Answer is Median Option(C.) since it divides RS into two equal halves i.e.RQ=QS=7
Hope this helps....
Answer:
Frequency is 2, 2 and 3.
Step-by-step explanation:
Temp in 101 to 106 - In this range only 2 temperature is there 104 and 104 in 2 cities. so frequency is 2.
Temp in 107 to 112 - In this range only 2 temperature is there 107 and 112 in 2 cities. so frequency is 2.
Temp in 113 to 118 - In this range only 3 temperature is there 114, 117 and 118 so frequency is 3.
Answer: (7x + 8)³ ↔ the cube of the sum of 7x and 8
7(x + 8)³ ↔ 7 times the cube of the sum of x and 8
7x³ + 8 ↔ 8 added to the product of 7 and x cubed
(7x)³ + 8 ↔ 8 added to the cube of 7x
Step-by-step explanation:
(7x + 8)³ ↔ the cube of the sum of 7x and 8: for this case you will find that the values are into the parenthesis, then for each value apply the same exponent because of that all the values into the parenthesis are elevated to the cube,
7(x + 8)³ ↔ 7 times the cube of the sum of x and 8: in this example the 7 is multiplying all the values into the parenthesis, then you can find six times the sum of x and 8 elevated to the cube, also keep in mind that exponent which is 3 only apply to the values which are into the parenthesis and this values are the x and 8
7x³ + 8 ↔ 8 added to the product of 7 and x cubed: for this equation, you will find that 7 is multiplying to the x elevated to the cube after that, the 8 is added to this product.
(7x)³ + 8 ↔ 8 added to the cube of 7x: this final we can see that the values into the parenthesis are elevated to cube, the values are 7 and x, after this equation the 8 is added to the product.
Answer:
1. The given series is
4 ⋅ 6 + 5 ⋅ 7 + 6 ⋅ 8 + ... + 4n( 4n + 2) = ![\frac{[4(4n+1)(8n+7)]}{6}](https://tex.z-dn.net/?f=%5Cfrac%7B%5B4%284n%2B1%29%288n%2B7%29%5D%7D%7B6%7D)
For n=1
L.H.S=4.6=24
R.H.S=[4×5×15]÷6
=300÷6
=50
So, for n=1,
L.H.S≠ R.H.S
Since the given expression is true for n=1 ,
So , the given series is untrue.
we should replace R.H.S by=4(n+1)(n+2)(4n-3)²
2.
12+42+72+.......+(3 n -2)2=
For n=1,
L.H.S=12
R.H.S=1×(6-3-1)/2
=2/2
=1
As L.H.S≠ R.H.S
We should Replace R.H.S by [(3 n-1)(3 n-2)]2
3.The given sequence is
2+4+6+....+2n=n(n+1)
L.H.S
P(1)=2
R.H.S
1×(1+1)
=1×2
=2
( b) L.H.S
P(n)=2+4+6+.....2 k
This is an A.P having n terms.
![S_{n}=[tex]\frac{n}{2}\times\text{[first term + last term]}](https://tex.z-dn.net/?f=S_%7Bn%7D%3D%5Btex%5D%5Cfrac%7Bn%7D%7B2%7D%5Ctimes%5Ctext%7B%5Bfirst%20term%20%2B%20last%20term%5D%7D)
tex]S_{n}=![\frac{n}{2}\text [{2+2n}]](https://tex.z-dn.net/?f=%5Cfrac%7Bn%7D%7B2%7D%5Ctext%20%5B%7B2%2B2n%7D%5D)
= n(n+ 1)
R.H.S=n(n+1)
So, P(k)=k(k+1)
(c) P(k+1)=2+4+6+.......+2(k+1)
This is an A.P having (k+1) terms.
![S_(k+1)=\frac{k+1}{2}[2+2k+2]](https://tex.z-dn.net/?f=S_%28k%2B1%29%3D%5Cfrac%7Bk%2B1%7D%7B2%7D%5B2%2B2k%2B2%5D)
=(k+1)(k+2)
So, P(k+1)= (k+1)(k+2)
