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ElenaW [278]
1 year ago
7

Kieran subscribes to a text messaging service on his cell phone. His monthly bill for the service is given by the equation b = 0

.25t, where b is the bill amount and t is the number of texts. The constant of proportionality in terms of cost per text is ?
Mathematics
2 answers:
liraira [26]1 year ago
6 0
B=0.25t

The cost of proportionality is 0.25
ipn [44]1 year ago
3 0
The question doesn't make a lot of sense, but I think the answer is $0.25, because In the equation b=0.25t, each text costs 25 cents
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Eight graders is the answer
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Last week, Lindsay earned $10 per hour plus a $60 bonus for good job performance. She spends StartFraction 1 Over 15 EndFraction
V125BC [204]

Answer:

StartFraction 1 Over 15 EndFraction left-parenthesis 10 h plus 60 right-parenthesis equals StartFraction 1 Over 10 Endfraction left-parenthesis 10 h right-parenthesis

Step-by-step explanation:

The desired equation equates 1/15 of the amount Lindsay earned with 1/10 the amount without the bonus:

StartFraction 1 Over 15 EndFraction left-parenthesis 10 h plus 60 right-parenthesis equals StartFraction 1 Over 10 Endfraction left-parenthesis 10 h right-parenthesis

_____

<em>Comment on the form of the answer</em>

If you don't like answers in this form, please do not post questions in this form. Ordinary math symbols are much appreciated.

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2 years ago
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Given matrix A below, and that A = B, find the value of the elements in B. A = 9 −2 3 2 17 0 3 22 8 b11 = b12 = b13 = b21 = b22
GuDViN [60]

Answer:

b_{11}=9,b_{12}=-2,b_{13}=3,b_{21}=2,b_{22}=17,b_{23}=0,b_{31}=3,b_{32}=22,b_{33}=8

Step-by-step explanation:

Consider the given matrix

A=\left[\begin{array}{ccc}9&-2&3\\2&17&0\\3&22&8\end{array}\right]

Let matrix B is

B=\left[\begin{array}{ccc}b_{11}&b_{12}&b_{13}\\b_{21}&b_{22}&b_{23}\\b_{31}&b_{32}&b_{33}\end{array}\right]

It is given that

A=B

\left[\begin{array}{ccc}9&-2&3\\2&17&0\\3&22&8\end{array}\right]=\left[\begin{array}{ccc}b_{11}&b_{12}&b_{13}\\b_{21}&b_{22}&b_{23}\\b_{31}&b_{32}&b_{33}\end{array}\right]

On comparing corresponding elements of both matrices, we get

b_{11}=9,b_{12}=-2,b_{13}=3

b_{21}=2,b_{22}=17,b_{23}=0

b_{31}=3,b_{32}=22,b_{33}=8

Therefore, the required values are b_{11}=9,b_{12}=-2,b_{13}=3,b_{21}=2,b_{22}=17,b_{23}=0,b_{31}=3,b_{32}=22,b_{33}=8.

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17

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