A tutor invited 2 students to a study group. Those 2 students each invited

1 answer:

8 0

Hello, CptnSkippy! The answer to your question is 14 students were invited in total! To get the answer, you can do 2 x 2=4 and 4 x 2 which is 8. now you can draw a picture to this too. good luck and hope you get a good grade for your homework

You might be interested in

Mrs. Bailey gives a test, and her students’ scores range from 30 to 70. She decides to curve the scores, so that they range from

Harman [31]

Answer:

f(x) = x*3/4 + 42.5

Step-by-step explanation:

The original difference between the pair is 70 - 30 = 40

The new difference between the pair is 95 - 65 = 30

Since the differences are not the same, Mrs Bailey must first perform a (slope) multiplication by a factor of 30/40 or 3/4

Then 30 * 3/4 = 22.5

Then she can shift the scores up by 65 - 22.5 = 42.5 in order to get the range from 65 to 95

Therefore, f(x) = x*3/4 + 42.5. We can test that

f(30) = 30*3/4 + 42.5 = 65

f(70) = 70*3/4 + 42.5 = 95

8 0

3 years ago

What is revenue from property taxes usually used for?

grandymaker [24]

Revenue from property taxes is used to fund all the services the municipal government provides us with, such as building a new park in the neighbourhood.

4 0

3 years ago

Read 2 more answers

) 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

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