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Вопрос по алгебре:
составтте уравнение прямой, проходящей через точку (7;4), перпендекулярную к прямой 3x-2y+4=0-
Автор:
mercedes3
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Ответ: 3x-2y+4=0⇒у=1,5*х+2. Перпендикулярная этой прямой прямая имеет угловой коэффициент к=-1/1,5=-2/3. Свободный член искомого уравнения ищем в виде 4=-2*7/3+b⇒b=4+14/3=26/3=8 2/3. Искомое уравнение имеет вид у=-2*х/3+8 2/3.
Ответ: у=-2*х/3+8 2/3.
Объяснение:
Еще 4 ненужных тебе вопроса, но это важно для поиска
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B) Rewrite the sentences in the passive.
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The bill ......... by Kerrie.
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Financial documents ........by the director and the accountant.
a) are usually signed b) usually sign
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The gold........on international markets.
a) is sold b) sells
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a) had worked b) have worked c) had been working
2. I........ my money into real estate.
a) invested b) have invested c) invest
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a) have shipped b) had shipped c) shipped
4. The invoice ........ by the end of the week so we sent them a strong reminder.
a) had arrived b) hasn't arrived c) hadn't arrived
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a) have bought b) didn't buy c) haven't bought
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Ответов: 1
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