Mathematics: (a) solve 7( k - 3 ) = 3k - 5 (b) expand and simplify (2x + 3 )( x - 8) (c) solve 7 - 3 f = 2 4

(a) solve 7( k - 3 ) = 3k - 5 (b) expand and simplify

Ответ:

sandraranger

30.12.2021

A) $7(k-3)=3k-5\\ 7k-21=3k-5\\ 7k-3k=-5+21\\ 4k=16\\ k=\frac { 16 }{ 4 } \\ k=4$

b) $(2x+3)(x-8)\\ 2{ x }^{ 2 }-16x+3x-24\\ 2{ x }^{ 2 }-13x-24$

c) $\frac { 7-3f }{ 4 } =2\\ 7-3f=4\cdot 2\\ 7-3f=8\\ -3f=8-7\\ -3f=1\\ f=-\frac { 1 }{ 3 }$

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Ответ:

860jay

30.12.2021

Mathematics

$(a)\\7(k-3)=3k-5\\7(k)+7(-3)=3k-5\\7k-21=3k-5\ \ \ \ |add\ 21\ to\ both\ sides\\7k=3k+16\ \ \ \ |subtract\ 3k\ from\ both\ sides\\4k=16\ \ \ \ \ |divide\ both\ sides\ by\ 4\\\boxed{k=4}$

$(b)\\(2x+3)(x-8)=(2x)(x)+(2x)(-8)+(3)(x)+3(-8)\\\\=2x^2-16x+3x-24=\boxed{2x^2-13x-24}$

$(c)\\\frac{7-3f}{4}=2\ \ \ \ |multiply\ both\ sides\ by\ 4\\\\\not4^1\cdot\frac{7-3f}{\not4_1}=4\cdot2\\\\7-3f=8\ \ \ \ \ |subtract\ 7\ from\ both\ sides\\\\-3f=1\ \ \ \ \ |divide\ both\ sides\ by\ (-3)\\\\\boxed{f=-\frac{1}{3}}$