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Calculus
The branch of mathematics that deals with the study of continuously changing quantities, with the use of limits and the differentiation and integration of functions of one or more variables, is called Calculus. Calculus analyzes aspects of change in processes or systems that can be modeled by functions. The English physicist, Isaac Newton, and the German mathematician, G. W. Leibniz, working independently, developed calculus during the 17th century.
25,068 Questions
How many real solution are in this equation 2x2 plus x-1 equals 0?
Use the discriminant to find out.
b^2 - 4ac
a = 2
b = 1
c = - 1
1^2 - 4(2)(- 1)
1 + 8 = 9 > 1
so, there are two real roots to this equation. ( two real solutions )
What are the order pairs for x plus y equals 6 and -x plus 3y equals -2?
The answer is - not are : (5, 1).
How do you solve 6x-2y equals -16 4x plus 4y equals -32?
x = -4 and y = -4.
These are known as simultaneous equations. We have to manipulate the equations in order to remove one of the variables. Then we can solve for the remaining variable. Once we have done this, we can substitute this value back into one of the original equations to solve for the other variable. So:
1. 6x - 2y = -16
2. 4x + 4y = -32
If we multiply 1. by 2 then we will end up with a -4y in one equation and 4y in the other. We will then be able to add these together which will cancel out and remove the y variable altogether.
So, 1. multiplied by 2 gives:
3. 12x - 4y = -32
Adding 2. & 3. gives:
4. 16x = -64
x = -64/16
x = -4
Now we have worked out the value of x we can substitute this value back into one of the original equations.
Substituting x = -4 back into 1. gives:
(6 * -4) - 2y = -16
-24 - 2y = -16
2y = -24 + 16
2y = -8
y = -4
We can double check we have solved this correctly by substituting both values into the other original equation.
Substituting x = -4 & y = -4 into 2. gives:
(4 * -4) + (4 * -4) = -32
-16 + -16 = -32
-32 = -32 (Yes this works, we have found the correct values of x & y!)
Which sequence of fx equals 2x plus 2 has integer input values?
f(x)=2x+2. Put in 0, 1/2, 1, 3/2, 2... and you will get integer values. That is for the domain. The numbers you get when you put that in are the range integral values.
What is the solution to 2x squared minus 3x squared plus 5x minus 4 when x equals negative 1?
This is not calculus. Please post this in the correct forum.
However, 2x2 - 3x2 +5x - 4 = -1
therefore -x2+4x-3 = 0
therefore x2-4x+3 = 0
therefore (x-3)(x-1)=0
therefore x = 3 or x = 1.
Put your questions in the correct category. Makes then more likely to be answered correctly.
Since there are 2 unknowns, it takes 2 independent equations to solve for unique values.
So far, from the one given equation, all we know is that [ x = 5 - y ] and [ y = 5 - x ].
There are an infinite supply of number-pairs ... solutions ... that satisfy those conditions.
What is the square root of pie x 3 x 6.55444 plus 64.79 x 60.5465565?
3,957.66372 (as determined by google calculator)
How come 1 cubic meter equals 1000 cubic centimeters isn't it 1 million cubic centimeters?
Whoever told you otherwise was wrong. One cubic meter equals 1,000,000 cubic comtimeters.
Does a trapezoid's opposite sides parallel?
One pair must be. If the other pair is also parallel it becomes a parallelogram.
lets say x equals 9
(9+4)*4*4*4*4=
13*4*4*4*4=
52*4*4*4=
208*4*4=
832*4=
3328 there. Easy!
How would you differentiate -3cosy times x squared?
((-3)cos(y))(x2)
Well first of all you need decide what rule you should use. Now, because they're multiplied together I'm going to use the product rule which is -
(dy/dx)= f'g +fg'
remember to write everything out -
f=((-3)cos(y))
f'=((3)sin(y))
g=x2
g'=2x
so from here it's pretty simple, you just sub in the values you have for each letter, remember to use brackets!
so....(dy/dx)= ((3)sin(y))(x2)+((-3cos(y))(2x)
Now you just need to simplify and gather like terms -
(dy/dx)=3x2siny-6xcosy
=3x(xsiny-cosy)
And there you have it!
Remember to check this yourself cause its been a while since I've studied this so I might have made a silly mistake somewhere. ^_^ Hope this helps!
