I think it is b that is the answer
We are given : m∠WYX=(2x−1)° and m∠WYZ=(4x+1)°.
∠WYX and ∠WYZ are complementary.
We know, sum of complementary angles is = 90°.
So, we need to add ∠WYX and ∠WYZ and set it equal to 90°.
m∠WYX + m∠WYZ = 90°.
Plugging values of ∠WYX and ∠WYZ in the above equation, we get
(2x−1)° + (4x+1)° = 90°.
Removing parentheses from both sides,
2x-1 + 4x+1 =90.
Combining like terms,
2x+4x= 6x and -1+1 =0
6x +0 =90.
6x=90.
Dividing both sides by 6.
6x/6 =90/6
x= 15.
Plugging value of x=15.
m∠WYX=(2x−1)° = 2*15 -1 = 30 -1 =29
m∠WYZ=(4x+1)° = 4*15 +1 = 60+1 = 61.
Therefore, ∠WYX=29° and ∠WYZ=61°.
An aritmetic sequence is like this
![a_n=a_1+d(n-1)](https://tex.z-dn.net/?f=a_n%3Da_1%2Bd%28n-1%29)
where a1=first term and d=common difference
geometric is
![a_n=a_1(r)^{n-1}](https://tex.z-dn.net/?f=a_n%3Da_1%28r%29%5E%7Bn-1%7D)
where a1=first term and r=common ratio
can it be both aritmetic and geometric
hmm, that means that the starting terms should be the same
therfor we need to solve
![d(n-1)=(r)^{n-1}](https://tex.z-dn.net/?f=d%28n-1%29%3D%28r%29%5E%7Bn-1%7D)
what values of d and r make all natural numbers of n true?
are there values that make all natural numbers for n true?
when n=1, then d(1-1)=0 and r^(1-1)=1, so already they are not equal
the answer is no, a sequence cannot be both aritmetic and geometric
Answer:
15h
Step-by-step explanation
-5g plus 15h is 10 h minus 25 is 15h