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yan [13]
2 years ago
10

ms. cox and her three friends are dining at a restaurant. the food they order cost %80. the bill includes an additional 15% serv

ice charge. how much does each person pay if they share the bill equally
Mathematics
1 answer:
zlopas [31]2 years ago
5 0

did i hear help i will help you soooooooo oooooooooo

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The equation y = kx represents a proportional relationship between x and y, where k is the constant of proportionality. For a mo
Natasha_Volkova [10]

Answer:

Step-by-step explanation:

for this equation

  d=st

we can represent s as d/t

we can represent t as d/s

and in real world we have a lot of this type of equations

some of them are,

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3 years ago
Let​ T: set of real numbers R Superscript nℝnright arrow→set of real numbers R Superscript mℝm be a linear​ transformation, and
Klio2033 [76]

Answer:

\{T(v_1), T(v_2), T(v_3)\} is linearly dependent set.

Step-by-step explanation:

Given:  \{v_1,v_2,v_3\} is a linearly dependent set in set of real numbers R

To show: the set \{T(v_1), T(v_2), T(v_3)\} is linearly dependent.

Solution:

If \{v_1,v_2,v_3,...,v_n\} is a set of linearly dependent vectors then there exists atleast one k_i:i=1,2,3,...,n such that k_1v_1+k_2v_2+k_3v_3+...+k_nv_n=0

Consider k_1T(v_1)+k_2T(v_2)+k_3T(v_3)=0

A linear transformation T: U→V satisfies the following properties:

1. T(u_1+u_2)=T(u_1)+T(u_2)

2. T(au)=aT(u)

Here, u,u_1,u_2∈ U

As T is a linear transformation,

k_1T(v_1)+k_2T(v_2)+k_3T(v_3)=0\\T(k_1v_1)+T(k_2v_2)+T(k_3v_3)=0\\T(k_1v_1+k_2v_2+k_3v_3)=0\\

As \{v_1,v_2,v_3\} is a linearly dependent set,

k_1v_1+k_2v_2+k_3v_3=0 for some k_i\neq 0:i=1,2,3

So, for some k_i\neq 0:i=1,2,3

k_1T(v_1)+k_2T(v_2)+k_3T(v_3)=0

Therefore, set \{T(v_1), T(v_2), T(v_3)\} is linearly dependent.

6 0
4 years ago
The CD Warehouse is having a clearance sale. A CD player that originally sells for $60 is now priced at $36. What is the percent
WINSTONCH [101]

Answer: 24

Step-by-step explanation:

i did the problem and got it right

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Total donations at the firemans ball failed to reach $940
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I’m not quite sure how to find the median..
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Step-by-step explanation:

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