A. 32°
:))))))))))))))))))))))
Given:
The graph of a line.
To find:
The y-intercept of the line.
Solution:
y-intercept of a line is the point at which the line intersect the y-axis.
We have, graph of the line.
From the given graph it is clear that the line intersect the y-axis at (0,4).
So, y-intercept is at point (0,4).
y-intercept = 4
Therefore, the y-intercept is 4.
For
y=a(x-h)^2+k
the axis of symmetry is x=h
so
y=2/3(x-2)^2-5
the axis of symmetry is x=2
answer is C
Answer:
∫
= ![\frac{1}{2}(x^6-2)^2+C](https://tex.z-dn.net/?f=%5Cfrac%7B1%7D%7B2%7D%28x%5E6-2%29%5E2%2BC)
Step-by-step explanation:
To find:
∫![6x^5(x^6-2)\,dx](https://tex.z-dn.net/?f=6x%5E5%28x%5E6-2%29%5C%2Cdx)
Solution:
Method of substitution:
Let ![x^6-2=t](https://tex.z-dn.net/?f=x%5E6-2%3Dt)
Differentiate both sides with respect to ![t](https://tex.z-dn.net/?f=t)
![6x^5\,dx=dt](https://tex.z-dn.net/?f=6x%5E5%5C%2Cdx%3Ddt)
[use
]
So,
∫
= ∫
=
where
is a variable.
(Use ∫
)
Put ![t=x^6-2](https://tex.z-dn.net/?f=t%3Dx%5E6-2)
∫
= ![\frac{1}{2}(x^6-2)^2+C_1](https://tex.z-dn.net/?f=%5Cfrac%7B1%7D%7B2%7D%28x%5E6-2%29%5E2%2BC_1)
Use ![(a-b)^2=a^2+b^2-2ab](https://tex.z-dn.net/?f=%28a-b%29%5E2%3Da%5E2%2Bb%5E2-2ab)
So,
∫
= ![\frac{1}{2}(x^6-2)^2+C_1=\frac{1}{2}(x^{12}+4-4x^6)+C_1=\frac{x^{12} }{2}-2x^6+2+C_1=\frac{x^{12} }{2}-2x^6+C](https://tex.z-dn.net/?f=%5Cfrac%7B1%7D%7B2%7D%28x%5E6-2%29%5E2%2BC_1%3D%5Cfrac%7B1%7D%7B2%7D%28x%5E%7B12%7D%2B4-4x%5E6%29%2BC_1%3D%5Cfrac%7Bx%5E%7B12%7D%20%7D%7B2%7D-2x%5E6%2B2%2BC_1%3D%5Cfrac%7Bx%5E%7B12%7D%20%7D%7B2%7D-2x%5E6%2BC)
where ![C=2+C_1](https://tex.z-dn.net/?f=C%3D2%2BC_1)
Without using substitution:
∫
= ∫
= ![\frac{6x^{12} }{12}-\frac{12x^6}{6}+C=\frac{x^{12} }{2}-2x^6+C](https://tex.z-dn.net/?f=%5Cfrac%7B6x%5E%7B12%7D%20%7D%7B12%7D-%5Cfrac%7B12x%5E6%7D%7B6%7D%2BC%3D%5Cfrac%7Bx%5E%7B12%7D%20%7D%7B2%7D-2x%5E6%2BC)
So, same answer is obtained in both the cases.