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Alika [10]
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?
Mathematics
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

#SPJ1

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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}

<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>D</u><u>E</u><u>:</u><u> </u></h3>

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>

<em>S</em><em>U</em><em>B</em><em> </em><em>I</em><em>N</em><em> </em><em>C</em><em>O</em><em>O</em><em>R</em><em>D</em><em>I</em><em>N</em><em>A</em><em>T</em><em>E</em><em>S</em><em> </em><em>O</em><em>F</em><em> </em><em>B</em><em> </em><em>A</em><em>N</em><em>D</em><em> </em><em>C</em>

<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>

6 0
3 years ago
What is the value of F(x)
pochemuha

Answer:2

Step-by-step explanation:

8 0
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.

The following first-order linear non-homogeneous differential equation is found:

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

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

\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

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

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

The initial concentration in the tank is:

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

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

Now, the integration constant is:

0.167 = 30 + C

C = -29.833

The solution of the differential equation is:

c(t) = 30 - 29.833\cdot e^{-\frac{t}{10} }

Now, the quantity of salt at time t is:

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.

7 0
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.


1 gallon= 16 cups


Multiply total gallons by cups per 1 gallon.

= 6 3/4 cups * 16 cups/gallon
convert mixed number to improper fraction

= 27/4 * 16
multiply numerators

= (27*16)/4
multiply in parentheses

= 432/4

= 108 cups


ANSWER: There are 108 cups

Hope this helps! :)
3 0
3 years ago
Solve for c <br><br> a= b+c/d
castortr0y [4]
B+c/d=a  subtract b from both sides

c/d=a-b multiply both sides by d

c=d(a-b) or if you prefer

c=ad-bd

Note: if you meant a=(b+c)/d, multiply both sides by d

b+c=ad  subtract b from both sides

c=ad-b
4 0
3 years ago
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