#3, #19, and #16 please!!

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lovegodest1 lovegodest1

03-12-2020
Mathematics

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#3, #19, and #16 please!!

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inspire123andcreate inspire123andcreate · Answer 1 · 2020-12-03T21:30:09+01:00

To answer number 3 all you have to do is put 15 times x <_ 12 (put a line directly underneath the sign)

To answer number 19, do 21-11= 10 This means that X=10

X(10) plus 11=21

21=21

[10 + 11] >_ 21

To solve 16.

Add 3 to 13 to get 16

Then you’ll have [3x +4] >_ 16

Once you have this, subtract 4 from both sides.

Then you’ll have 3x>_ 12

Now you can divide each side by 3 to get x>_4. X=4

So if we plug that in [3x +4] >_ 13, you’ll get

[3(4) +4] >_ 13

And then you’ll get

[16] >_ 13

I hope this helps!!