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vladimir2022 [97]
3 years ago
15

Instructions:

Mathematics
1 answer:
satela [25.4K]3 years ago
5 0

Answer:

The volume of a container is the amount of space in it.

The number of cubes on the edges are: 15, 12 and 13

The number of fish cubes in the box is: 2340

From the complete question (see attachment), we have the following parameters

Fish cubes

Shipping box

(a) The number of cubes on the edges

To do this, we simply the dimensions of the boxes, by the length of the fish cube

So, we have:

Hence, the number of cubes on the edges are: 15, 12 and 13

(b) The total number of cubes in the shipping box

To do this, we multiply the results of the computation in (a)

So, we have:

Hence, the number of fish cubes in the box is: 2340

Step-by-step explanation:

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A point K is on the perpendicular bisector of a segment with endpoints at H and J. What must be true about point K? It is equidi
kykrilka [37]

Answer:

K is equidistant from H and J.

Step-by-step explanation:

Given that the point K which is on the perpendicular bisector of the line segment having endpoints at H and J.

The given situation can be represented as the diagram as attached in the answer area.

Referring to the \triangle HOK, \triangle JOK:

\angle HOK = \angle JOK=90^\circ (As it is the perpendicular bisector)

OH = OJ (As it is the perpendicular bisector)

Also, the side OK is the common side.

Therefore by S-A-S congruence, \triangle HKO\cong \triangle JKO

As per the properties of congruent triangles:

Side HK = Side JK

HK and JK are nothing but the distance of the point K from the end points H and J which are proved to be equal to each other.

Therefore, we can conclude that:

K is equidistant from H and J.

5 0
3 years ago
For the point P(6,8) and Q(11,11), find the distance d(P,Q) and the coordinates of the
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Answer:

The distance of PQ =  \sqrt{34}

Step-by-step explanation:

<u><em>Explanation:-</em></u>

Given that the points are P(6,8) and Q(11,11)

Distance formula

              PQ = \sqrt{x_{2}-x_{1} )^{2} +(y_{2} - y_{1})^{2}   }

                    =  \sqrt{11-6)^{2} +(11-8)^{2}   }

                   = \sqrt{(5)^{2}+(3)^{2}}   }

                  =  \sqrt{25+9}

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4 0
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What is the radius of a circle whose equation is x2 + y2 – 10x + 6y + 18 = 0? how many units?
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(x-a)^2+(y-b)^2=r^2
x^2 + y^2 - 10x + 6y + 18 = 0\\\\ x^2-10x+y^2+6y+18=0 \\\\ (x-5)^2-25+(y+3)^2-9+18=0\\\\ (x-5)^2+(y+3)^2=16\\\\ r^2=16
r = \sqrt{16} \\\\ r=4 
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4 years ago
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Find the area of the figure.
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Answer:

Step-by-step explanation:

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Step-by-step explanation:

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