The best method that will work for any quadratic equation is to use the quadratic formula: x = (-b±√(b² - 4ac))/2a, this will work for any quadratic of the form ax² + bx + c = 0.
As for the last equation in the attachment, that is a cubic equation, these are much trickier to solve and as such the formula is much longer and very complicated. Therefore it is easier to see if it can be broken down into a linear term and a quadratic. This can be done by substituting integer values of x into the equation to see if it holds true. If both sides of the equation are equal for a given value of x then the equation ax³ + bx² + cx + d can be rewritten as (Ax + B)(px² + qx + r). This can then be put into the quadratic formula mentioned above.
Answer: 50 minutes
Step-by-step explanation: a quarter to 5 is always 15 minutes from 5, 4:45. So 5:35-4:45 is 50.
Answer:
y = - 6x - 2
Step-by-step explanation:
The straight line equation is y = mx + c
First step is to bring the equation given to this form, where you make y the subject.
Now you can easily note down the gradient(m1). You will need it to find the gradient(m2) of the perpendicular line, because m1 × m2 = -1
After you have that substitute the value and the values of x and y given to find c .
Lastly, substitute the value of m2 and c into y = mx + c to find the line that is perpendicular .
Hope I was able to help:)))
Y - y1 = m(x - x1)
slope(m) = 3
(4,7)....x1 = 4 and y1 = 7
now we sub
y - 7 = 3(x - 4) <==
a,37
if you need help in math i do turring
