find a
∵ a + 2(7) = 5
∴ a + 14 = 5
- Subtract 14 from both sides
∴ a = -9
The twenty fourth term means n = 24
∵ a = -9 and d = 7
- Substitute the values of a and d in the formula of the nth term
∴ 
∴ 
∴ 
∴ 
The 24th term is 152
Learn more:
You can learn more about the arithmetic sequence in brainly.com/question/7221312
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The two dot plots are missing, so i have attached it.
Answer:
The mean at the beginning of the school year was 9.5 miles and the mean at the end of the school year was 10.2 miles
Step-by-step explanation:
From the attached image, we are told to compare the means for each plot to the nearest tenth.
Mean = Σx/n
Now, from the image, total number of miles run by the 14 students at the beginning of the school year is;
(1 × 7) + (2 × 8) + (4 × 9) + (4 × 10) + (2 × 11) + (1 × 12) = 133
Mean of miles run at the beginning of the school year = 133/14 = 9.5 miles
Again, from the table, total miles run at the end of the school year = (2 × 8) + (2 × 9) + (4 × 10) + (3 × 11) + (3 × 12) = 143
Mean of miles run at the end of the school year = 143/14 = 10.2 miles
Thus;
The mean at the beginning of the school year was 9.5 miles and the mean at the end of the school year was 10.2 miles