Which equation is represented by the table?

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skad [1K]

Answer:

maybe lol

Step-by-step explanation:

user and pass?

6 0

2 years ago

Read 2 more answers

The inches of rainfall in the rainforest is represented by the function f(x) = 3(x-2)+ 4. What is the average rate of change bet

nata0808 [166]

Answer:

The average rate of change of rainfall in the rainforest between 2nd year and 6th year = <u>3 inches</u>

Step-by-step explanation:

Given function representing inches of rainfall:

$f(x)=3(x-2)+4$

To find the average rate of change between the 2nd year and the 6th year.

Solution:

The average rate of change between interval $[a,b]$ is given as :

$\frac{f(b)-f(a)}{b-a}$

For the given function we need to find the average rate of change between 2nd year and 6th year. $[2,6]$

So, we have:

$f(2)=3(2-2)+4=3(0)+4=4$

$f(6)=3(6-2)+4=3(4)+4=12+4=16$

Thus, average rate of change will be:

$\frac{f(6)-f(2)}{6-2}$

⇒ $\frac{16-4}{6-2}$

⇒ $\frac{12}{4}$

⇒ $3$

Thus, the average rate of change of rainfall in the rainforest between 2nd year and 6th year = 3 inches

8 0

3 years ago

Is 3x+4y=8 parallel or perpendicular

daser333 [38]

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6 0

3 years ago

A house on the land measures 45 feet by 38 feet,a wooded area covers 118 feet by 60 feet,and the front yard is 78 feet by 40 fee

Mekhanik [1.2K]

The figure is composed of 2 shapes. A rectangle and a right triangle.

Area of the Rectangle = 600 ft * 200 ft = 120,000 ft²
Area of the right triangle = (600 ft * 450 ft)/2 = 270,000 / 2 = 135,000 ft²

Total area = 120,000 ft² + 135,000 ft² = 260,000 ft²

Area of the house = 45 ft * 38 ft = 1,710 ft²
Area of the woods = 118 ft * 60 ft = 7,080 ft²
Area of the front yard = 78 ft * 40 ft = 3,120 ft²

Area of farmed land = 260,000 - 1,710 - 7,080 - 3,120 = 248,090 ft²

248,090 ft² * 1acre/43,560 ft² = 5.695 acres

7 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

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

$=\displaystyle\int_0^1\int_0^1(6-6v,2v-2uv,-8+6v+4uv)\cdot(20v,8v,4v)\,\mathrm du\,\mathrm dv$

$=\displaystyle8\int_0^1\int_0^1(11v-10v^2)\,\mathrm du\,\mathrm dv=\boxed{\frac{52}3}$

8 0

3 years ago