Write the equation of the line below in slope-intercept form.

Sally wants to fill ten 8-inch tea glasses.how much tea does she need

ra1l [238]

More info is needed. If you were able to provide the radius of the base of the tea glasses as well as the height (8 inches), then it'd be possible to calculate the volume of one such glass. Are these short and fat tea glasses? or tall and slim tea glasses? My point: We don't know. Try to supply the missing info.

There are 10 such 8-inch tea glasses. Of course you could multiply 10 and 8 to get 80 inches, but 80 inches is NOT a suitable measure of the volume of a liquid.

3 0

4 years ago

How do I factor by grouping this expression 32u-uv+8u^2-4v

tangare [24]

Bring u outside the bracket so u (32 -v + 8u)

3 0

3 years ago

A cardboard box has a square base and an open top. the four sides are made of wood that costs 2 dollars per square foot, while t

serious [3.7K]

Answer:

$300

Step-by-step explanation:

Let x represent the side length of the square base in feet. Then the height of each side is ...

... h = (50 ft³)/(x ft)² = (50/x²) ft

The cost of the sides of the box is then ...

... (4 sides) × (x ft)(50/x² ft)/side × $2/ft² = $400/x

The cost of the bottom is ...

... (x ft)² × $25/ft² = $25x²

So, the total dollar cost is

... C = 400/x + 25x²

This will be a minimum where its derivative with respect to x is zero.

... 0 = -400/x² +50x

... 400/50 = 8 = x³ . . . . . add 400/x²; multiply by x²/50

... x = ∛8 = 2

For this value of x, the minimum cost is ...

... C = 400/2 + 25·2² = 300

The minimum possible cost is $300.

_____

<em>Comments on the problem</em>

1) Cardboard boxes are usually made of cardboard. They are rarely made of wood and alumninum.

2) The cost of the bottom is half the cost of the sides. When the dimensions are unconstrained, you will find (as here) the cost is shared equally between the bottom and pairs of opposite sides—each being 1/3 the total cost.

4 0

3 years ago

Use the substitution x = et to transform the given Cauchy-Euler equation to a differential equation with constant coefficients.

Anika [276]

Answer:

$\boxed{\sf \ \ \ ax^2+bx^{-10} \ \ \ }$

Step-by-step explanation:

Hello,

let's follow the advise and proceed with the substitution

first estimate y'(x) and y''(x) in function of y'(t), y''(t) and t

$x(t)=e^t\\\dfrac{dx}{dt}=e^t\\y'(t)=\dfrac{dy}{dt}=\dfrac{dy}{dx}\dfrac{dx}{dt}=e^ty'(x)y'(x)=e^{-t}y'(t)\\y''(x)=\dfrac{d^2y}{dx^2}=\dfrac{d}{dx}(e^{-t}\dfrac{dy}{dt})=-e^{-t}\dfrac{dt}{dx}\dfrac{dy}{dt}+e^{-t}\dfrac{d}{dx}(\dfrac{dy}{dt})\\=-e^{-t}e^{-t}\dfrac{dy}{dt}+e^{-t}\dfrac{d^2y}{dt^2}\dfrac{dt}{dx}=-e^{-2t}\dfrac{dy}{dt}+e^{-t}\dfrac{d^2y}{dt^2}e^{-t}\\=e^{-2t}(\dfrac{d^2y}{dt^2}-\dfrac{dy}{dt})$

Now we can substitute in the equation

$x^2y''(x)+9xy'(x)-20y(x)=0\\ e^{2t}[ \ e^{-2t}(\dfrac{d^2y}{dt^2}-\dfrac{dy}{dt}) \ ] + 9e^t [ \ e^{-t}\dfrac{dy}{dt} \ ] -20y=0\\ \dfrac{d^2y}{dt^2}-\dfrac{dy}{dt}+ 9\dfrac{dy}{dt}-20y=0\\ \dfrac{d^2y}{dt^2}+ 8\dfrac{dy}{dt}-20y=0\\$

so the new equation is

$y''(t)+ 8y'(t)-20y(t)=0$

the auxiliary equation is

$x^2+8x-20=0\\ x^2-2x+10x-20=0\\x(x-2)+10(x-2)=0\\(x+10)(x-2)=0\\ x=-10\text{ or }x=2$

so the solutions of the new equation are

$y(t)=ae^{2t}+be^{-10t}$

with a and b real

as

$x(t)=e^t\\ t(x)=ln(x)$

$y(x)=ae^{2ln(x)}+be^{-10ln(x)}=ax^2+bx^{-10}$

hope this helps

do not hesitate if you have any questions

8 0

3 years ago

There were 292 tickets sold for a high school basketball game. An adult ticket cost $3. A student ticket cost $1. Ticket sales w

77julia77 [94]

Best way to start this is set up your ground work...
Adult= A Student =S

A + S = 292

A3 + S1 = 470$

this will be your system of equations now set up your matrix and have A or S at the bottom stand alone.

4 0

4 years ago