<span>The general form of quadratic equation with real
coefficients and leading coefficient 1, has x = -bi as a root
=> x = <u>-b + </u></span><u>√ b^2 – 4 ac</u><span>
2a
It is also written as:
=> ax^2 + bx + c = 0
Quadratic equation involves unknown numbers which is x, the numbers which a, b
and c are called coeffecients.
There are also quadratic factorization where you factor the polynomial give to
be able to get the value of the equation.</span>
Answer:
Step-by-step explanation:Total hours:2 hours
Total distance:120 miles
speed: 120miles/2hours equal to 60miles/hour
Hey there! :)
Answer:
(5, -2), or x = 5 and y = -2.
Step-by-step explanation:
We can solve the two equations algebraically by eliminating a variable:
2x + 5y = 0
3x - 4y = 23
Eliminate the x variable by finding the least common multiple and multiplying both equations:
3(2x + 5y = 0)
2(3x - 4y = 23)
Distribute and subtract the bottom equation from the top:
6x + 15y = 0
6x - 8y = 46
------------------
0x + 23y = -46
23y = -46
y = -2.
Plug in y into an equation to solve for x:
2x + 5(-2) = 0
2x - 10 = 0
2x = 10
x = 5. Therefore:
The solution to this equation is (5, -2), or x = 5 and y = -2.
X² - 7x + 10 = 0
Let's factor out the equation first.
x² - 7x + 10
x -2
x -5
This factoring of (x-2)(x-5) fit the equation.
(x-2)(x-5) = 0
One of the factors must equal 0 as to equal zero.
Either x - 2 = 0 or x - 5 = 0
x - 2 = 0
x = 2
x - 5 = 0
x = 5
The two answers are x = 2 or x = 5. Hope this helps!
Let Ch and C denote the events of a student receiving an A in <u>ch</u>emistry or <u>c</u>alculus, respectively. We're given that
P(Ch) = 88/520
P(C) = 76/520
P(Ch and C) = 31/520
and we want to find P(Ch or C).
Using the inclusion/exclusion principle, we have
P(Ch or C) = P(Ch) + P(C) - P(Ch and C)
P(Ch or C) = 88/520 + 76/520 - 31/520
P(Ch or C) = 133/520
