What is the greatest integer a, such that a^2 + 3b is less than (2b)^2, assuming that b is 5? Please answer and have a nice day!

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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

9

Step-by-step explanation:

I just did the question on AOPS, please see the attachment down below.

Hope this helped! :)

given that triangle abc is transformed two times to get triangle

a"b"c"

first abc is transformed in to a'b'c' as follows:

the coordinates of a are (1,-1) and transformed into a'(-1,1)

similarly b (4,-2) became b'(-4,2) and

c(7,-2) became c'(-7,2)

i.e. (x,y) becomes (-x,-y)

this is nothing but reflection about a point here origin.

thus first transformation is reflection on the origin.

next is exactly shifting 3 units down

source:

step-by-step explanation: