Round $34.7691 to the nearest cent.

Paint sells in metal cans one gallon each. How many gallons of paint must be purchased to paint a room containing 840 square fee

Kisachek [45]

Answer:

5 Gallons of paint

Step-by-step explanation:

840 square ft one Gallons covers 190 square ft

6 0

3 years ago

Follow me on inst.a sarinagaytan123

kati45 [8]

ok Step-by-step explanation:

5 0

3 years ago

The length of the shorter base in an isosceles trapezoid is 4 in, its altitude is 5 in, and the measure of one of its obtuse ang

siniylev [52]

The sum of angles in any quadrilateral, including trapezoid, is 360⁰.
Because we have an isosceles trapezoid, we have 2 angles with measure 135⁰,
and we have 2 equal acute angles with measure x⁰.
So, we can find value of acute angle,
135*2 +2x =360⁰
270+2x=360
2x=360-270
2x=90
x=45⁰

So, acute angles in trapezoid = 45⁰.
From triangle ABC,
angle ACB =90⁰
angle A=45⁰,
so angle ABC= 180-(90-45)=45⁰
Triangle ABC is isosceles triangle,so |AC| = |CB|= 5 in.

So, longer base AA' = 5+4+5= 14 in

Now, we can find area of trapezoid.
shorter base = 4 in
longer base = 14 in
altitude =h = 5 in
Area of trapezoid =(1/2)(base1+base2)*h
Area of trapezoid = (1/2)(4+14)*5= 9*5=45 in²
Answer is 45 in².

5 0

3 years ago

A metal cylinder can with an open top and closed bottom is to have volume 4 cubic feet. Approximate the dimensions that require

Aleksandr-060686 [28]

Answer:

$r\approx 1.084\ feet$

$h\approx 1.084\ feet$

$\displaystyle A=11.07\ ft^2$

Step-by-step explanation:

Optimizing With Derivatives

The procedure to optimize a function (find its maximum or minimum) consists in :

Produce a function which depends on only one variable
Compute the first derivative and set it equal to 0
Find the values for the variable, called critical points
Compute the second derivative
Evaluate the second derivative in the critical points. If it results positive, the critical point is a minimum, if it's negative, the critical point is a maximum

We know a cylinder has a volume of 4 $ft^3$ . The volume of a cylinder is given by

$\displaystyle V=\pi r^2h$

Equating it to 4

$\displaystyle \pi r^2h=4$

Let's solve for h

$\displaystyle h=\frac{4}{\pi r^2}$

A cylinder with an open-top has only one circle as the shape of the lid and has a lateral area computed as a rectangle of height h and base equal to the length of a circle. Thus, the total area of the material to make the cylinder is

$\displaystyle A=\pi r^2+2\pi rh$

Replacing the formula of h

$\displaystyle A=\pi r^2+2\pi r \left (\frac{4}{\pi r^2}\right )$

Simplifying

$\displaystyle A=\pi r^2+\frac{8}{r}$

We have the function of the area in terms of one variable. Now we compute the first derivative and equal it to zero

$\displaystyle A'=2\pi r-\frac{8}{r^2}=0$

Rearranging

$\displaystyle 2\pi r=\frac{8}{r^2}$

Solving for r

$\displaystyle r^3=\frac{4}{\pi }$

$\displaystyle r=\sqrt[3]{\frac{4}{\pi }}\approx 1.084\ feet$

Computing h

$\displaystyle h=\frac{4}{\pi \ r^2}\approx 1.084\ feet$

We can see the height and the radius are of the same size. We check if the critical point is a maximum or a minimum by computing the second derivative

$\displaystyle A''=2\pi+\frac{16}{r^3}$