Answer:
The first '5' terms of the given sequence
![\frac{7}{2} , 4 ,\frac{9}{2} ,5,\frac{11}{2}](https://tex.z-dn.net/?f=%5Cfrac%7B7%7D%7B2%7D%20%2C%204%20%2C%5Cfrac%7B9%7D%7B2%7D%20%2C5%2C%5Cfrac%7B11%7D%7B2%7D)
Step-by-step explanation:
<u><em>Explanation:-</em></u>
Given that the
aₙ = ![\frac{N}{2} + 3](https://tex.z-dn.net/?f=%5Cfrac%7BN%7D%7B2%7D%20%2B%203)
<em>first term</em>
Put n=1 ⇒ ![\frac{1}{2} + 3 = \frac{7}{2}](https://tex.z-dn.net/?f=%5Cfrac%7B1%7D%7B2%7D%20%2B%203%20%3D%20%5Cfrac%7B7%7D%7B2%7D)
<em>second term</em>
Put n=2 ⇒ ![\frac{2}{2} + 3 = 1+3=4](https://tex.z-dn.net/?f=%5Cfrac%7B2%7D%7B2%7D%20%2B%203%20%3D%201%2B3%3D4)
<em>Third term</em>
Put n=3 ⇒ ![\frac{3}{2} + 3 = \frac{3+6}{2}=\frac{9}{2}](https://tex.z-dn.net/?f=%5Cfrac%7B3%7D%7B2%7D%20%2B%203%20%3D%20%5Cfrac%7B3%2B6%7D%7B2%7D%3D%5Cfrac%7B9%7D%7B2%7D)
<em>Fourth term</em>
Put n=4 ⇒ ![\frac{4}{2} + 3 = 2+3 =5](https://tex.z-dn.net/?f=%5Cfrac%7B4%7D%7B2%7D%20%2B%203%20%3D%202%2B3%20%3D5)
<em>Fifth term</em>
Put n=5 ⇒ ![\frac{5}{2} + 3 = \frac{5+6}{2}=\frac{11}{2}](https://tex.z-dn.net/?f=%5Cfrac%7B5%7D%7B2%7D%20%2B%203%20%3D%20%5Cfrac%7B5%2B6%7D%7B2%7D%3D%5Cfrac%7B11%7D%7B2%7D)
The first '5' terms of the given sequence
![\frac{7}{2} , 4 ,\frac{9}{2} ,5,\frac{11}{2}](https://tex.z-dn.net/?f=%5Cfrac%7B7%7D%7B2%7D%20%2C%204%20%2C%5Cfrac%7B9%7D%7B2%7D%20%2C5%2C%5Cfrac%7B11%7D%7B2%7D)
D.
The unary postfix decrement operator, when used in an expression,
decrements the value of its operand but has the old value of the
operand.
The answer to this question is simple you just and on a decimal point 13.6
Answer:
20 seconds
Step-by-step explanation:
The least common multiple of 4 and 5 is 20. So the two lights will blink at the same time every 20 seconds.
The simplified product of
and -3
is -30
.
Given Two expressions: -3
and 2
.
We have to multiply both the expressions and it can be done as under:
-3
*2![\sqrt{5x^{3} }](https://tex.z-dn.net/?f=%5Csqrt%7B5x%5E%7B3%7D%20%7D)
Firstly we have to multiply -3 with 2 to get
=-6
Then we have to find square root of x cube and x square which is x to the power 3/2 and x to the power 1.
=![-6x^{3/2} x\sqrt{10}\sqrt{5}](https://tex.z-dn.net/?f=-6x%5E%7B3%2F2%7D%20x%5Csqrt%7B10%7D%5Csqrt%7B5%7D)
Now we have to multiply both the numbers in the root to get the answer;
=-6![x^{5/2} \sqrt{50}](https://tex.z-dn.net/?f=x%5E%7B5%2F2%7D%20%5Csqrt%7B50%7D)
Square root of 50 is 5 root 2.
=-6*5![\sqrt{2}](https://tex.z-dn.net/?f=%5Csqrt%7B2%7D)
![x^{5/2}](https://tex.z-dn.net/?f=x%5E%7B5%2F2%7D)
=-30![\sqrt{2}](https://tex.z-dn.net/?f=%5Csqrt%7B2%7D)
![x^{5/2}](https://tex.z-dn.net/?f=x%5E%7B5%2F2%7D)
Hence the simplified product is -30![\sqrt{2}](https://tex.z-dn.net/?f=%5Csqrt%7B2%7D)
.
Learn more about product here brainly.com/question/10873737
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