The 15th term in the given A.P. sequence is a₁₅ = 33.
According to the statement
we have given that the A.P. Series with the a = 5 and the d is 2.
And we have to find the 15th term of the sequence.
So, for this purpose we know that the
An arithmetic progression or arithmetic sequence is a sequence of numbers such that the difference between the consecutive terms is constant.
And the formula is a
an = a + (n-1)d
After substitute the values in it the equation become
an = 5 + (15-1)2
a₁₅ = 5 + 28
Now the 15th term is a₁₅ = 33.
So, The 15th term in the given A.P. sequence is a₁₅ = 33.
Learn more about arithmetic progression here
brainly.com/question/6561461
#SPJ1
Let 
. The tangent plane to the surface at (0, 0, 8) is
The gradient is
so the tangent plane's equation is
The normal vector to the plane at (0, 0, 8) is the same as the gradient of the surface at this point, (1, 1, 1). We can get all points along the line containing this vector by scaling the vector by 
, then ensure it passes through (0, 0, 8) by translating the line so that it does. Then the line has parametric equation
or 
, 
, and 
.
(See the attached plot; the given surface is orange, (0, 0, 8) is the black point, the tangent plane is blue, and the red line is the normal at this point)
Answer:
see below
Step-by-step explanation:
19-3 will be greater
We are multiplying 1/2 *(19-3) which is multiplying by a number less than 1 it will be less than 1
18-3 = 16
1/2 * (19-3)
Using PEMDAS
We do parentheses first
1/2 ( 16)
Then multiply
8
Answer:
relating to or expressed as a number or numbers.
Step-by-step explanation:
Answer:
Option A
Step-by-step explanation:
According to the distance formula , for any 2 points P & Q whose coordinates are (x¹ , y¹) and (x² , y²) respectively. So ,
NOTE = HERE x² , y² DOESN'T MEAN THE SQUARES OF x & y . THEY ARE JUST COORDINATES.
According to the question ,
Coordinate of C = (2 , -1)
Coordinate of D = (5 , 3)
Using distance formula ,
