Answer:
144,148,152
Step-by-step explanation:
Here we will have, 4x + 4(x+1) + 4(x+2) = 444
=> x + (x+1) + (x+2) = 111 (dividing both sides by 4)
=>x + x+1 + x+2 =111
=> 3x + 3 = 111
=> 3x = 108
=> x = 36.
So, the three desired numbers 4x, 4(x+1) and 4(x+2) are,
4*36, 4*(36+1) and 4*(36+2)
That means the numbers are, 144, 148 and 152.
The equation factors as
.. (x +3)(x +4) = 0
By the zero-product rule, the roots are
.. x1 = -4
.. x2 = -3
3ab...when a = 2 and b = 3
(3)(2)(3) = 6(3) = 18
4y^2+4x+1-4y^2+2
4y+1+2
4y+3
To solve this we use trigonometric functions that would relate the hypotenuse y and the given values. For this case we use cosine function which is expressed as:
cosine theta = adjacent side / hypotenuse
cosine 52 = 35 / y
y = 35 / cos 52
y = 56.85
