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sattari [20]
3 years ago
11

Visualise 5.96 on the number line up to 43 decimmal paces

Mathematics
1 answer:
Firlakuza [10]3 years ago
7 0

Answer:

It's legit just 5.9600000000000000000000000000000000000000000.

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What is the product of 5,330,000 and 0.22 x 10^-10?
Marizza181 [45]

Answer:0.00011726

Step-by-step explanation:

8 0
3 years ago
Micheal buys two bags or chips and three boxes of pretzels for $5.13 . He then buys another bag of chips and 2 boxes of pretzels
trapecia [35]

Answer:

So a bag of chip costs $2.39

Step-by-step explanation:

Let x = cost of a bag of chip

Let y = cost of a box of pretzel

Micheal buys two bags of chips and three boxes of pretzels for $5.13

This means

2x + 3y = 5.13 - - - - - - - - - -1

He then buys another bag of chips and 2 boxes of pretzels for $3.09

This means

x + 2y = 3.09 - - - - - - - - - - - -2

Solving equation 1 and equation 2 simultaneously and using the elimination method,

Multiply equation 1 by 1 and equation 2 by 2

2x + 3y = 5.13

2x + 6y = 6.18

Subtracting,

-3y = -1.05

y = -1.05/-3= $0.35

Put y= 0.35 in equation, x + 2y = 3.09

x + 2(0.35)= 3.09

x + 0.7= 3.09

x = 3.09-0.7 = $ 2.39

So a bag of chip costs $2.39

5 0
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
3 years ago
How many solutions are there to this system of equations?
Zinaida [17]

Answer:

one

Step-by-step explanation:

5 0
3 years ago
7 is what percent of 8
11Alexandr11 [23.1K]

Answer:

87.5%

Step-by-step explanation:

.

..

...

....

......

......

....

....

7 0
3 years ago
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