You’re going it plug in the inputs for every input for the letter x. So the first one, fx= -4(-2)+1= 9. The second one fx= -4(0)+1=1. The third one fx= -4(5)+1=19.
Question 11a)
We are given side BC equals to side CE and angle CBA equals to angle CED
We also know that angle ACB equals to angle ECD are equal (opposite angles properties)
We have enough information to deduce that triangle ABC and triangle CDE are equal by postulate Angle-Side-Angle (ASA)
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Question 11b)
We are given side AB equal to side ED, side BC equals to side EF, and side AC equals to side DF
We have enough information to deduce that triangle ABC and triangle DEF congruent by postulate Side-Side-Side (SSS)
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Question 11c)
We are given side AC equals to side DF, angle ABC equals to angle DEF, and angle BAC equals to angle EDF
We have enough information to deduce that triangle ABC congruent to triangle DEF by postulate Angle-Side-Angle (ASA)
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Question 11d)
We do not have enough information to tell whether this shape congruent or not
Answer:

Step-by-step explanation:
First, note that

And using the chain rule in one variable

Now remember that the chain rule in several variables sates that

Therefore the chain rule in several variables would look like this.

Answer:
1/3 , only one solution
Step-by-step explanation:
3x-1 = 1-3x
3x-1 = -3x+1
6x-1 = 1
6x = 2
x = 1/3
Answer:
Step-by-step explanation:
a) We have 15! as the product of 1 to 15 natural numbers. Since 17 is prime there will be no factor common to these
By actual division we find
15! (mod 17) =16
From this we deduce
even 16! mod 17 = 16 = -1
According to Wilson theorem
(17-1)! = -1 mod 17
Thus verified 17 is prime
Hence 15! (mod 17) =-1=16
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b) 2(26!) is divided by 29
Since 29 is prime
(29-1)! = -1 mod 29
28! = -1 mod 29 = 28
When divided this gives 25 as remainder