What is the quotient ? StartFraction 7 Superscript negative 6 Over 7 squared EndFraction StartFraction 1 Over 7 Superscript 8 En
2 answers:
The quotient of the start fraction 7 superscript 7 negative 6 over 7 squared end fractions is,
Quotient is the resultant number which is obtained by dividing a number with another. Let a number<em> a</em> is divided by number <em>b.</em> Then the quotient of these two number will be,
Here, (<em>a, b</em>) are the real numbers.
The number given in the problem is,
Let the quotient of these number is n. Therefore,
Now if the power of numerator is negative, then it can be written in denominator with positive power of the same number. Therefore, the above equation written as,
Hence, the quotient of the start fraction 7 superscript 7 negative 6 over 7 squared end fractions is,
Learn more about the quotient here;
You might be interested in
See attached. The trick is to use similar triangles.
Answer:
The Line integral is π/2.
Step-by-step explanation:
We have to find a funtion f such that its gradient is (ycos(xy), x(cos(xy)-ze^(yz), -ye^(yz)). In other words:
we can find the value of f using integration over each separate, variable. For example, if we integrate ycos(x,y) over the x variable (assuming y and z as constants), we should obtain any function like f plus a function h(y,z). We will use the substitution method. We call u(x) = xy. The derivate of u (in respect to x) is y, hence
(Remember that c is treated like a constant just for the x-variable).
This means that f(x,y,z) = sen(x,y)+C(y,z). The derivate of f respect to the y-variable is xcos(xy) + d/dy (C(y,z)) = xcos(x,y) - ye^{yz}. Then, the derivate of C respect to y is -ze^{yz}. To obtain C, we can integrate that expression over the y-variable using again the substitution method, this time calling u(y) = yz, and du = zdy.
Where, again, the constant of integration depends on Z.
As a result,
if we derivate f over z, we obtain
That should be equal to -ye^(yz), hence the derivate of K(z) is 0 and, as a consecuence, K can be any constant. We can take K = 0. We obtain, therefore, that f(x,y,z) = cos(xy) - e^(yz)
The endpoints of the curve are r(0) = (0,0,1) and r(1) = (1,π/2,0). FOr the Fundamental Theorem for Line integrals, the integral of the gradient of f over C is f(c(1)) - f(c(0)) = f((0,0,1)) - f((1,π/2,0)) = (cos(0)-0e^(0))-(cos(π/2)-π/2e⁰) = 0-(-π/2) = π/2.