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bonufazy [111]
2 years ago
9

a teacher has three packages of crayons. one package contains 52 crayons, another contains 12 crayons, and the third package con

tains 96 crayons. how many crayons will each student receive if the teacher divides them equally among her class of 22 students?
Mathematics
1 answer:
kotegsom [21]2 years ago
5 0

Answer: 7 crayons

Step-by-step explanation:

To solve we first must find the total number of crayons available by adding up how many crayons are in each container.

         56+12+96 = 160 crayons total

Next we take this number and divide it by 22 (the number of students in the class) to determine how many crayon each student gets

         160 crayons/22 students = 7.272727 crayons per student

However, it's important to note that this number is not a whole number. We must then round down to the nearest whole number because students aren't given fractions of crayons.

         7.272727 ==> 7 crayons per student

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The local planetarium is selling tickets for the new laser light show at the price of 2 tickets for $5.00 which graph shows the
Minchanka [31]

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Step-by-step explanation:

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-4(x+9)+-16 PLEASE ANSWER FAST
PIT_PIT [208]

Answer:

-4x-52

Step-by-step explanation:

-4(x+9)+(-16)

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3 0
3 years ago
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erastovalidia [21]

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6 0
3 years ago
Based on the information marked in the diagram, ABC and DEF must be congruent.
drek231 [11]

Answer:

Option A is correct

True, ΔABC and ΔDEF must be congruent.

Explanation:

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AB = DE              [Leg]                [Given in the figure]

\angle BAC = \angle EDF = 90^{\circ}     [Given]

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Then, by the LA theorem or Postulates;

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3 0
4 years ago
Read 2 more answers
A cylinder is inscribed in a right circular cone of height 4.5 and radius (at the base) equal to 5.5 . What are the dimensions o
cluponka [151]

Answer:

r = 3.667

h = 1.5

Step-by-step explanation:

Given:-

- The base radius of the right circular cone, R = 5.5

- The height of the right circular cone, H = 4.5

Solution:-

- We will first define two variables that identifies the volume of a cylinder as follows:

                                r: The radius of the cylinder

                                h: The height of cylinder

- Now we will write out the volume of the cylinder ( V ) as follows:

                                V = \pi*r^2h

- We see that the volume of the cylinder ( V ) is a function of two variables ( don't know yet ) - ( r,h ). This is called a multi-variable function. However, some multi-variable functions can be reduced to explicit function of single variable.

- To convert a multi-variable function into a single variable function we need a relationship between the two variables ( r and h ).

- Inscribing, a cylinder in the right circular cone. We will denote 5 points.

              Point A: The top vertex of the cone

              Point B: The right end of the circular base ( projected triangle )

              Point C: The center of both cylinder and base of cone.

              Point D: The top-right intersection point of cone and cylinder

              Point E: Denote the height of the cylinder on the axis of symmetry of both cylinder and cone.  

- Now, we will look at a large triangle ( ABC ) and smaller triangle ( ADE ). We see that these two triangles are "similar". Therefore, we can apply the properties of similar triangles as follows:

                              \frac{AC}{AE} = \frac{BC}{DE}  \\\\\frac{H}{H-h} = \frac{R}{r}

- Now we can choose either variable variable to be expressed in terms of the other one. We will express the height of cylinder ( h ) in term of radius of cylinder ( r ) as follows:

                             H- h = r\frac{H}{R} \\\\h = \frac{H}{R}*(R-r)

- We will use the above derived relationship and substitute into the formula given above:

                            V = \pi r^2 [ \frac{H}{R}*(R - r )]\\\\V = \frac{\pi H}{R}.r^2.(R-r)

- Now our function of volume ( V ) is a single variable function. To maximize the volume of the cylinder we need to determine the critical points of the function as follows:

                            \frac{dV}{dr} =  \frac{\pi H}{R}*(2rR-2r^2 - r^2 )\\\\\frac{dV}{dr} =  \frac{\pi H}{R}*(2rR-3r^2 ) = 0\\\\(2rR-3r^2 ) = 0\\\\2R -3r = 0\\\\r = \frac{2}{3}*R

- We found the limiting value of the function. The cylinder volume maximizes when the radius ( r ) is two-thirds of the radius of the right circular cone.

- We can use the relationship between the ( r ) and ( h ) to determine the limiting value of height of cylinder as follows:

                          h = \frac{H}{R} * ( R - \frac{2}{3}R)\\\\h = \frac{H}{3}

- The dimension of the inscribed cylinder with maximum volume are as follows:

                         r = \frac{2}{3}*5.5 = 3.667\\\\h = \frac{4.5}{3} = 1.5

Note: When we solved for the critical value of radius ( r ). We actually had two values: r = 0 , r = 2R/3. Where, r = 0 minimizes the volume and r = 2R/3 maximizes. Since the function is straightforward, we will not test for the nature of critical point ( second derivative test ).

7 0
4 years ago
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