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3 years ago

A hockey puck has a diameter of 3 inches and rolls on its edge for 24 rotations. How far did it roll before falling flat?

Mathematics

1 answer:

scoundrel [369]3 years ago

4 0

Find the circumference of the puck using circumference = pi x diameter

Circumference = 3.14 x 3 = 9.42 inches.

For every rotation the puck would travel 9.42 inches.

Now multiply y by 24 rotations:

9.42 x 24 = 226.08 inches.

Answer: 226.08 inches

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In quadrilateral ABCD, AD is congruent to BC, and AD is parallel to BC. Andre has written a

Dahasolnce [82]

Answer:

AD is Congruent to BC and it's given, (given just means that it was already said or stated, and you don't need to do the work to find it) Angle DAC would be congruent to angle BAC (if that doesn't work, rearrange them to look like angle CAB). AC would be congruent to DB. Triangle ADC would be congruent to triangle BCD by (since i don't know exactly which way the letters are arranged is would either be SAS or SSA) and that's because you know two of the sides are congruent to each other and one angle.

I tried hard to sketch out what the shape looked like based on the information given, and that's because I need a visual of what the shape looks like. Sorry, this took so long to answer.

5 0

1 year ago

If x,y and z are positive integers and 3x=4y=7z,

Pani-rosa [81]

3x=4y=7z
so
GCF:
3 * 4 * 7 = 84

3x = 84
x = 28

4y = 84
y = 21

7z = 84
z = 12

least possible value for
x + y + z = 28 + 21 + 12 = 61

answer
(d) 61

5 0

3 years ago

Translate this sentence into an equation.

ElenaW [278]

Answer:

The product of Victor's age and 7 is 84 in equation form is 7v=84

Step-by-step explanation:

Let v be the Victor's age.

We are supposed to Translate this sentence into an equation

The product of Victor's age and 7 is 84

Victor's age = v

Product of victor's age and 7 = $7 \times v$

We are given that their product is 84

So, $7 \times v =84$

Hence The product of Victor's age and 7 is 84 in equation form is 7v=84

7 0

3 years ago

SA=2(3.14)r(2)+2(3.14)rh Solve for h

uranmaximum [27]

Answer:

I think but I might have done it the wrong way

2πr(r+h)

Step-by-step explanation:

7 0

4 years ago

You are creating an open top box with a piece of cardboard that is 16 x 30“. What size of square should be cut out of each corne

Arada [10]

Answer:

$\frac{10}{3} \ inches$ of square should be cut out of each corner to create a box with the largest volume.

Step-by-step explanation:

Given: Dimension of cardboard= 16 x 30“.

As per the dimension given, we know Lenght is 30 inches and width is 16 inches. Also the cardboard has 4 corners which should be cut out.

Lets assume the cut out size of each corner be "x".

∴ Size of cardboard after 4 corner will be cut out is:

Length (l)= $30-2x$

Width (w)= $16-2x$

Height (h)= $x$

Now, finding the volume of box after 4 corner been cut out.

Formula; Volume (v)= $l\times w\times h$

Volume(v)= $(30-2x)\times (16-2x)\times x$

Using distributive property of multiplication

⇒ Volume(v)= $4x^{3} -92x^{2} +480x$

Next using differentiative method to find box largest volume, we will have $\frac{dv}{dx}= 0$

$\frac{d (4x^{3} -92x^{2} +480x)}{dx} = \frac{dv}{dx}$

Differentiating the value

⇒ $\frac{dv}{dx} = 12x^{2} -184x+480$

taking out 12 as common in the equation and subtituting the value.

⇒ $0= 12(x^{2} -\frac{46x}{3} +40)$

solving quadratic equation inside the parenthesis.

⇒ $12(x^{2} -12x-\frac{10x}{x} +40)$ =0

Dividing 12 on both side

⇒ $[x(x-12)-\frac{10}{3} (x-12)]$ = 0

We can again take common as (x-12).

⇒ $x(x-12)[x-\frac{10}{3} ]$ =0

∴ $(x-\frac{10}{3} ) (x-12)= 0$

We have two value for x, which is $12 and \frac{10}{3}$

12 is invalid as, w= $(16-2x)= 16-2\times 12$

∴ 24 inches can not be cut out of 16 inches width.

Hence, the cut out size from cardboard is $\frac{10}{3}\ inches$

Now, subtituting the value of x to find volume of the box.

Volume(v)= $(30-2x)\times (16-2x)\times x$

⇒ Volume(v)= $(30-2\times \frac{10}{3} )\times (16-2\times \frac{10}{3})\times \frac{10}{3}$

⇒ Volume(v)= $(30-\frac{20}{3} ) (16-\frac{20}{3}) (\frac{10}{3} )$

∴ Volume(v)= 725.93 inches³

6 0

3 years ago