(5)(5)(5)(5)(5)(5)(5)(5)(5)(5)(5)
5^11
That is two of the three but I'm not quite sure if there is a third.
Answer:
6. 6x-21=33+9x 7. -15x=-5(3x+7) 8. 16x-19=113-6x 9. -19x-34=56-x 10. -6(4x+3)=6(4x-3)?
6. 6x-21 = 33 +9x
+21 + 21
6x = 54 + 9x
-9x -9x
-3x = 54
/3 /3
x = -18
7. -15x = -5(3x+7)
First multiply the -5 to the other numbers.
-15x = -15x - 35
0x = -35
7 is impossible I believe.
8. 16x-19 = 113-6x
+6 +6
22x - 19 = 113
+ 19 +19
22x = 132
/22 /22
x = 6
9. -19x - 34 = 56 - x
+19x +19x
-34 = 56 + 18x
-56 -56
-90 = 18x
/18 /18
x = -5
10. -6(4x+3) = 6[4x-3]
-24x - 18 = 24x - 18
+18 +18
-24x= 24x
x = 0, since only zero can make the equation true.
Answer:
Step-by-step explanation:
First let us write the given polynomial as in descending powers of x with 0 coefficients for missing items
F(x) = x^3-3x^2+0x+0
We have to divide this by x-2
Leading terms in the dividend and divisor are
x^3 and x
Hence quotient I term would be x^3/x=x^2
x-2) x^3-3x^2+0x+0(x^2
x^3-2x^2
Multiply x-2 by x square and write below the term and subtract
We get
x-2) x^3-3x^2+0x+0(x^2
x^3-2x^2
---------------
-x^2+0x
Again take the leading terms and find quotient is –x
x-2) x^3-3x^2+0x+0(x^2-x
x^3-2x^2
---------------
-x^2+0x
-x^2-2x
Subtract to get 2x +0 as remainder.
x-2) x^3-3x^2+0x+0(x^2-x-2
x^3-2x^2
---------------
-x^2+0x
-x^2+2x
-------------
-2x-0
-2x+4
------------------
-4
Thus remainder is -4 and quotient is x^2-x-2
There’s two ways to substitute, but i think you mean if you know x but don’t know y. if that is true, then yes use substitution
