0
If z = 3, what is 5 x (6 – z)?
What else can I help you with?
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.
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
6 plus z equals 3 solve for z?
6+z = 3 z = 3-6 z = -3
X varies directly with y and z. x 120 when y 5 and z 8. Find x when y 6 and z 2.?
Since ( x ) varies directly with ( y ) and ( z ), we can express this relationship as ( x = k \cdot y \cdot z ) for some constant ( k ). Given that ( x = 120 ) when ( y = 5 ) and ( z = 8 ), we can find ( k ) as follows: [ 120 = k \cdot 5 \cdot 8 \implies k = \frac{120}{40} = 3. ] Now, to find ( x ) when ( y = 6 ) and ( z = 2 ): [ x = 3 \cdot 6 \cdot 2 = 36. ] Thus, ( x = 36 ).
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.
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.
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.
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.
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.
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
What is the answer for x(-3y - 5z) if x 14 y 5 and z 6?
To find the answer for ( x(-3y - 5z) ) given ( x = 14 ), ( y = 5 ), and ( z = 6 ), first substitute the values into the expression. Calculate ( -3y - 5z ): [ -3(5) - 5(6) = -15 - 30 = -45 ] Now, substitute this result back into the expression: [ x(-3y - 5z) = 14 \times (-45) = -630 ] So, the answer is (-630).
