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3 years ago

Plz someone help me, this is due in 45mins

Mathematics

2 answers:

Sergeu [11.5K]3 years ago

6 0

Answer:

Stem Leaf

2 8

3 4 4

5 0 4

So we need to put all the tens values in the "stem" column and all the ones values in the "leaf" column, sorted from least to greatest.

We have three tens values: 2, 3, and 5. So those are our three stems, in order.

We have to put the second number, the ones value, in the leaf column, least to greatest. So with two values of 34, we need two 4 leaves.

ipn [44]3 years ago

5 0

Answer:

Step-by-step explanation:

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23/40 as a percentage

Leviafan [203]

23/40* 100%= 57.5%
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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.

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3 years ago

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Answer:

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Step-by-step explanation:

77 x .24% (0.0024) = .1848

Round to the nearest hundredth...

.18!

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