Find the product, write the product as a mixed number 3/10x42

Find the volume of the composite solid. Round your answer to the nearest tenth.

EastWind [94]

Answer:

646.04 cm³

Step-by-step explanation:

Cube volume formula:

V=l*w*h

V=8*8*8

V=512

Hemisphere volume formula:

V=(4/3πr³)1/2

V=(4/3π4³)1/2

V=128/3

Add both together

128/3 + 512 ≈ 646.04cm³

4 0

3 years ago

what is the slope of the line that passes through the points (-4,-4) and (-4,-9)? Write your answer in simplest form

Aneli [31]

Answer:

undefined

Step-by-step explanation:

We can find the slope of a line using two points by

m = (y2-y1)/(x2-x1)

= (-9- -4)/(-4 - -4)

= (-9+4)/(-4+4)

= -5/0

When we divide by zero, our solutions is undefined

The slope is undefined

3 0

3 years ago

Shea wants to wrap a package that is 8 inches by 11 inches by 2 inches. Shea purchased a roll of wrapping paper that is 1 foot w

nekit [7.7K]

First we need to know the surface area of the package:
2(lw) + 2(lh) + 2(wh)
2(8*11) + 2(11*2) + 2(8*2)
2(88) + 2(22) + 2(16)
176 + 44 + 32
252 square inches needed

Wrapping paper is in feet, so convert it to inches before finding the area.
1' = 12"; 2' = 24"
12* 24 = 288 square inches.

So, yes there is enough paper to cover all sides of the gift if there are no overlaps.

6 0

3 years ago

A punch glass is in the shape of a hemisphere with a radius of 5 cm. If the punch is being poured into the glass so that the cha

Galina-37 [17]

Answer:

28.27 cm/s

Step-by-step explanation:

Though Process:

The punch glass (call it bowl to have a shape in mind) is in the shape of a hemisphere
the radius $r=5cm$
Punch is being poured into the bowl
The height at which the punch is increasing in the bowl is $\frac{dh}{dt} = 1.5$
the exposed area is a circle, (since the bowl is a hemisphere)
the radius of this circle can be written as $'a'$

what is being asked is the rate of change of the exposed area when the height $h = 2 cm$
the rate of change of exposed area can be written as $\frac{dA}{dt}$ .
since the exposed area is changing with respect to the height of punch. We can use the chain rule: $\frac{dA}{dt} = \frac{dA}{dh} . \frac{dh}{dt}$

and since $A = \pi a^2$ the chain rule above can simplified to $\frac{da}{dt} = \frac{da}{dh} . \frac{dh}{dt}$ -- we can call this Eq(1)

Solution:

the area of the exposed circle is

$A =\pi a^2$

the rate of change of this area can be, (using chain rule)

$\frac{dA}{dt} = 2 \pi a \frac{da}{dt}$ we can call this Eq(2)

what we are really concerned about is how $a$ changes as the punch is being poured into the bowl i.e $\frac{da}{dh}$

So we need another formula: Using the property of hemispheres and pythagoras theorem, we can use:

$r = \frac{a^2 + h^2}{2h}$

and rearrage the formula so that a is the subject:

$a^2 = 2rh - h^2$

now we can derivate a with respect to h to get $\frac{da}{dh}$

$2a \frac{da}{dh} = 2r - 2h$

simplify

$\frac{da}{dh} = \frac{r-h}{a}$

we can put this in Eq(1) in place of $\frac{da}{dh}$

$\frac{da}{dt} = \frac{r-h}{a} . \frac{dh}{dt}$

and since we know $\frac{dh}{dt} = 1.5$

$\frac{da}{dt} = \frac{(r-h)(1.5)}{a}$

and now we use substitute this $\frac{da}{dt}$ . in Eq(2)

$\frac{dA}{dt} = 2 \pi a \frac{(r-h)(1.5)}{a}$

simplify,

$\frac{dA}{dt} = 3 \pi (r-h)$

This is the rate of change of area, this is being asked in the quesiton!

Finally, we can put our known values:

$r = 5cm$

$h = 2cm$ from the question

$\frac{dA}{dt} = 3 \pi (5-2)$

$\frac{dA}{dt} = 9 \pi cm/s// or//\frac{dA}{dt} = 28.27 cm/s$

5 0

3 years ago

I need an answer and I’m offering a lot!!

ss7ja [257]

Line t. It cuts directly into the angle. Don’t get it confused with bisect because that comes directly from the angle and not just anywhere on it. Hope this helps

5 0

2 years ago