Given:
The function is
![f(x)=2x+1](https://tex.z-dn.net/?f=f%28x%29%3D2x%2B1)
To find:
The range value if the domain value is 1.75.
Solution:
Domain is the set of input values and range is the set of output values.
Here, range value if the domain value is 1.75. It means the value of function f(x) at x=1.75.
Substitute x=1.75 in the given function.
![f(1.75)=2(1.75)+1](https://tex.z-dn.net/?f=f%281.75%29%3D2%281.75%29%2B1)
![f(1.75)=3.5+1](https://tex.z-dn.net/?f=f%281.75%29%3D3.5%2B1)
![f(1.75)=4.5](https://tex.z-dn.net/?f=f%281.75%29%3D4.5)
Therefore, the required range value is 4.5.
Answer:
See Explanation
Step-by-step explanation:
<em>Question like this are better answered if there are list of options; However, I'll simplify as far as the expression can be simplified</em>
Given
![sec^4 x + sec^2 x tan^2 x - 2 tan^4 x](https://tex.z-dn.net/?f=sec%5E4%20x%20%2B%20sec%5E2%20x%20tan%5E2%20x%20-%202%20tan%5E4%20x)
Required
Simplify
![(sec^2 x)^2 + sec^2 x tan^2 x - 2 (tan^2 x)^2](https://tex.z-dn.net/?f=%28sec%5E2%20x%29%5E2%20%2B%20sec%5E2%20x%20tan%5E2%20x%20-%202%20%28tan%5E2%20x%29%5E2)
Represent
with a
Represent
with b
The expression becomes
![a^2 + ab- 2 b^2](https://tex.z-dn.net/?f=a%5E2%20%2B%20ab-%202%20b%5E2)
Factorize
![a^2 + 2ab -ab- 2 b^2](https://tex.z-dn.net/?f=a%5E2%20%2B%202ab%20-ab-%202%20b%5E2)
![a(a + 2b) -b(a+ 2 b)](https://tex.z-dn.net/?f=a%28a%20%2B%202b%29%20-b%28a%2B%202%20b%29)
![(a -b) (a+ 2 b)](https://tex.z-dn.net/?f=%28a%20-b%29%20%28a%2B%202%20b%29)
Recall that
![b = tan^2x](https://tex.z-dn.net/?f=b%20%3D%20tan%5E2x)
The expression
becomes
![(sec^2x -tan^2x) (sec^2x+ 2 tan^2x)](https://tex.z-dn.net/?f=%28sec%5E2x%20-tan%5E2x%29%20%28sec%5E2x%2B%202%20tan%5E2x%29)
..............................................................................................................................
In trigonometry
![sec^2x =1 +tan^2x](https://tex.z-dn.net/?f=sec%5E2x%20%3D1%20%20%2Btan%5E2x)
Subtract
from both sides
![sec^2x - tan^2x =1 +tan^2x - tan^2x](https://tex.z-dn.net/?f=sec%5E2x%20-%20tan%5E2x%20%3D1%20%20%2Btan%5E2x%20-%20tan%5E2x)
![sec^2x - tan^2x =1](https://tex.z-dn.net/?f=sec%5E2x%20-%20tan%5E2x%20%3D1)
..............................................................................................................................
Substitute 1 for
in ![(sec^2x -tan^2x) (sec^2x+ 2 tan^2x)](https://tex.z-dn.net/?f=%28sec%5E2x%20-tan%5E2x%29%20%28sec%5E2x%2B%202%20tan%5E2x%29)
![(1) (sec^2x+ 2 tan^2x)](https://tex.z-dn.net/?f=%281%29%20%28sec%5E2x%2B%202%20tan%5E2x%29)
Open Bracket
------------------This is an equivalence
![(secx)^2+ 2 (tanx)^2](https://tex.z-dn.net/?f=%28secx%29%5E2%2B%202%20%28tanx%29%5E2)
Solving further;
................................................................................................................................
In trigonometry
![secx = \frac{1}{cosx}](https://tex.z-dn.net/?f=secx%20%3D%20%5Cfrac%7B1%7D%7Bcosx%7D)
![tanx = \frac{sinx}{cosx}](https://tex.z-dn.net/?f=tanx%20%3D%20%5Cfrac%7Bsinx%7D%7Bcosx%7D)
Substitute the expressions for secx and tanx
................................................................................................................................
becomes
![(\frac{1}{cosx})^2+ 2 (\frac{sinx}{cosx})^2](https://tex.z-dn.net/?f=%28%5Cfrac%7B1%7D%7Bcosx%7D%29%5E2%2B%202%20%28%5Cfrac%7Bsinx%7D%7Bcosx%7D%29%5E2)
Open bracket
![\frac{1}{cos^2x}+ 2 (\frac{sin^2x}{cos^2x})](https://tex.z-dn.net/?f=%5Cfrac%7B1%7D%7Bcos%5E2x%7D%2B%202%20%28%5Cfrac%7Bsin%5E2x%7D%7Bcos%5E2x%7D%29)
![\frac{1}{cos^2x}+ \frac{2sin^2x}{cos^2x}](https://tex.z-dn.net/?f=%5Cfrac%7B1%7D%7Bcos%5E2x%7D%2B%20%5Cfrac%7B2sin%5E2x%7D%7Bcos%5E2x%7D)
Add Fraction
------------------------ This is another equivalence
................................................................................................................................
In trigonometry
![sin^2x + cos^2x= 1](https://tex.z-dn.net/?f=sin%5E2x%20%2B%20cos%5E2x%3D%201)
Make
the subject of formula
![sin^2x= 1 - cos^2x](https://tex.z-dn.net/?f=sin%5E2x%3D%201%20%20-%20cos%5E2x)
................................................................................................................................
Substitute the expressions for
for ![sin^2x](https://tex.z-dn.net/?f=sin%5E2x)
![\frac{1 + 2(1 - cos^2x)}{cos^2x}](https://tex.z-dn.net/?f=%5Cfrac%7B1%20%2B%202%281%20%20-%20cos%5E2x%29%7D%7Bcos%5E2x%7D)
Open bracket
![\frac{1 + 2 - 2cos^2x}{cos^2x}](https://tex.z-dn.net/?f=%5Cfrac%7B1%20%2B%202%20%20-%202cos%5E2x%7D%7Bcos%5E2x%7D)
---------------------- This is another equivalence
Answer:
18ft
Step-by-step explanation:
just add all the sides. since it’s a rectangle, you there is 2 sides that’s 3ft long and that are 6ft long. so you would do 3+3+6+6
Well using elimination is easy because they basically did it for you.
4x-5y=-6
2x+5y=12
The postive 5y and negative 5y cancel out because -5y plus 5y is zero.
so you are left with
4x=-6
2x=12
Add these together
6x=18
divide both sides by 6 and Solve to find your x
x=3
Now we have to eliminate x to find y.
we multiply the second equation by -2 so it can cancel out so
(2x+5y=12)-2
-4x-10y=24
Add the equations together
4x-5y=-6
-4x-10y=24
You are left with -15y = 30
Solve for y
y = -2
So your final answer for the elimination would be
(3,-2)