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6z2 plus 12z plus 4?
24z+4 or 6(4z+1)
Does this problem correctly demonstrate the identity property 24z1 equals 24z?
Yes, the equation ( 24z_1 = 24z ) demonstrates the identity property, which states that any number multiplied by one remains unchanged. In this case, if ( z_1 ) is equal to ( z ), then both sides of the equation represent the same value, confirming that the identity property holds true. Thus, the equation illustrates that multiplying by 24 does not affect the equality as long as ( z_1 ) and ( z ) are equivalent.
How do you factor 24w plus 24z?
24(w + z)
When is the schedule for line 24z?
The schedule for line 24z can be found on the public transportation website or by contacting the transportation authority.
6z2 plus 12z plus 4?
24z+4 or 6(4z+1)
24zx1 equals 24z does this problem correctly demonstrate the identity property?
Yes, of multiplication.
What does polynomial have in it?
A polynomial has 3 or more terms. It can have exponents and variables. Example: 24x + 74y - 24z 24x = term 1 74y = term 2 24z = term 3 EDIT: Polynomials can have 1 or 2 terms as well. But those are special, they are monomials and binomials.
What are the valve clearances on a 1995 Nissan 24Z engine?
Both Exhaust & Inlet Clearances are 0.30mm = 0.012"...or..12 thou HOT
How is going to be the rock at the game?
well actually how you beat the rock at the game is like say I'm the rock and I'm at the game its like multiplying 24z=5/7 a pint of silver
Solve the system of equations 3x plus y plus 2z equal 1 also 2x minus y plus z equals negative 3 also x plus y minus 4z equals negative 3?
Given: 3x + y + 2z = 1 2x - y + z = -3 x + y - 4z = -3 Take any one of the equations (we'll use the first one), and solve for any one of the variables (we'll use y): y = 1 - 2z - 3x Now plug that value of y into the latter two equations: 2x - (1 - 2z - 3x) + z = -3 x + (1 - 2z - 3x) - 4z = -3 Now take either of those (again, we'll use the first one), and solve it for either of the remaining variables (we'll go for x): 2x - 1 + 2z + 3x + z = -3 ∴ 5x = -2 - 3z ∴ x = (3z + 2) / -5 Now take that value, and plug it into our other equation that uses x and z: (3z + 2) / -5 + 1 - 2z - 3(3z + 2) / -5 - 4z = -3 Then solve for z: ∴ 2(3z + 2) / 5 - 6z = -4 ∴ (6z + 4 - 30z) / 5 = -4 ∴ 4 - 24z = -20 ∴ 24z = 24 ∴ z = 1 Now we can take that value for z, and plug it back into our previous equation for x and z: x = (3z + 2) / -5 ∴ x = (3 + 2) / -5 ∴ x = -1 Finally, we can take those two values, and plug them into our equation for y: y = 1 - 2z - 3x ∴ y = 1 - 2 + 3 ∴ y = 2 So x = -1, y = 2, and z = 1 You can test these values by plugging them into each of the original three equations, and seeing if they solve correctly: 3x + y + 2z = 1 ∴ 3(-1) + 2 + 2(1) = 1 ∴ 2 + 2 - 3 = 1 ∴ 1 = 1 2x - y + z = -3 ∴ 2(-1) - 2 + 1 = -3 ∴ -2 - 2 + 1 = -3 ∴ -3 = -3 x + y - 4z = -3 ∴ -1 + 2 - 4 = -3 ∴ -3 = -3 which shows our answer to be correct.
