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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?
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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?
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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.
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.
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.
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).
