I hope this helps you
Circumference=2pi.r
36pi.=2pi.r
r=36/2
r=18
Answer:
6.
Step-by-step explanation:
Answer:
23m-18
Step-by-step explanation:
2(7m−5)+(−8+9m)
Remove parenthesis
2(7m-5)-8+9m
expand
14m-10
14m-10-8+9
simplify
23m-18
Ask me for more questions brainliest please.
For problems like this, change the wording into simple algebraic
formulas. We know that the area of a rectangle is its width times its
length, or A=LW. The problem tells you that the area is 48, so LW=48.
Also, if the length = twice the width minus 4, then L=2W-4.
From here, substitute your new L value for the L in the first equation:
48=LW --> 48=(2W-4)W
Now solve this equation by factoring it out into a polynomial:
48=2W^2-4W --> 24=W^2-2W --> W^2-2W-24=0 --> (W+4)(W-6)=0.
Solving this equation gives us W values of 6 and -4, but since the width
of a rectangle cannot be negative, the width must be 6.
Since L=2W-4, then L=2(6)-4 --> L=12-4 = 8. Therefore, the dimensions are 6X8, or 6 feet by 8 feet.
To check our work: The length equals four feet less than twice the width: 8=2(6)-4 --> 8=12-4 --> 8=8. This checks out.
Also, the area is 48 ft^2, and (6)(8) = 48, so this also checks out.
Hopefully this helps. If you need any more help or if I went over something too quickly, just let me know.
Answer:
B, Josef is confused about the range and the interquartile range.
Step-by-step explanation:
You have to make the box plot using the numbers given
Find the IQR *interquartile range* and the difference between the max and the minimum numbers.
For the cars the IQR is: 6
The range for the cars is: 12
The IQR of the minivan is: 3
The range of the minivan is: 12
The ranges are the same but IQR of the cars is bigger, that's where Josef got confused.
