Answer:
.3 is greater.
Step-by-step explanation:
.13
.3 or .30
These are the same, and obviously .30 is larger than .13.
![\mathfrak{\huge{\orange{\underline{\underline{AnSwEr:-}}}}}](https://tex.z-dn.net/?f=%5Cmathfrak%7B%5Chuge%7B%5Corange%7B%5Cunderline%7B%5Cunderline%7BAnSwEr%3A-%7D%7D%7D%7D%7D)
Actually Welcome to the Concept of the Linear equations.
Let the number of bicycles = x
and number of tricycles = y
the relationship given is,
x = 5y-1
number of wheels of bicycle = 2x and of tricycles=3y
and other relation is, 2x+3y = 154
so now substituting we get as,
==> 2(5y-1) +3y= 154
10y-2+3y = 154
13y = 154+2
=> y = 12
=> 12*3 = 36 wheels of tricycles
so, 154-36 => 118 wheels of the bicycle.
so the number of bicycles = 118/2 = 59
hence,
==> number of bicycles = 59 and tricycles = 12
I'm pretty sure it should be 2h+3
There are a lot of them
3/4 , 2/3 any fraction greater than .5 will fall in this category.
Answer:
B. ![x = \frac{9}{2}](https://tex.z-dn.net/?f=%20x%20%3D%20%5Cfrac%7B9%7D%7B2%7D%20)
Step-by-step explanation:
On our left, we have 7 "x's" and 7 "1's". Let this be 7x + 7.
On our right, we have 5 "x's" and 16 "1's". Let this be 5x + 16.
We would have the equation:
![7x + 7 = 5x + 16](https://tex.z-dn.net/?f=%207x%20%2B%207%20%3D%205x%20%2B%2016%20)
Solve for x. Collect like terms.
![7x - 5x = 16 - 7](https://tex.z-dn.net/?f=%207x%20-%205x%20%3D%2016%20-%207%20)
![2x = 9](https://tex.z-dn.net/?f=%202x%20%3D%209%20)
Divide both sides by 2
![\frac{2x}{2} = \frac{9}{2}](https://tex.z-dn.net/?f=%20%5Cfrac%7B2x%7D%7B2%7D%20%3D%20%5Cfrac%7B9%7D%7B2%7D%20)
![x = \frac{9}{2}](https://tex.z-dn.net/?f=%20x%20%3D%20%5Cfrac%7B9%7D%7B2%7D%20)