Which one is larger 1.0 or 1.03

Mathematics

2 answers:

lilavasa [31]4 years ago

5 0

Answer: 1.0 is the larger one.

lara [203]4 years ago

5 0

Answer:

1.03

Step-by-step explanation:

1.0 < 1.03

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(xsinA-ucosA)^2+(xcosA+ysinA)^2=x^2+y^2

Artist 52 [7]

Answer:

u = x tan(A) - sec(A) sqrt(x^2 + y^2 - (x cos(A) + y sin(A))^2) or u = sec(A) sqrt(x^2 + y^2 - (x cos(A) + y sin(A))^2) + x tan(A)

Step-by-step explanation:

Solve for u:

(x sin(A) - u cos(A))^2 + (x cos(A) + y sin(A))^2 = x^2 + y^2

Subtract (x cos(A) + y sin(A))^2 from both sides:

(x sin(A) - u cos(A))^2 = x^2 + y^2 - (x cos(A) + y sin(A))^2

Take the square root of both sides:

x sin(A) - u cos(A) = sqrt(x^2 + y^2 - (x cos(A) + y sin(A))^2) or x sin(A) - u cos(A) = -sqrt(x^2 + y^2 - (x cos(A) + y sin(A))^2)

Subtract x sin(A) from both sides:

-u cos(A) = sqrt(x^2 + y^2 - (x cos(A) + y sin(A))^2) - x sin(A) or x sin(A) - u cos(A) = -sqrt(x^2 + y^2 - (x cos(A) + y sin(A))^2)

Divide both sides by -cos(A):

u = x tan(A) - sec(A) sqrt(x^2 + y^2 - (x cos(A) + y sin(A))^2) or x sin(A) - u cos(A) = -sqrt(x^2 + y^2 - (x cos(A) + y sin(A))^2)

Subtract x sin(A) from both sides:

u = x tan(A) - sec(A) sqrt(x^2 + y^2 - (x cos(A) + y sin(A))^2) or -u cos(A) = -x sin(A) - sqrt(x^2 + y^2 - (x cos(A) + y sin(A))^2)

Divide both sides by -cos(A):

Answer: u = x tan(A) - sec(A) sqrt(x^2 + y^2 - (x cos(A) + y sin(A))^2) or u = sec(A) sqrt(x^2 + y^2 - (x cos(A) + y sin(A))^2) + x tan(A)

7 0

3 years ago

Read 2 more answers

Let divf = 6(5 − x) and 0 ≤ a, b, c ≤ 12. (a) find the flux of f out of the rectangular solid 0 ≤ x ≤ a, 0 ≤ y ≤ b, and 0 ≤ z ≤

dusya [7]

Continuing from the setup in the question linked above (and using the same symbols/variables), we have

$\displaystyle\iint_{\mathcal S}\mathbf f\cdot\mathrm d\mathbf S=\iiint_{\mathcal R}(\nabla\cdot f)\,\mathrm dV$
$=\displaystyle6\int_{z=0}^{z=c}\int_{y=0}^{y=b}\int_{x=0}^{x=a}(5-x)\,\mathrm dx\,\mathrm dy\,\mathrm dz$
$=\displaystyle6bc\int_0^a(5-x)\,\mathrm dx$
$=6bc\left(5a-\dfrac{a^2}2\right)=3abc(10-a)$

The next part of the question asks to maximize this result - our target function which we'll call $g(a,b,c)=3abc(10-a)$ - subject to $0\le a,b,c\le12$ .

We can see that $g$ is quadratic in $a$ , so let's complete the square.

$g(a,b,c)=-3bc(a^2-10a+25-25)=3bc(25-(a-5)^2)$

Since $b,c$ are non-negative, it stands to reason that the total product will be maximized if $a-5$ vanishes because $25-(a-5)^2$ is a parabola with its vertex (a maximum) at (5, 25). Setting $a=5$ , it's clear that the maximum of $g$ will then be attained when $b,c$ are largest, so the largest flux will be attained at $(a,b,c)=(5,12,12)$ , which gives a flux of 10,800.