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3 years ago

Joe ate twice as many pieces of Halloween candy as Ed. Together they ate 48 pieces.

Mathematics

1 answer:

gulaghasi [49]3 years ago

8 0

Answer: 16

Step-by-step explanation:

Let the number of candy Ed took be denoted as "x"

Since Joe ate twice the amount of candy Ed ate, then the number of candies joe ate will be denoted as "2x"

Since all together they ate a total of 48 candies, then the sum of the candies eaten by Joe and Ed is what gives a total of 48

This means:

2x + x = 48

3x = 48

X = 16.

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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.

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Answer:

Step-by-step explanation:

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Hope this helped ·ω·

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Step-by-step explanation:

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