The answer says 42,185.4
Hope this helps!
Hope this helps!
Round the number 45,812 to the hundreds place
2 answers:
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Answer:
Isosceles triangle
Step-by-step explanation:
It is assumed that the triangle is a right angle triangle, implying that Pythagoras theorem becomes applicable.
If the first side is represented by x and the second side by y,
x + y = 5
The hypotenuse is always the longest side hence it can be presumed to be 4 since both x and y cannot be more than 5.
Therefore using pythagoras equation:
hyp^2 = sqrt(opp^2 + adj^2)
this implies that
therefore
16 = sqrt (x^2 + y^2) ----------------- (i)
since x + y = 5, y = 5 - x ------------------- (ii)
substituting (ii) into eqn (i)
16 = sqrt ( x^2 + (5-X)^2)
16 = sqrt( x^2 + 25 -10x + x^2)
16 = sqrt( x^2 + x^2 - 10x + 25)
16 = sqrt( 2x^2 - 10x + 25)
squaring both sides
256 = 2x^2 - 10x + 25
this can be refined as
2x^2 - 10x - 231 = 0
Solving using quadratic equation:
where a = 2, b = -10 and c = -231
x = 13.53 or -8.53.
Since x is a side of a triangle that cannot be negative,
x = 13.53 but x can never be greater than 5 since x + y = 5
This implies that the triangle is not a right angle triangle; it is most likely an isosceles triangle with two sides of equal length such that x = y = 2.5 inches while the longest side; the hypotenuse is 4 inches.
substituting for x in
Answer:
Indefinite integration acts as a tool to solve many physical problems.
There are many type of problems that require an indefinite integral to solve.
Basically indefinite integration is required when we deal with quantities that vary spatially or temporally.
As an example consider the following example:
Suppose that we need to calculate the total force on a object placed in a non- uniform field.
As an example let us consider a rod of length L that posses an charge 'q' per meter length and suppose that we place it in a non uniform electric field which is given by
Now in order to find the total force on the rod we cannot use the similar procedure as we can see that the force on the rod varies with the position of the rod.
But if w consider an element 'dx' of the rod at a distance 'x' from the origin the force on this element will be given by
Now to find the whole force on the rod we need to sum this quantity over the whole length of the rod requiring integration, as shown
Similarly there are numerous problems considering motion of particles that require applications of indefinite integration.