Question

ребята пожалуйста ПОМОГИТЕ очень СРОЧНО надо номер 3.5

Answer 1

$tt 1) sqrt{122^2-22^2}= sqrt{(122+22)(122-22)}=sqrt{144cdot100}=\\ {} =sqrt{144}cdot sqrt{100} =12cdot10=120\\\ 2) sqrt{6.8^2-3.2^2}= sqrt{(6.8+3.2)(6.8-3.2)}=sqrt{10cdot3.6}=\\ {} =sqrt{36}=6$

$displaystylett 3) sqrt{45.8^2-44.2^2}= sqrt{(45.8+44.2)(45.8-44.2)}=sqrt{90cdot1.6}=\\ {} =sqrt{144}=12\\\ 4) sqrt{3.13^2-3.12^2}= sqrt{(3.13+3.12)(3.13-3.12)}=sqrt{6.25cdot 0.01}=\\ {} =sqrt{frac{625}{100} }cdotsqrt{frac{1}{100} }=frac{25}{10}cdot frac{1}{10}= frac{1}{4}= 0.25$

Answer 2

Спасибо

Answer 3

Пожалуйста

Answer 4

a²-b²=(a+b)(a-b)

√(122²-22²)=√(122+22)(122-22)=√(144.100)=√144.√100=12.10=120

√(6,8²-3,2²)=√(6,8+3,2)(6,8-3,2)=√(10.3,6)=√36=6

√(45,8²-44,2²)=√(45,8+44,2)(45,8-44,2)=√(90.1,6)=√(9.16)=3.4=12

√(3,13²-3,12²)=√(6,25.0,01)=√6,25.√0,01=2,5.0,1=0,25