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Bingel [31]
3 years ago
5

Below is a proof showing that the sum of a rational number and an irrational number is an irrational number.

Mathematics
2 answers:
ankoles [38]3 years ago
4 0

Answer:

Step-by-step explanation:

To prove: The sum of a rational number and an irrational number is an irrational number.

Proof: Assume that a + b = x and that x is rational.

Then b = x – a = x + (–a).

Now, x + (–a) is rational because addition of  two rational numbers is rational (Additivity property).

However, it was stated that b is an irrational number. This is a contradiction.

Therefore, the assumption that x is rational in the equation a + b = x must be incorrect, and x should be an irrational number.

Hence,  the sum of a rational number and an irrational number is irrational.

Goryan [66]3 years ago
4 0

Answer:

I believe the answer is B

Step-by-step explanation:

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How do you write 5.9 in expanded form
tresset_1 [31]
<span><u><em>Answer:</em></u>
(5 * 1) + (9 * 0.1).<span>

<u><em>Explanation:</em></u>
To write a number in expanded form you have each digit multiplied by its numbers place and add each digit together. 5 is in the ones place so it becomes (5 * 1).
9 is in the tenths place so it becomes (9 * 0.1).
add them together to get (5 * 1) + (9 * 0.1)</span></span>
3 0
3 years ago
Read 2 more answers
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
Identify the range of the function shown in the graph.
Masteriza [31]
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6 0
2 years ago
What is the length x of the right triangle, rounded to the nearest<br> tenth?
Zarrin [17]

Answer:

109.5; B

Step-by-step explanation:

From your identity,

CosA = adjacent/ hypothenus

A represent an arbitrary angle between the sides in question.

In the question above, A=64

Hypothenus is the longest side and adjacent is the side just below the angle .

In the above case,

Hypothenus= X

adjacent =48

This means;

Cos64 = 48 /X

X = 48 / cos64; [ from cross multiplication and diving through by cos64]

X = 48 /0.4383 [ cos64 in radian = 0.4383]

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= 109.5 to the nearest tenth.

Note( do your calculation of angle in radian or else, you won't get the answer)

6 0
3 years ago
The diagonals of a parallelogram are 24 meters and 40 meters and
Naddika [18.5K]

Answer:

  28 m

Step-by-step explanation:

The longer side can be found using the Law of Cosines. The semi-diagonals are of length 12 and 20 m, and the angle between them facing the long side is 180° -60° = 120°.

  c^2 = a^2 +b^2 -2ab·cos(C)

  c^2 = 12^2 +20^2 -2·12·20·cos(120°) = 144 +400 +240 = 784

  c = √784 = 28

The longer side is 28 meters.

8 0
3 years ago
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