Ihave a few algebra problems that i have no clue how to solve. would be greatly appreciated! 37= -3+5(k+6)-2= -(w-8)42= 8x+13x8= 8c-4(c+8)-11-5z= 6(5z+4)0=-5y-2y

I'm just going to number all of the problems so life get's easier. :D

JSYK: A negative and a negative multiplied make a positive. Like that... a + and a + is +. And a + and a - is a - .

1) 37= -3+5(k+6)

1. Use Distributive Property to take out the parentheses. Positive 5 times k and then 5 times 6... 37=-3+5k+30. 2. Take all the normal number to either side of the equal sign. 37-3-30=5k 3. Subtract all the numbers on the left side.... 4=5k 4. Now divide both sides by 5 to isolate k.... k=[tex] \frac{4}{5} [/text] ----------------------------------------------------------------------------------------------- 2) -2= -(w-8)

1. Use Distributive Property to take out the parentheses. "Distribute the negative sign... -2= -w+8 2. Put the -w to the left side and the -2 to the right side, so that the variable won't be negative... w= 2+8 3. Add everything on the right side... w= 10 ----------------------------------------------------------------------------------- 3) 42= 8x+13x

1. Since both of the numbers have the same variables (in this case, x), add them up... 42=21x 2. Divide each side by 21 to isolate x... x=42 ------------------------------------------------------------------------------------ 4) 8= 8c-4(c+8)

1. Use Distributive Property. Multiply negative 4 with c and 8... 8=8c-4c-32 2. On the right side, 8 and 4 both have the same variable, c , solve it!...8=4c-32 3. Put -32 to the left side... 8+32=4c 4. Add on the left... 40=4c 5. Divide both sides by 4 to isolate c... c=10 ------------------------------------------------------------------------------------- 5) -11-5z= 6(5z+4) 1. Use Distributive Property; multiply positive 6 with 5z and 4... -11-5z=30z+24 2. Bring -5z to the right and positive 24 to the left... -11-24= 30z+5z 3. Subtract the integers on the left and add the numbers with same variable on the right... -35=35z 4. Divide both sides by 35 to isolate z... z= -1 ----------------------------------------------------------------------------- 6) 0= -5y-2y

1. Subtract the negative number with the same variable, y ... 0= -7y 2. Divide both sides by -7, to isolate y... y=0

And for the last two, I don't know either... my apologies! I hope my answers help! :D

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step-by-step explanation: vsfdvsfvd

step-by-step explanation:

given that there are two triangles j'k'l' and jkl>

the former is got by dilating the second one.

when drawn on a grid, (graph paper) coordinates of vertices of first triangle is

(6,12)j, (9,6)k and (12,9)l

side jk has length = distance between j and k

=

side j'k'=distance between j' and k'

=

ie j'k'l' has sides 1/3 of that of corresponding sides of jkl.

hence scale factor of 1/3 is used to get j'k'l'from jkl