5 days ago, hope this still helps though!
Alright 21 miles per hour. We know that there are 60 minutes in an hour. So this is basically saying 21 miles per 60 minutes. To find the rate for miles per minute, you divide 21 by 60 which is 0.35.
In one minute the cyclist rides 0.35 miles
In two minutes the cyclist rides 0.7 miles
Answer: 4) 157
Step-by-step explanation:
We know that there is no association between the grade level and the andedness, then we should find that the ratio between left handeds and right handed is the same for both grades.
In 7-th grade we have:
Left handed : 11
Right handed: 72
The ratio is 11/72 = 0.14
Then, the ratio for the 8-th graders must be about the same:
Left handed: 24
Right handed: X
Ratio: 24/X
Let's start with the bigger option, X = 157.
24/157 = 0.15
Ok, we now see that with the bigger option we obtained almost the same ratio (if we use the smaller values for X, we will get a ratio bigger than 0.15, so 0.15 is the better aproximation that we can find to the 0.14 of the 7-th graders)
Then the correct option is 4) 157
1. Change both numbers to common base.
2. Use quotient rule 
e)
You can write 9 as 3², then both numbers have the same base.
, where x is the exponent
I cannot really see the exponents because picture is bad quality.
You want to replace x and y with actual numbers, but as I said the picture is too bad.
f)
Convert to base 4, then quotient rule.
g)
Use the quotient rule, you already have same base.
h)
Convert to base 
, then quotient rule.
He needs 6 litres of the 20% and 2 litres of the 60% a simple trial and error method will do in this type of equation
All definitions can be read as a conditional statement. They can be read forward (conditional) and backward (its converse). ex) perpendicular lines<span> conditional </span>If two lines are perpendicular<span>, </span>then<span> they </span>intersect<span> to form </span>right angles<span>. converse </span>If two lines intersect<span> to form </span>right angles<span>, </span>then<span> they are </span>perpendicular<span>.</span>
