First, let's find the x and y intercepts
In the first equation
y - 4x = -1
Put x =0
y= - 1
(0, -1)
Put y=0 and the n solve for x
0 - 4x = -1
-4x = -1
x=0.25
(0.25 , 0)
The points for the first equation is (0, -1 ) and (0.25, 0)
Next is to find the intercts for the second equation
y + x = 4
put x=0
y = 4
(0, 4)
Put y =0
0 + x = 4
x = 4
( 4, 0)
The points for the second equation are;
(0, 4) and (4, 0)
Below is the graph
So here is what you would do:
lets just pretend that there is no x in the equation
so this is what the equation would look like:
first, is 4 to the power of 2=16 and now we are going to put the x back into the equation so: 4x to the power of 2, but we did the exponents so, now x is the missing number. We can use the 32 because it is the solution to the equation. So we would do 32-16=16 so the missing number is 16 so x=16
Answer: x=16
Given:
The expression is:
To find:
The integration of the given expression.
Solution:
We need to find the integration of .
Let us consider,
![[\because 1+\cos 2x=2\cos^2x,1-\cos 2x=2\sin^2x]](https://tex.z-dn.net/?f=%5B%5Cbecause%201%2B%5Ccos%202x%3D2%5Ccos%5E2x%2C1-%5Ccos%202x%3D2%5Csin%5E2x%5D)
![\left[\because \tan \theta =\dfrac{\sin \theta}{\cos \theta}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbecause%20%5Ctan%20%5Ctheta%20%3D%5Cdfrac%7B%5Csin%20%5Ctheta%7D%7B%5Ccos%20%5Ctheta%7D%5Cright%5D)
It can be written as:
![[\because 1+\tan^2 \theta =\sec^2 \theta]](https://tex.z-dn.net/?f=%5B%5Cbecause%201%2B%5Ctan%5E2%20%5Ctheta%20%3D%5Csec%5E2%20%5Ctheta%5D)
Therefore, the integration of is .
Answer:
he whent up and not down and left and not right
Step-by-step explanation:we just is messed up in life
Answer:
18b+12
Step-by-step explanation:
6x3=18,6x2=12 so it's 18b(keep the variable)+12
