The location of four buildings on a coordinate grid are as follows:

Mathematics

2 answers:

Ainat [17]3 years ago

5 0

Answer:

The answer is 3,400

Explanation:

I took the test this is right! Have a good day!

nataly862011 [7]3 years ago

3 0

1) From building 1 to 2: 5 - (-3) = 8

2) From 2 to 3: 4 - (-5) = 9

3) From 3 to 4: 5 - (-3) = 8

4) From 4 to 1: 4 - (-5) = 9

Total: 8 + 9 + 8 + 9 = 34 units = 34*100 feet = 3400 feet.

Answer: 3400 feet

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Last question and then I’ll take care of the rest on my own ^^^^

Bond [772]

Answer:

Step-by-step explanation:

The applicable rules of exponents are ...

(a^b)(a^c) = a^(b+c)

(a^b)/(a^c) = a^(b-c)

a^-b = 1/a^b

Using these rules, we can simplify the expression as follows:

(5^-3)(2^2)(5^6)/((2^3)(5^2)) = (5^(-3+6-2))(2^(2-3)) = (5^1)(2^-1)

= 5/2

6 0

2 years ago

) Set up a double integral for calculating the flux of F=3xi+yj+zk through the part of the surface z=−5x−2y+2 above the triangle

Fynjy0 [20]

The surface (call it $S$ ) is a triangle with vertices at the points

$x=0,y=0\implies z=2\implies(0,0,2)$

$x=0,y=2\implies z=-2\implies(0,2,-2)$

$x=2,y=0\implies z=-8\implies(2,0,-8)$

Parameterize $S$ by

$\vec s(u,v)=(1-v)(2,0,-8)+v\bigg((1-u)(0,2,-2)+u(0,0,2)\bigg)=(2-2v,2v-2uv,-8+6v+4uv)$

with $0\le u\le1$ and $0\le v\le1$ . Take the normal vector to $S$ to be

$\vec s_v\times\vec s_u=(20v,8v,4v)$

Then the flux of $\vec F$ across $S$ is

$\displaystyle\iint_S\vec F(x,y,z)\cdot\mathrm d\vec S=\int_0^1\int_0^1\vec F(x(u,v),y(u,v),z(u,v))\cdot(\vec s_v\times\vec s_u)\,\mathrm du\,\mathrm dv$