Answer:
2910 ….....................….......
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.
Area=pi x radius squared so (3.14)(10)squared = 314
Answer: The value of k for which one root of the quadratic equation kx2 - 14x + 8 = 0 is six times the other is k = 3.
Let's look into the solution step by step.
Explanation:
Given: A quadratic equation, kx2 - 14x + 8 = 0
Let the two zeros of the equation be α and β.
According to the given question, if one of the roots is α the other root will be 6α.
Thus, β = 6α
Hence, the two zeros are α and 6α.
We know that for a given quadratic equation ax2 + bx + c = 0
The sum of the zeros is expressed as,
α + β = - b / a
The product of the zeros is expressed as,
αβ = c / a
For the given quadratic equation kx2 - 14x + 8 = 0,
a = k, b = -14, c = 8
The sum of the zeros is:
α + 6α = 14 / k [Since the two zeros are α and 6α]
⇒ 7α = 14 / k
⇒ α = 2 / k --------------- (1)
The product of the zeros is:
⇒ α × 6α = 8 / k [Since the two zeros are α and 6α]
⇒ 6α 2 = 8 / k
⇒ 6 (2 / k)2 = 8 / k [From (1)]
⇒ 6 × (4 / k) = 8
⇒ k = 24 / 8
⇒ k = 3
1) Use the distributive property to eliminate parentheses.
.. 3(6x) -3(5) -7(3x) -7(10) = 0
.. 18x -15 -21x -70 = 0 . . . . . . finish multiplying terms
.. -3x -85 = 0 . . . . . . . . . . . . . collect like terms
.. -85 = 3x . . . . . . . . . . . . . . . .add 3x
.. -85/3 = x . . . . . . . . . . . . . . .divide by 3
.. -28 1/3 = x . . . . . . . . . . . . . write as mixed number
2) 5 -(6 +9x) = 9 -(4x -1)
.. 5 -6 -9x = 9 -4x +1 . . . . . eliminate parentheses using the distributive property
.. -1 -9x = 10 -4x . . . . . . . . . collect like terms
.. -1 = 10 +5x . . . . . . . . . . . . add 9x
.. -11 = 5x . . . . . . . . . . . . . . . subtract 10
.. -11/5 = x . . . . . . . . . . . . . . divide by 5
.. -2 1/5 = x . . . . . . . . . . . . . write as mixed number
