Item 3 find the sum. −1/2(7z+4)+1/5(5z−15)

We are given : −1/2(7z+4)+1/5(5z−15)

Solution: We have -1/2 in front of parenthesis (7z+4)  and 1/5 in front of parenthesis (5z-15).

That represents multiplication of -1/2 by (7z+4) and multiplication of 1/5 by (5z-15).

So, we need to distribute -1/2 over (7z+4) and 1/5 over (5z-15).

Therefore, −1/2(7z+4)+1/5(5z−15) = -1/2 * 7z -1/2 *4 +1/5*5z + 1/5* -15.

On simplifying this step, we get

-7/2 z - 2 + z -3.

Now, we need to combine like terms.

We have like term of -7/2 z is z and like term of -2 is -3.

Let us add those like terms.

-7/2 z + z:  z could be written as z/1 to add with fraction -7/2 z.

Therefore, -7/2 z + z/1 = -7/2 z + 2z/2     (we took common denomiator here)

= (-7+2)z/2

= -5/2 z.

-2 -3 = -5.

Therefore,

-7/2 z - 2 + z -3 = -7/2 z +z -2-3

= -5/2 z -5.

So, the final answer is  -5/2 z -5.

answer rather than having to add and subtract the measurements of the semi-circle, notice that the semicircle to the left of ad is actually identical in area to the blank semicircle-shaped space to the left of bc. so if we transferred the red semicircle into the blank space, we would end up with a square of side length 24 cm. the area of a square is s^2 = 24^2 = 676 cm^2.

u also answered this so i think u would know but i am here to

put in or use mine

6mm^3 or .6cm^3

      216

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  36     6

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6     6   6

factor trees are nice.