The values of 3 over 4 are: 3/4, 0.75 and 75%
Given: 2x - 3y = 2 3x + 2y = 3 Take the first equation, and solve for x: x = (2 + 3y) / 2 Now plug it into the second equation: ∴ 3(2 + 3y) / 2 + 2y = 3 ∴ 3 + 9y/2 + 2y = 3 ∴ 9y/2 + 2y = 0 ∴ 22y = 0 ∴ y = 0 Then you can take that value of y, and plug it into either of our first equations to find x; 2x - 3y = 2 ∴ 2x - 3(0) = 2 ∴ 2x = 2 ∴ x = 1 So x is equal to one, and y is equal to zero.
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.
let y= xsqrt(x) -1 y= x^(3/2) -1 ---- since xsqrt(x) is the same as x^(3/2) y' = (3/2) x^(3/2-1) y' = (3/2) x^(1/2) y'' = (3/2) (1/2) x^(1/2-1) y'' =(3/2)(1/2) x^(-1/2) y'' = 3/4x^(1/2) y'' = 3 / 4sqrt(x)
log33+log3x +2=3 log33+log3x=1 log3(3x)=1 3x=3 x=1 Other interpretation: log33+log3(x+2)=3 log3(3(x+2))=3 3(x+2)=27 x+2=9 x=7
2 apples because if WE TAKE 2 apples then WE HAVE 3 apples.
5 apples take away 3 apples leaves you with 2 apples. 5 - 3 = 2
You will have 3 apples, and there will be 2 apples left.
You would have 2 apples 5 - 3 = 2
2 apples, (this is the amount I took, therefore it is what I have)
2 apples.
1 apple. No, 2 apples because you took 2.
3 apples
You would have 2 apples.
5 - 3 = 2 Therefore, you would have two apples.
2 because that's how many you took.
YOU would have 3 apples. You see, there are 5 apples, but you don't have those apples, if you take 3 away, then you are taking 3 of those apples, which then you have 3 apples, and you leave 2 left over.