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If z = 3, what is 5 x (6 – z)?
Shellyca Medica
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X - y z when x -3 y -7 z 5?
-32
If 15x's equals 20y's and 16y's equals 10z's then 6x's equals how many z's?
15x = 20y16y = 10z6x = ? Reducing equation #1 says that x = 4/3yDoing the same to #2 says that y = 5/8zIf x=(4/3)y and y=(5/8)z then x must be (4/3)(5/8) = (4/8)(5/3) = (1/2)(5/3) = (5/6)If x=(5/6) then 6x = 5 ■
If you throw 3 coins one with x on 1 side and y on other x on one side and z on other other with y on 1 side and z on other what is theoretical prob of getting 2 that match I don't understand?
3 out of 4. 8 possiableaties------------------ coins 1--- 2--- 3--- 4--- 5--- 6--- 7--- 8 x y--- x--- x--- x--- x--- y--- y---- y--- y x z--- x--- x--- z--- z--- x--- x----z--- z y z--- y----z---y--- z--- y--- z----y-----z There are 8 possiabilities for the three coins to land, you count the matches, there 6 out of 8 that match.
6 plus z over 9 equals 9?
6 + z/9 = 9 Therefore, z/9 = 3 3 x 9 = 27 Therefore, z = 27
In the coordinate What gos first y your z?
your x goes first. for example : 3, 6 3 is your x and 6 is your y
What compounds can be formed from element X with oxidation number 3 plus and 5 plus and element Z with oxidation numbers 2- and 3- write their formulas?
X(3+) and Z(2-) will give X2Z3. X(3+) and Z(3-) will give XZ. X(5+) and Z(2-) will give X2Z5. X(5+) and Z(3-) will give X3Z5.
Can anyone code this in C plus plus output 1 5 2 4 3 3 2 4 1 5?
int i, x, y, z; x=1; y=5; z=1; for (i=0; i<5; ++i) { printf ("%d %d ", x, y); if (x==3) z=z*(-1); x=x+z; y=y-z; } Output: 1 5 2 4 3 3 4 2 5 1
What compounds can be formed from element X with oxidation number 3 plus and 5 plus and element Z with oxidation numbers 2 and 3 write their formulas?
X(3+) and Z(2-) will give X2Z3. X(3+) and Z(3-) will give XZ. X(5+) and Z(2-) will give X2Z5. X(5+) and Z(3-) will give X3Z5.
X - y z when x -3 y -7 z 5?
-32
If you throw 3 coins one with x on 1 side and y on other x on one side and z on other other with y on 1 side and z on other what is theoretical prob of getting 2 that match I don't understand?
3 out of 4. 8 possiableaties------------------ coins 1--- 2--- 3--- 4--- 5--- 6--- 7--- 8 x y--- x--- x--- x--- x--- y--- y---- y--- y x z--- x--- x--- z--- z--- x--- x----z--- z y z--- y----z---y--- z--- y--- z----y-----z There are 8 possiabilities for the three coins to land, you count the matches, there 6 out of 8 that match.
If 15x's equals 20y's and 16y's equals 10z's then 6x's equals how many z's?
15x = 20y16y = 10z6x = ? Reducing equation #1 says that x = 4/3yDoing the same to #2 says that y = 5/8zIf x=(4/3)y and y=(5/8)z then x must be (4/3)(5/8) = (4/8)(5/3) = (1/2)(5/3) = (5/6)If x=(5/6) then 6x = 5 ■
In a normal distribution with mu equals 25 and sigma equals 6 what number corresponds to z equals 3?
Z = (x-mu)/sigma or 3 = (x - 25)/6 or 18 = x - 25 or x = 42. So, for Z = 3, x = 42.
How do you convert an equation into a system?
you can convert Fermat's equation into a system. Assuming z^3=x^3+y^3. Mean [z(z+1)/2]^2 - [1^3+2^3+3^3+4^3+5^3+6^3+7^3+....+(a-1)^3+(a+1)^3+.......+z^3]+[z(z+1)/2]^2 - [1^3+2^3+3^3+4^3+5^3+6^3+7^3+....+(a-1)^3+(a+1)^3+.......+x^3+y^3]=[x(x+1)/2]^2 - [1^3+2^3+3^3+4^3+5^3+6^3+7^3+....+(a-1)^3+(a+1)^3+......+x^3]+[y(y+1)/2]^2 - [1^3+2^3+3^3+4^3+5^3+6^3+7^3+....+(a-1)^3+(a+1)^3+.......+y^3] after simple, x^3 and y^3 have been lost , just rest z^3. number (a) can change unlimited This complex system have no solution integer.
Can you understand my proof about Fermat?
Pierre De Fermat 's last Theorem. The conditions: x,y,z,n are the integers and >0. n>2. Proof: z^n=/x^n+y^n. We have; z^3=[z(z+1)/2]^2-[(z-1)z/2]^2 Example; 5^3=[5(5+1)/2]^2-[5(5-1)/2]^2=225-100=125 And z^3+(z-1)^3=[z(z+1)/2]^2-[(z-2)(z-1)/2]^2 Example; 5^3+4^3=[5(5+1)/2]^2-[(5-2)(5-1)/2]^2=225-36=189 And z^3+(z-1)^3+(z-2)^3=[z(z+1)/2]^2-[(z-3)(z-2)/2]^2 Example 5^3+4^3+2^3=[5(5+1)/2]^2-[(5-3)(5-2)/2]^2=225-9=216 And z^3+(z-1)^3+(z-2)^3+(z-3)^3=[z(z+1)/2]^2-[(z-4)(z-3)/2]^2 Example 5^3+4^3+3^3+2^3=[5(5+1)/2]^2-[(5-4)(5-3)/2]^2=225-1=224 General: z^3+(z-1)^3+....+(z-m)^3=[z(z+1)/2]^2-[(z-m-1)(z-m)/2]^2 We have; z^3=z^3+(z-m-1)^3 - (z-m-1)^3. Because: z^3+(z-m-1)^3=[z^3+(z-1)^3+....+(z-m-1)^3] - [(z-1)^3+....+(z-m)^3] So z^3=[z(z+1)/2]^2-[(z-m-2)(z-m-1)/2]^2 - [z(z-1)/2]^3+[(z-m-1)(z-m)/2]^2 - (z-m-1)^3. Similar: z^3=z^3+(z-m-2)^3 - (z-m-2)^3. So z^3=[z(z+1)/2]^2-[(z-m-3)(z-m-2)/2]^2 - [z(z-1)/2]^3+[(z-m-2)(z-m-1)/2]^2 - (z-m-2)^3. .... .... Suppose: z^n=x^n+y^n So z^(n-3)*z^3=x^(n-3)^n*x^3+y^(n-3)*y^3. So z^(n-3)*{[z(z+1)/2]^2-[(z-m-2)(z-m-1)/2]^2 - [z(z-1)/2]^3+[(z-m-1)(z-m)/2]^2 - (z-m-1)^3}=x^(n-3)*{[x(x+1)/2]^2-[(x-m-2)(x-m-1)/2]^2 - [x(x-1)/2]^3+[(x-m-1)(x-m)/2]^2 - (x-m-1)^3}+y^(n-3)*{[y(y+1)/2]^2-[(y-m-2)(y-m-1)/2]^2 - [y(y-1)/2]^3+[(y-m-1)(y-m)/2]^2 - (y-m-1)^3} Similar: z^(n-3)*{[z(z+1)/2]^2-[(z-m-3)(z-m-2)/2]^2 - [z(z-1)/2]^3+[(z-m-2)(z-m-1)/2]^2 - (z-m-2)^3=x^(n-3)*{[x(x+1)/2]^2-[(x-m-3)(x-m-2)/2]^2 - [x(x-1)/2]^3+[(x-m-2)(x-m-1)/2]^2 - (x-m-2)^3+y^(n-3)*{[y(y+1)/2]^2-[(y-m-3)(y-m-2)/2]^2 - [y(y-1)/2]^3+[(y-m-2)(y-m-1)/2]^2 - (y-m-2)^3. .... .... Because it is codified . So Impossible all are the integers. So: z^n=/x^n+y^n. ISHTAR.
How is newest proof Fermat?
Pierre De Fermat 's last Theorem. The conditions: x,y,z,n are the integers and >0. n>2. Proof: z^n=/x^n+y^n. We have; z^3=[z(z+1)/2]^2-[(z-1)z/2]^2 Example; 5^3=[5(5+1)/2]^2-[5(5-1)/2]^2=225-100=125 And z^3+(z-1)^3=[z(z+1)/2]^2-[(z-2)(z-1)/2]^2 Example; 5^3+4^3=[5(5+1)/2]^2-[(5-2)(5-1)/2]^2=225-36=189 And z^3+(z-1)^3+(z-2)^3=[z(z+1)/2]^2-[(z-3)(z-2)/2]^2 Example 5^3+4^3+2^3=[5(5+1)/2]^2-[(5-3)(5-2)/2]^2=225-9=216 And z^3+(z-1)^3+(z-2)^3+(z-3)^3=[z(z+1)/2]^2-[(z-4)(z-3)/2]^2 Example 5^3+4^3+3^3+2^3=[5(5+1)/2]^2-[(5-4)(5-3)/2]^2=225-1=224 General: z^3+(z-1)^3+....+(z-m)^3=[z(z+1)/2]^2-[(z-m-1)(z-m)/2]^2 We have; z^3=z^3+(z-m-1)^3 - (z-m-1)^3. Because: z^3+(z-m-1)^3=[z^3+(z-1)^3+....+(z-m-1)^3] - [(z-1)^3+....+(z-m)^3] So z^3=[z(z+1)/2]^2-[(z-m-2)(z-m-1)/2]^2 - [z(z-1)/2]^3+[(z-m-1)(z-m)/2]^2 - (z-m-1)^3. Similar: z^3=z^3+(z-m-2)^3 - (z-m-2)^3. So z^3=[z(z+1)/2]^2-[(z-m-3)(z-m-2)/2]^2 - [z(z-1)/2]^3+[(z-m-2)(z-m-1)/2]^2 - (z-m-2)^3. .... .... Suppose: z^n=x^n+y^n So z^(n-3)*z^3=x^(n-3)^n*x^3+y^(n-3)*y^3. So z^(n-3)*{[z(z+1)/2]^2-[(z-m-2)(z-m-1)/2]^2 - [z(z-1)/2]^3+[(z-m-1)(z-m)/2]^2 - (z-m-1)^3}=x^(n-3)*{[x(x+1)/2]^2-[(x-m-2)(x-m-1)/2]^2 - [x(x-1)/2]^3+[(x-m-1)(x-m)/2]^2 - (x-m-1)^3}+y^(n-3)*{[y(y+1)/2]^2-[(y-m-2)(y-m-1)/2]^2 - [y(y-1)/2]^3+[(y-m-1)(y-m)/2]^2 - (y-m-1)^3} Similar: z^(n-3)*{[z(z+1)/2]^2-[(z-m-3)(z-m-2)/2]^2 - [z(z-1)/2]^3+[(z-m-2)(z-m-1)/2]^2 - (z-m-2)^3=x^(n-3)*{[x(x+1)/2]^2-[(x-m-3)(x-m-2)/2]^2 - [x(x-1)/2]^3+[(x-m-2)(x-m-1)/2]^2 - (x-m-2)^3+y^(n-3)*{[y(y+1)/2]^2-[(y-m-3)(y-m-2)/2]^2 - [y(y-1)/2]^3+[(y-m-2)(y-m-1)/2]^2 - (y-m-2)^3. .... .... Because it is codified . So Impossible all are the integers. So: z^n=/x^n+y^n. ISHTAR.
What are the efforts of Africa to solve the global news flow disequilibrium?
Pierre De Fermat 's last Theorem. The conditions: x,y,z,n are the integers and >0. n>2. Proof: z^n=/x^n+y^n. We have; z^3=[z(z+1)/2]^2-[(z-1)z/2]^2 Example; 5^3=[5(5+1)/2]^2-[5(5-1)/2]^2=225-100=125 And z^3+(z-1)^3=[z(z+1)/2]^2-[(z-2)(z-1)/2]^2 Example; 5^3+4^3=[5(5+1)/2]^2-[(5-2)(5-1)/2]^2=225-36=189 And z^3+(z-1)^3+(z-2)^3=[z(z+1)/2]^2-[(z-3)(z-2)/2]^2 Example 5^3+4^3+2^3=[5(5+1)/2]^2-[(5-3)(5-2)/2]^2=225-9=216 And z^3+(z-1)^3+(z-2)^3+(z-3)^3=[z(z+1)/2]^2-[(z-4)(z-3)/2]^2 Example 5^3+4^3+3^3+2^3=[5(5+1)/2]^2-[(5-4)(5-3)/2]^2=225-1=224 General: z^3+(z-1)^3+....+(z-m)^3=[z(z+1)/2]^2-[(z-m-1)(z-m)/2]^2 We have; z^3=z^3+(z-m-1)^3 - (z-m-1)^3. Because: z^3+(z-m-1)^3=[z^3+(z-1)^3+....+(z-m-1)^3] - [(z-1)^3+....+(z-m)^3] So z^3=[z(z+1)/2]^2-[(z-m-2)(z-m-1)/2]^2 - [z(z-1)/2]^3+[(z-m-1)(z-m)/2]^2 - (z-m-1)^3. Similar: z^3=z^3+(z-m-2)^3 - (z-m-2)^3. So z^3=[z(z+1)/2]^2-[(z-m-3)(z-m-2)/2]^2 - [z(z-1)/2]^3+[(z-m-2)(z-m-1)/2]^2 - (z-m-2)^3. .... .... Suppose: z^n=x^n+y^n So z^(n-3)*z^3=x^(n-3)^n*x^3+y^(n-3)*y^3. So z^(n-3)*{[z(z+1)/2]^2-[(z-m-2)(z-m-1)/2]^2 - [z(z-1)/2]^3+[(z-m-1)(z-m)/2]^2 - (z-m-1)^3}=x^(n-3)*{[x(x+1)/2]^2-[(x-m-2)(x-m-1)/2]^2 - [x(x-1)/2]^3+[(x-m-1)(x-m)/2]^2 - (x-m-1)^3}+y^(n-3)*{[y(y+1)/2]^2-[(y-m-2)(y-m-1)/2]^2 - [y(y-1)/2]^3+[(y-m-1)(y-m)/2]^2 - (y-m-1)^3} Similar: z^(n-3)*{[z(z+1)/2]^2-[(z-m-3)(z-m-2)/2]^2 - [z(z-1)/2]^3+[(z-m-2)(z-m-1)/2]^2 - (z-m-2)^3=x^(n-3)*{[x(x+1)/2]^2-[(x-m-3)(x-m-2)/2]^2 - [x(x-1)/2]^3+[(x-m-2)(x-m-1)/2]^2 - (x-m-2)^3+y^(n-3)*{[y(y+1)/2]^2-[(y-m-3)(y-m-2)/2]^2 - [y(y-1)/2]^3+[(y-m-2)(y-m-1)/2]^2 - (y-m-2)^3. .... .... Because it is codified . So Impossible all are the integers. So: z^n=/x^n+y^n. ISHTAR.
If x equals -3 Andy equals 6 and z equals 8 what is 6y over x?
x = -3; y = 6; 6y/x = 6(6)/-3 = 36/-3 = -12
