Use the method pemdas and you get x=28
Step-by-step explanation:
b^2-4b+3=0
b²-3x-b+3=0
b(b-3)-1(b-3)=0
(b-3)(b-1)=0
either
b=3 or b=1
.
2n^2 + 7 = -4n + 5
2n²+4n+7-5=0
2n²+4n+2=0
2(n²+2n+1)=0
(n+1)²=0/2
:.n=-1
.
x - 3x^2 = 5+ 2x - x^2
0=5+ 2x - x^2-x +3x^2
0=5+x+2x²
2x²+x+5=0
comparing above equation with ax²+bx +c we get
a=2
b=1
c=5
x={-b±√(b²-4ac)}/2a ={-1±√(1²-4×2×5)}/2×1
={-1±√-39}/2
Read the question carefully: it costs 4 tokens to park in a garage for an hour.
We will apply the unitary method to solve this question
It costs 4 tokens to park in a garage for 1 hour
Find how many hours can park in a garage for 1 token
If it costs 4 token to park in a garage for 1 hour
Then it will cost 1 token to park in a garage for 1/4 hour
Step2:
With 20 token we can park in a garage for (1/4) * 20
= 5 hours
So, we can park for 5 hours with 20 tokens.
Another method
If we take twenty tokens and divide them into groups of four, we will find that we are left with five groups of tokens. Each group of tokens represents an hour of parking time. This will give us five groups, or five hours, total.
So, we can park for 5 hours with 20 tokens
Answer:
D
Step-by-step explanation:
Using the cosine ratio in the right triangle.
cos66° = 
= 
( multiply both sides by x )
x × cos66° = 13 ( divide both sides by cos66° )
x = 
≈ 32.0 ( to the nearest tenth )
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