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1 year ago

9

This is a cross-sectional view of candy bar ABC. A candy company wants to create a cylindrical container for candy bar ABC so th

at it is circumscribed about the
candy bar. If AD = 2.5 cm, what is the smallest diameter of wrapper that will fit the candy bar?

1 answer:

Rzqust [24]1 year ago

3 0

TheThe smallest diameter of wrapper that will fit the candy bar is 5cm.

<h3>Smallest diameter</h3>

Given:

AD = DC

AC = AD + DC

Segment AD = 2.5cm

Hence:

AC = AD + AD (AD = DC)

AC = 2AD

Where:

AD=2.5

So,

AC = 2(2.5cm)

AC = 5cm

Therefore the smallest diameter of wrapper that will fit the candy bar is 5cm.

Learn more about smallest diameter here: brainly.com/question/12944899

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How to solve this problem step by step. Make brainliest answer if its done. Please

sladkih [1.3K]

D is halfway between A and B

so the coordinates of D are (2,2)

E is halfway between A and C so the coordinates of E are (-1,1)

now you need to find the gradient/slope of DE and BC using the formula:

$\frac{y2 - y1}{x2 - x1}$

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SUB IN COORDINATES OF D AND E

$\frac{1 - 2}{ - 1 - 2}$

therefore the gradient of DE is 1/3.

<h3><u>G</u><u>r</u><u>a</u><u>d</u><u>i</u><u>e</u><u>n</u><u>t</u><u> </u><u>o</u><u>f</u><u> </u><u>B</u><u>C</u><u>:</u></h3>

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<em> $\frac{ - 2 - 0}{ - 3 - 3}$ </em>

therefore the gradient of BC is -2/-6 which simplifies to 1/3.

<h3>therefore, BC and DE are parallel as they both have a gradient/slope of 1/3 and parallel lines have the same gradient</h3>

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3 years ago

What is the value of F(x)

pochemuha

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3 years ago

Problem 10: A tank initially contains a solution of 10 pounds of salt in 60 gallons of water. Water with 1/2 pound of salt per g

AysviL [449]

Answer:

The quantity of salt at time t is $m_{salt} = (60)\cdot (30 - 29.833\cdot e^{-\frac{t}{10} })$ , where t is measured in minutes.

Step-by-step explanation:

The law of mass conservation for control volume indicates that:

$\dot m_{in} - \dot m_{out} = \left(\frac{dm}{dt} \right)_{CV}$

Where mass flow is the product of salt concentration and water volume flow.

The model of the tank according to the statement is:

$(0.5\,\frac{pd}{gal} )\cdot \left(6\,\frac{gal}{min} \right) - c\cdot \left(6\,\frac{gal}{min} \right) = V\cdot \frac{dc}{dt}$

Where:

$c$ - The salt concentration in the tank, as well at the exit of the tank, measured in $\frac{pd}{gal}$ .

$\frac{dc}{dt}$ - Concentration rate of change in the tank, measured in $\frac{pd}{min}$ .

$V$ - Volume of the tank, measured in gallons.

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$V \cdot \frac{dc}{dt} + 6\cdot c = 3$

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$\frac{dc}{dt} + \frac{1}{10}\cdot c = 3$

This equation is solved as follows:

$e^{\frac{t}{10} }\cdot \left(\frac{dc}{dt} +\frac{1}{10} \cdot c \right) = 3 \cdot e^{\frac{t}{10} }$

$\frac{d}{dt}\left(e^{\frac{t}{10}}\cdot c\right) = 3\cdot e^{\frac{t}{10} }$

$e^{\frac{t}{10} }\cdot c = 3 \cdot \int {e^{\frac{t}{10} }} \, dt$

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$c = 30 + C\cdot e^{-\frac{t}{10} }$

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$c_{o} = \frac{10\,pd}{60\,gal}$

$c_{o} = 0.167\,\frac{pd}{gal}$

Now, the integration constant is:

$0.167 = 30 + C$

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$m_{salt} = V_{tank}\cdot c(t)$

$m_{salt} = (60)\cdot (30 - 29.833\cdot e^{-\frac{t}{10} })$

Where t is measured in minutes.

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3 years ago

A pail holds 6 3 4 gallons of water. How much is this in cups? Write your answer as a whole number or a mixed number in simplest

storchak [24]

I'm assuming the pail holds 6 3/4 (not 634) gallons of water.

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Hope this helps! :)

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3 years ago

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castortr0y [4]

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3 years ago