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SashulF [63]
2 years ago
7

613x47 What is the answer

Mathematics
2 answers:
Monica [59]2 years ago
7 0

Answer:

28,811

Step-by-step explanation:

Eddi Din [679]2 years ago
5 0
28,811 you can easily calculate or solve the problem yourself...
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The price for each notebook is p. The number of the
ArbitrLikvidat [17]

Answer:  price of total notebooks = p(a+b+c) or pa+pb+pc= price of total notebooks

Step-by-step explanation:

you have to add all the number of notebooks or a b and c and then multiply it by the price of each notebook. there are multiple ways to do this but those are the ways I could think of.

 if a=2, b=3 and c=6 then you have to add 2+3+6 to equal 11 then multiply 11 by the price of each notebook, say p=$4.50  your equation would be 4.50*11 which is $49.50, and that would be the total of all the notebooks. these numbers metaphorical of course, but hope that puts it into perspective.

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3 years ago
Identify the distance between points (3,7,1) and (9,6,7), and identify the midpoint of the segment for which these are the endpo
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Step-by-step explanation:

range = (3,7,1)  -  (9,6,7) \\  = ( - 6,1, - 6) \\ distance =  \sqrt{ {x}^{2} +  {y}^{2} +  {z}^{2}   }  \\  =  \sqrt{ {( - 6)}^{2} +  {1}^{2} +  {( - 6)}^{2}   }  \\  =  \sqrt{73}  \\  = 8.5 \: units

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2 years ago
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What term best describes the set of all points in a plane for which the sum of the distances to two fixed points equals a certai
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You are referring to an ELLIPSE.

In mathematics, an ellipse<span> is a curve in a plane surrounding two focal points such that the sum of the distances to the two focal points is constant for every point on the curve. As such, it is a generalization of a circle, which is a special type of an </span>ellipse<span>having both focal points at the same location.
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Gre4nikov [31]
73 I believe I’m not sure
7 0
2 years ago
I need help with this problem
cupoosta [38]
\bf \begin{cases}&#10;(n^4)^p=n^{12}\to &n^{4\cdot p}=n^{12}\to 4p=12&#10;\\&#10;&\qquad \uparrow \\&#10;&\textit{same bases, thus}\\&#10;&\qquad \downarrow &#10;\\&#10;n^3\cdot n^q=n^6\to &n^{3+q}=n^6\to 3+q=6&#10;\end{cases}&#10;\\\\\\&#10;thus\implies &#10;\begin{cases}&#10;4p=12\implies p=\frac{12}{4}&#10;\\\\&#10;3+q=6\implies q=6-3&#10;\end{cases}&#10;\\\\&#10;&#10;\\\\&#10;then\implies p\cdot q=\boxed{?}
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2 years ago
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