19) 67x²+115x+48
20) (w+8)²
21) (d+11)²
22) (r+7)(r-7)
Explanation:
19) The area of the ceiling is given by (10x+9)(7x+7). Multiplying we have:
10x*7x+7*10x+9*7x+9*7
=70x²+70x+63x+63
=70x²+133x+63
The area of the skylight is given by (x+5)(3x+3). Multiplying we have:
x*3x+3*x+5*3x+5*3
=3x²+3x+15x+15
=3x²+18x+15
To find the remaining area of the ceiling we subtract:
(70x²+133x+63)-(3x²+18x+15)
=70x²+133x+63-3x²-18x-15
=67x²+115x+48
20) To factor, we want to find factors of c, 64, that sum to b, 16. 8*8=64 and 8+8=16; therefore we have:
(w+8)(w+8)
Since this is the same binomial twice, we write it as (w+8)²
21) Again we look for factors of c, 121, that sum to b, 22. 11*11=121 and 11+11=22, so:
(d+11)(d+11)
Since this is the same binomial twice, we have (d+11)²
22) Factors of -49 that sum to 0 (there is no r term, just r²): -7*7=-49 and -7+7=0, so:
(r-7)(r+7)
Answer:
Integration of I= 
=![[\frac{logx}{n+1} x^{(n+1)}]-[\frac{1}{(n+1)^{2}}x^{(n+1)}]](https://tex.z-dn.net/?f=%5B%5Cfrac%7Blogx%7D%7Bn%2B1%7D%20x%5E%7B%28n%2B1%29%7D%5D-%5B%5Cfrac%7B1%7D%7B%28n%2B1%29%5E%7B2%7D%7Dx%5E%7B%28n%2B1%29%7D%5D)
Step-by-step explanation:
Given integral is I=
Take logx=t
I= 
I= 
Using integration by part,
![I= (t)\int [e^{(n+1)t}]\, dt-\int[\frac{d}{dt}{t}\times\int (e^{(n+1)t})]\\\\I= (t) [\frac{1}{n+1}e^{(n+1)t}]-\int[1\times\frac{1}{n+1}e^{(n+1)t}]\,dt\\\\I=[\frac{t}{n+1}e^{(n+1)t}]-[\frac{1}{(n+1)^{2}}e^{(n+1)t}]](https://tex.z-dn.net/?f=I%3D%20%28t%29%5Cint%20%5Be%5E%7B%28n%2B1%29t%7D%5D%5C%2C%20dt-%5Cint%5B%5Cfrac%7Bd%7D%7Bdt%7D%7Bt%7D%5Ctimes%5Cint%20%28e%5E%7B%28n%2B1%29t%7D%29%5D%5C%5C%5C%5CI%3D%20%28t%29%20%5B%5Cfrac%7B1%7D%7Bn%2B1%7De%5E%7B%28n%2B1%29t%7D%5D-%5Cint%5B1%5Ctimes%5Cfrac%7B1%7D%7Bn%2B1%7De%5E%7B%28n%2B1%29t%7D%5D%5C%2Cdt%5C%5C%5C%5CI%3D%5B%5Cfrac%7Bt%7D%7Bn%2B1%7De%5E%7B%28n%2B1%29t%7D%5D-%5B%5Cfrac%7B1%7D%7B%28n%2B1%29%5E%7B2%7D%7De%5E%7B%28n%2B1%29t%7D%5D)
Writing in terms of x
I=![[\frac{t}{n+1}e^{(n+1)t}]-[\frac{1}{(n+1)^{2}}e^{(n+1)t}]](https://tex.z-dn.net/?f=%5B%5Cfrac%7Bt%7D%7Bn%2B1%7De%5E%7B%28n%2B1%29t%7D%5D-%5B%5Cfrac%7B1%7D%7B%28n%2B1%29%5E%7B2%7D%7De%5E%7B%28n%2B1%29t%7D%5D)
I=![[\frac{logx}{n+1}e^{(n+1)logx}]-[\frac{1}{(n+1)^{2}}e^{(n+1)logx}]](https://tex.z-dn.net/?f=%5B%5Cfrac%7Blogx%7D%7Bn%2B1%7De%5E%7B%28n%2B1%29logx%7D%5D-%5B%5Cfrac%7B1%7D%7B%28n%2B1%29%5E%7B2%7D%7De%5E%7B%28n%2B1%29logx%7D%5D)
I=![[\frac{logx}{n+1}e^{logx^{(n+1)}}]-[\frac{1}{(n+1)^{2}}e^{logx^{(n+1)}}]](https://tex.z-dn.net/?f=%5B%5Cfrac%7Blogx%7D%7Bn%2B1%7De%5E%7Blogx%5E%7B%28n%2B1%29%7D%7D%5D-%5B%5Cfrac%7B1%7D%7B%28n%2B1%29%5E%7B2%7D%7De%5E%7Blogx%5E%7B%28n%2B1%29%7D%7D%5D)
I=
Thus,
Integration of I= =
Answer:
a) (6;6)
b) Yes
d) The intersection is on the vertical axis.
Step-by-step explanation:
a) A(2,4)
B (2,2)
Vecto BA (4, 2)
Vecto n (-4,2)
We have:
(2 - 4) x (X - (-2)) + (2 - (-2)) x (Y - 2) = 0
=> -2X + 4Y - 12 = 0
=> Y = X/2 +3
Let's find the value of y if (6,y) lies on the plane:
Put X = 6 => Y = 6
b)
X/2 + 3 = 3 - X
=> X/2 = -X
=> X =0 -> Y = 3
C (0;3)
d) The intersection is on the vertical axis.
I hope my answer will help you.
Answer:
x2 - 4x - 4
———————————
x - 2
Step-by-step explanation:
Answer:
No, it is not a perfect square.
