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s344n2d4d5 [400]
3 years ago
5

62784/16 use long division

Mathematics
1 answer:
Scorpion4ik [409]3 years ago
8 0

Answer:

3924

Step-by-step explanation:

what you have to do is see how many times sixteen goes into 6 first and since that is zero you look to the left so it is now  62. now you see how many times 16 goes into 62 which is 3 times and 16x3=48. Now you place ur 48 below ur 62 and subtract. u will get 14 obviously. now you carry down ur 7 and see how many times 16 goes into 147. Do this until u have no numbers left

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solniwko [45]
<h3>Answer: Largest value is a = 9</h3>

===================================================

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.

------------------

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 a \le 9

So the greatest integer possible for 'a' is a = 9.

------------------

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

5 0
3 years ago
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Natasha_Volkova [10]

Answer:

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Step-by-step explanation:

7 0
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nasty-shy [4]

Answer:

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Step-by-step explanation:

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guapka [62]

Answer:

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Step-by-step explanation:

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4 0
3 years ago
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