Calculus

It could equal almost anything.

The validity or invalidity of a function are not abstract but depend on its domain and codomain or range.

If for any point, A, in the domain there is a unique point, B, in the range such that f(A) = B then the function is valid at A.

The validity of a function can change from point to point.

For example, f(x) = sqrt(x) is not a function from the set of Real Numbers to the set of Real Numbers because any negative number in the domain is not mapped to any value in the range. This can be corrected either by changing the domain to the set of non-negative Real Numbers or (if you are a more advanced mathematician) change the range to the set of Complex Numbers.

Similarly the reciprocal function, f(x) = 1/x is valid everywhere except for x = 0.

Or f(x) = tan(x) is valid except for x = 90+k*180 degrees for all integer values of k - so it is not valid at an infinite number of points.

0.3183 (approx)

-1

The equation is -2y + 16 = 10

-2x2 + 4x - 3 when x = -3 is

-2*(-3)2 + 4*(-3) - 3

= -2*9 + 4*(-3) - 3

= -18 -12 -3

= -33

The letter X negative take away seven take away five take away one is an equation. This equation equals .500.

(6g)(4qts/g)(2pts/qt)(2cups/pt)=(6g)16cups/g)=96 cups

it 1.6

By unit of length and distance and conversion ,we can say that

1 dm=100 mm

2X + 5Y = 6

5Y = - 2X + 6

Y = - 2/5X + 6/5

----------------------------------and

5X - 2Y = 8

2Y = - 5X + 8

Y = - 5/2 + 4

-------------------

Look like intersecting lines, though not perpendicular

If x squared = 5 then x = sqrt(5) = 2.2361 (approx).

The "double prime", or second derivative of y = 5x, equals zero. The first derivative is 5, a constant. Since the derivative of any constant is zero, the derivative of 5 is zero.

To do this equation, you collect like terms. 7x-4x+x-15 is your original equation, you have 3 terms with "'x'" in them, and they are 7x, -4x, and x. 7x + (-4x) +x simplifies to 4x, because 7-4 = 3 + 1 = 4 (the x is really 1x....). After this, you get 4x-15, which is as simplified as this equation can be.

Take a few (minimum of two, take 2 only after prior practice) values of x as desired and substitute in the equation to get the y values. Now take a graph and have a scale (usually 1m = 1 unit) and point the x and y values on the graph. Join these points to form a line. This is the required graph

When x equals 8, the value of 5x would be 5 times 8, which is 40.

all reals

(12, 1)

6.4 x 10-9

It's because the diagonal line on each trapezoid cuts down on the error of your area estimation. It is the average of the left and right rules.

Y3

This quadratic equation has no solutions because the discriminant is less than zero.