Legend1111111
13.07.2021 21:25

Hi, Sergei, I hope you're well. Guess where I was last weekend! I decided to go to the Crazy Ride Adventure Park with my friends. The weather was great. There were so many rides I couldn't choose which one to go on. I tired the Mega roller coaster. It was great. We wanted to go to the Hall of Mirrors but we didn't because there was a very long queue. We tired candyfloss but i didn't like it. We stayed until late in the evening. What about you? Did you visit your grandparents? Write back soon. Sam. напишите свой текст по этому образцу .

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Ответ:
olykharkov
25.01.2023 03:50

а) предложения, в которых ing-форма является подлежащим

The driving wheel of the machine is broken.

These happenings are remarkable.

Travelling is a pleasant way of improving one’s education.

Asking him for help is useless.

в) предложения, в которых ing-форма является обстоятельством

Every trust arranges for the marketing of its products.

It is no use crying over spilt milk.

We have every chance of passing our examinations well.

You don’t know what you miss, not having the desire to listen to good music.

с) предложения, в которых ing-форма является частью сказуемого.

Happily we escaped being delayed on our way.

Having been knocked down by a passing car, the poor man was at once taken to hospital.

I was told of a great friendship existing between two captains.

Driving in a motor car, we passed many villages.

Объяснение:

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Ответ:
mashkakornilova
25.01.2023 03:50
Сначала формулы . затем формулы двойного угла sin2α=2sinαcosα cos2α=cos²α-sin²α=cos²α-(1-cos²α)=2cos²α-1⇒  2cos²α=1+cos2α cos2α=cos²α-sin²α=1-sin²α-sin²α=1-2sin²α⇒  2sin²α=1-cos2α а) сos75°cos105°=cos(90°-15°)·cos(90°+15°)== sin15°(-sin15°)=-sin²15°=-(1-cos30°)/2=(cos30°-1)/2= ((√3/2)-1)/2=0,25√3-0,5б) sin75°sin15°=°sin(90°-15°)sin15°=cos15° sin15°=sin30°/2=1/4=0,25в) sin105°cos15°=sin(180°-75°)cos15°=sin75°cos15°=sin(90°-15°)cos15°=cos15°cos15°=(1+cos30°)/2=(1+(√3/2))/2=0,5 +0,25√3 2 способ формулы cosα·cosβ=0,5cos(α-β)+0,5cos(α+β) sinα·sinβ=0,5cos(α-β)-0,5cos(α+β) sinα·cosβ=0,5sin(α+β)+0,5sin(α-β) а) сos75°cos105°=0,5cos(75°-105°)+0,5cos(75°+105°)=0,5cos(-30°)+0,5 cos180°==0,5· √3/2+0,5·(-1)=0,25√3-0,5б) sin75°sin15°=0,5cos(75°-15°)-0,5cos(75°+15° )=0,5cos60°-0,5 cos90°=0,5·0,5=0,25в) sin105°cos15°=0,5sin(105°+15°)+0,5sin(105°-15°)= =0,5sin120°+0,5sin90°==0,5 sin(180°-60°)+0,5·1=0,5 sin 60°+0,5=0,25√3+0,5
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