One number is 4 less than a second number. Twice the second number is 16 less than 5 times the first.

Mathematics

1 answer:

andrew-mc [135]3 years ago

7 0

Answer:

1st num=x

sec num=y

4-x<y

2y<16-5x

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wlad13 [49]

Answer:

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Step-by-step explanation:

6 0

3 years ago

I have hard prolems on subtract 4.000 -6.000

anyanavicka [17]

6.000 - 4.000 = 2.000

It is a very simple equation. The easiest way to remember it would be to look at the first two numbers. 6 subtract 4 equals 2. So when you see this equation, simply for starters do 6 subtract 4, which equals 2, and then add a decimal place and the 3 zeros at the end. Does that make sense? Please let me know whether this was useful to you or not, and if you have any further questions please feel free to ask so on comments. Thank you!

8 0

3 years ago

Let ????C be the positively oriented square with vertices (0,0)(0,0), (2,0)(2,0), (2,2)(2,2), (0,2)(0,2). Use Green's Theorem to

bonufazy [111]

Answer:

-48

Step-by-step explanation:

Lets call L(x,y) = 10y²x, M(x,y) = 4x²y. Green's Theorem stays that the line integral over C can be calculed by computing the double integral over the inner square of Mx - Ly. In other words

$\int\limits_C {L(x,y)} \, dx + M(x,y) \, dy = \int\limits_0^2\int\limits_0^2 (M_x - L_y ) \, dx \, dy$

Where Mx and Ly are the partial derivates of M and L with respect to the x variable and the y variable respectively. In other words, Mx is obtained from M by derivating over the variable x treating y as constant, and Ly is obtaining derivating L over y by treateing x as constant. Hence,

M(x,y) = 4x²y
Mx(x,y) = 8xy
L(x,y) = 10y²x
Ly(x,y) = 20xy
Mx - Ly = -12xy

Therefore, the line integral can be computed as follows

$\int\limits_C {10y^2x} \, dx + {4x^2y} \,dy = \int\limits_0^2\int\limits_0^2 -12xy \, dx \, dy$

Using the linearity of the integral and Barrow's Theorem we have

$\int\limits_0^2\int\limits_0^2 -12xy \, dx \, dy = -12 \int\limits_0^2\int\limits_0^2 xy \, dx \, dy = -12 \int\limits_0^2\frac{x^2y}{2} |_{x = 0}^{x=2} \, dy = -12 \int\limits_0^22y \, dy \\= -24 ( \frac{y^2}{2} |_0^2) = -24*2 = -48$