WILL GIVE BRAINIEST ANSWER

A publisher needs to send many books to a local book retailer and will send the books in a combination of small and large boxes.

mote1985 [20]

Answer:

5 large boxes

3 small boxes

Step-by-step explanation:

Create two equations to represent the problem.

let "a" be the number of small boxes

let "b" be the number of large boxes

20a + 40b = 260 This equation shows the numbers of books

a + b = 8 This equations shows the number of boxes

Solve the system of equations (solve for a and b). We can solve using the substitution method.

Rearrange a + b = 8 to isolate one of the variables.

a = 8 - b New equation that represents "a"

Since we know an equation for "a", we can substitute what "a" equals into the other equation. There will only be one variable in the equation, so we can solve by isolating. Isolate by doing reverse operations.

Substitute "a" for 8 - b

20a + 40b = 260

20(8 - b) + 40b = 260 Distribute 20 over the brackets by multiplying

160 - 20b + 40b = 260 Collect like terms (numbers with same variables)

160 + 20b = 260 Start isolating "b". Subtract 160 from both sides

20b = 260 - 160

20b = 100 Divide both sides by 20

b = 5 Number of large boxes

Substitute "b" for 5 in the simplest equation

a + b = 8

a + 5 = 8 Subtract 5 from both sides to isolate "a"

a = 8 - 5

a = 3 Number of small boxes

Therefore there were 5 large boxes and 3 small boxes sent.

3 0

2 years ago

The height, h, of a falling object t seconds after it is dropped from a platform 300 feet above the ground is modeled by the fun

Svetllana [295]

H(x)=-16t^2+300

The average rate is the change in h divided by the change in t, mathematically:

r=(h(3)-h(0))/(t2-t1), in this case:

r=(-16*9+300-0-300)/(3-0)

r=-144/3

r= -48 ft/s

5 0

3 years ago

Read 2 more answers

Evaluate the surface integral S F · dS for the given vector field F and the oriented surface S. In other words, find the flux of

tresset_1 [31]

Because I've gone ahead with trying to parameterize $S$ directly and learned the hard way that the resulting integral is large and annoying to work with, I'll propose a less direct approach.

Rather than compute the surface integral over $S$ straight away, let's close off the hemisphere with the disk $D$ of radius 9 centered at the origin and coincident with the plane $y=0$ . Then by the divergence theorem, since the region $S\cup D$ is closed, we have

$\displaystyle\iint_{S\cup D}\vec F\cdot\mathrm d\vec S=\iiint_R(\nabla\cdot\vec F)\,\mathrm dV$

where $R$ is the interior of $S\cup D$ . $\vec F$ has divergence

$\nabla\cdot\vec F(x,y,z)=\dfrac{\partial(xz)}{\partial x}+\dfrac{\partial(x)}{\partial y}+\dfrac{\partial(y)}{\partial z}=z$

so the flux over the closed region is

$\displaystyle\iiint_Rz\,\mathrm dV=\int_0^\pi\int_0^\pi\int_0^9\rho^3\cos\varphi\sin\varphi\,\mathrm d\rho\,\mathrm d\theta\,\mathrm d\varphi=0$

The total flux over the closed surface is equal to the flux over its component surfaces, so we have

$\displaystyle\iint_{S\cup D}\vec F\cdot\mathrm d\vec S=\iint_S\vec F\cdot\mathrm d\vec S+\iint_D\vec F\cdot\mathrm d\vec S=0$

$\implies\boxed{\displaystyle\iint_S\vec F\cdot\mathrm d\vec S=-\iint_D\vec F\cdot\mathrm d\vec S}$

Parameterize $D$ by

$\vec s(u,v)=u\cos v\,\vec\imath+u\sin v\,\vec k$

with $0\le u\le9$ and $0\le v\le2\pi$ . Take the normal vector to $D$ to be

$\vec s_u\times\vec s_v=-u\,\vec\jmath$

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

$\displaystyle\iint_D\vec F\cdot\mathrm d\vec S=\int_0^{2\pi}\int_0^9\vec F(x(u,v),y(u,v),z(u,v))\cdot(\vec s_u\times\vec s_v)\,\mathrm du\,\mathrm dv$