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3 years ago

7

HELP

2 answers:

Viktor [21]3 years ago

6 0

Answer:

An average resulting from the multiplication of each component by a factor reflecting it's importance.

Step-by-step explanation:

Assuming only one is selectable. Only possible answer doesn't need much explanation I believe.

romanna [79]3 years ago

3 0

Answer: an average resulting from the multiplication of each component by a factor reflecting its importance.

Step-by-step explanation:

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4 years ago

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givi [52]

Given: (a) $2(x+3)=2x+6$ and (b) $4(y-3)=4y-12$ .

To find: The expanded form of the given expressions.

(c) $4(m+n)=4m+4n$ (using $a(b+c)=ab+ac$ )

(d) $3(5-q)=15-3q$ (using $a(b-c)=ab-ac$ )

(e) $5(2c+1)=10c+5$ (using $a(b+c)=ab+ac$ )

(f) $3(2x-5)=6x-15$ (using $a(b-c)=ab-ac$ )

(g) $7(4b-1)=28b-7$ (using $a(b-c)=ab-ac$ )

(h) $3(2x+y-5)=3((2x+y)-5)$

⇒ $3(2x+y-5)=3(2x+y)-15$ (using $a(b-c)=ab-ac$ )

⇒ $3(2x+y-5)=6x+3y-15$ (using $a(b+c)=ab+ac$ )

(i) $2(6a-4b+3)=2((6a-4b)+3)$

⇒ $2(6a-4b+3)=2(6a-4b)+6$ (using $a(b+c)=ab+ac$ )

⇒ $2(6a-4b+3)=12a-8b+6$ (using $a(b-c)=ab-ac$ )

(j) $6(m+n+p)=6((m+n)+p)$

⇒ $6(m+n+p)=6(m+n)+6p$ (using $a(b+c)=ab+ac$ )

⇒ $6(m+n+p)=6m+6n+6p$ (using $a(b+c)=ab+ac$ )

(k) $y(y+2)=y^2+2y$ (using $a(b+c)=ab+ac$ )

(l) $g(g-3)=g^2-3g$ (using $a(b-c)=ab-ac$ )

(m) $n(4-n)=4n-n^2$ (using $a(b-c)=ab-ac$ )

(n) $a(b+c)=ab+ac$ (using $a(b+c)=ab+ac$ )

(o) $s(3s-4)=3s^2-4s$ (using $a(b-c)=ab-ac$ )

(p) $2x(x+5)=2x^2+10x$ (using $a(b+c)=ab+ac$ )

(q) $4y(x-3)=4xy-12y$ (using $a(b-c)=ab-ac$ )

(r) $5a(2b-5)=10ab-25a$ (using $a(b-c)=ab-ac$ )

(s) $4a(3b+2c)=12ab+8ac$ (using $a(b+c)=ab+ac$ )

(t) $5p(4p-5q)=20p^2-25pq$ (using $a(b-c)=ab-ac$ )

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3 years ago