Divide y^7/y^5. Write the quotient as one power. The variable is not equal to zero.

Mathematics

1 answer:

daser333 [38]3 years ago

7 0

Pls need to answer this I need point ,a make brainliest

You might be interested in

8x-3y=6-4x in linear equation

castortr0y [4]

It would be y=4x-2 because:
8x-3y=6-4x
-8x -8x
-3y=6-12x
/-3 /-3 /-3
y=-2+4x
y=4x-2

4 0

3 years ago

Todea what did one pencil say to another mathematics algebra order of operations answer

Serjik [45]

Part 1

$6- \frac{18-3^2}{4+(2-3)} =6- \frac{18-9}{4+(-1)} \\ \\ =6- \frac{9}{4-1} =6- \frac{9}{3} =6-3 \\ \\ =3$

Part 2

$7-8\div4(3^2-1)=7-8\div4(9-1) \\ \\ =7-8\div4(8)=7-8\div32=7- \frac{1}{2} \\ \\ =6 \frac{1}{2} = \frac{13}{2}$

Part 3

$2-2(2-5)^2+3^2=2-2(-3)^2+9 \\ \\ =2-2(9)+9=2-18+9=-7$

Part 4

$(5^2-9\cdot3)^2-11=(25-27)^2-11 \\ \\ =(-2)^2-11=4-11=-7$

Part 5

$[(2-3)^2-5]\cdot4=[(-1)^2-5]\cdot4 \\ \\ =(1-5)\cdot4=-4\cdot4=-16$

Part 6

$12-11\cdot2+16\div8=12-22+2 \\ \\ =-8$

Part 7

$-5+1\cdot3-(7-2^3)=-5+3-(7-8) \\ \\ =-2-(-1)=-2+1=-1$

Part 8

$8+2\cdot3-14\div7=8+6-2 \\ \\ =12$

Part 9

$(8-2)^2+\frac{1}{4}[4-3(6-10)]=6^2+\frac{1}{4}[4-3(-4)] \\ \\ =36+\frac{1}{4}[4-(-12)]=36+\frac{1}{4}(4+12)=36+\frac{1}{4}(16) \\ \\ =36+4=40$

Part 10

$7+(2-3^2)\cdot5=7+(2-9)\cdot5 \\ \\ =7+(-7)\cdot5=7+(-35)=7-35 \\ \\ =-28$

Part 11

$3-2[2^3+(-1)^3]=3-2[8+(-1)] \\ \\ =3-2(8-1)=3-2(7)=3-14 \\ \\ -11$

Part 12

$18+6\div3(2^2+5)=18+6\div3(4+5) \\ \\ =18+6\div3(9)=18+6\div27=18+ \frac{2}{9} \\ \\ =18 \frac{2}{9}$