Answer:
9.577 mm.
Step-by-step explanation:
We have been given that the diameter of a quarter is 24.26 mm and the diameter of a nickel is 21.21 mm.
Let us find the circumference of quarter and nickel using circumference of circle formula.
, where r represents radius of the circle.
Since we know that diameter of circle is 2 times its radius, so let us find radius of quarter and nickel by dividing their diameter by 2.
Therefore, circumference of quarter is 76.1764 mm.
Let us find circumference of nickel.
Let us subtract circumference of nickel from circumference of quarter.
Therefore, the distance around a quarter is 9.577 millimeters bigger than a nickel.
Answer:
12×0.25 is definitely right
×0.25 is another way of finding a quarter of something. I looked at the other answers and I think this one is the only correct one
Answer:
4(8 + 160)
Step-by-step explanation:
Sorry if i'm wrong i suck at translating expressions
hope i helped you in some way tho!
U can simplify each number to the bottom. Then determine which number pair u can take out. Put those outside of the square too sign. Numbers with out a pair must stay inside the square room
<h3>
Answer: Largest value is a = 9</h3>
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Work Shown:
b = 5
(2b)^2 = (2*5)^2 = 100
So we want the expression a^2+3b to be less than (2b)^2 = 100
We need to solve a^2 + 3b < 100 which turns into
a^2 + 3b < 100
a^2 + 3(5) < 100
a^2 + 15 < 100
after substituting in b = 5.
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Let's isolate 'a'
a^2 + 15 < 100
a^2 < 100-15
a^2 < 85
a < sqrt(85)
a < 9.2195
'a' is an integer, so we round down to the nearest whole number to get 
So the greatest integer possible for 'a' is a = 9.
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Check:
plug in a = 9 and b = 5
a^2 + 3b < 100
9^2 + 3(5) < 100
81 + 15 < 100
96 < 100 .... true statement
now try a = 10 and b = 5
a^2 + 3b < 100
10^2 + 3(5) < 100
100 + 15 < 100 ... you can probably already see the issue
115 < 100 ... this is false, so a = 10 doesn't work
