Answer:
The statements are incorrect as: The sum of even numbers from 1 to 100(i.e. 2550) is not double\twice of the sum of odd numbers from 1 to 100(i.e. 2500).
Step-by-step explanation:
We know that sum of an Arithmetic Progression(A.P.) is given by:
where 'n' denotes the "number" of digits whose sum is to be determined, 'a' denotes the first digit of the series and '' denote last digit of the series.
Now the sum of even numbers i.e. 2+4+6+8+....+100 is given by the use of sum of the arithmetic progression since the series is an A.P. with a common difference of 2.
image with explanation
Hence, sum of even numbers from 1 to 100 is 2550.
Also the series of odd numbers is an A.P. with a common difference of 2.
sum of odd numbers from 1 to 100 is given by: 1+3+5+....+99
.
Hence, the sum of all the odd numbers from 1 to 100 is 2500.
Clearly the sum of even numbers from 1 to 100(i.e. 2550) is not double of the sum of odd numbers from 1 to 100(i.e. 2500).
Hence the statement is incorrect.
Step-by-step explanation:
Considering the least common multiple of the denominators, it is found that the result of the expression is given by:
<h3>How do we add fractions?</h3>
We place all the terms of the addition in "equivalent" fractions, with the same denominator, found from the last common multiple of all the denominators.
In this problem, the denominators are as follows: 28, 70, 130, 208, 304. Using a calculator, their lcm is of 138,320.
Considering equivalent fractions(the numerators are the division of the lcm by the previous denominator), the expression is:
More can be learned about the addition of fractions at brainly.com/question/78672
#SPJ1
Answer:
Both fireworks will explode after 1 seconds after firework b launches.
Step-by-step explanation:
Given:
Speed of fire work A= 300 ft/s
Speed of Firework B=240 ft/s
Time before which fire work b is launched =0.25s
To Find:
How many seconds after firework b launches will both fireworks explode=?
Solution:
Let t be the time(seconds) after which both the fireworks explode.
By the time the firework a has been launched, Firework B has been launch 0.25 s, So we can treat them as two separate equation
Firework A= 330(t)
Firework B=240(t)+240(0.25)
Since we need to know the same time after which they explode, we can equate both the equations
330(t) = 240(t)+240(0.25)
300(t)= 240(t)+60
300(t)-240(t)= 60
60(t)=60
t=1
A center line of a rectangle divides it into 2 equal pieces. Therefore, the chance that the point will be on the left, or the right side for that matter, side of the rectangle is 50%.
