Two angles are said to be complementary, if the sum of the measures of the to angles is equal to 90 degrees.
Thus, given that HFG is complementary to ACB, them mHFG + mACB = 90 degrees.
From the figure, given that the line from point F meats line CE at point P, then HFG = CFP.
But mCFE = 90 degrees and mCFE = mCFP + mPFE
Also PFE = DFH
Thus, mCFE = mCFP + mPFE = mHFG + mDFH = 90 degrees
Recall that mHFG + mACB = 90 degrees
Thus, mHFG + mACB = mHFG + mDFH
Therefore, mACB = mDFH.
I'm sorry you need to be more clear I don't understand the question.
Answer: The required solution is 
Step-by-step explanation:
We are given to solve the following differential equation :

where k is a constant and the equation satisfies the conditions y(0) = 50, y(5) = 100.
From equation (i), we have

Integrating both sides, we get
![\int\dfrac{dy}{y}=\int kdt\\\\\Rightarrow \log y=kt+c~~~~~~[\textup{c is a constant of integration}]\\\\\Rightarrow y=e^{kt+c}\\\\\Rightarrow y=ae^{kt}~~~~[\textup{where }a=e^c\textup{ is another constant}]](https://tex.z-dn.net/?f=%5Cint%5Cdfrac%7Bdy%7D%7By%7D%3D%5Cint%20kdt%5C%5C%5C%5C%5CRightarrow%20%5Clog%20y%3Dkt%2Bc~~~~~~%5B%5Ctextup%7Bc%20is%20a%20constant%20of%20integration%7D%5D%5C%5C%5C%5C%5CRightarrow%20y%3De%5E%7Bkt%2Bc%7D%5C%5C%5C%5C%5CRightarrow%20y%3Dae%5E%7Bkt%7D~~~~%5B%5Ctextup%7Bwhere%20%7Da%3De%5Ec%5Ctextup%7B%20is%20another%20constant%7D%5D)
Also, the conditions are

and

Thus, the required solution is 
Its 2 i guess but im not sure
Answer:The answer is It has no slope because x2 - x1 in the formula m=y2-y1-x2-x1 is zero, and the denominator of a fraction cannot be zero
Step-by-step explanation:
I did the quiz