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I am Lyosha [343]
3 years ago
11

-7/9 greater or less than -5/8 help me quick!!

Mathematics
1 answer:
ch4aika [34]3 years ago
4 0
-7/9 is less than -5/8! (hopefully this helps ^^ )
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Express the solution of, x^3 + 2x^2 < x + 2, using interval notation.
Rudiy27

x³ + 2x² - x - 2 < 0

x³ - x + 2x² - 2 < 0

x(x² - 1) - 2(x² - 1) < 0

(x - 2)(x² - 1) < 0

(x - 2)(x - 1)(x + 1) < 0

by chacking we get the solution

x ∈ (-∞,-1) ∪ (-1,1)

6 0
3 years ago
Find X and Y in each figure
svlad2 [7]
Sooooooooooooooo X= 28 and Y= 47
7 0
3 years ago
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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.

6 0
3 years ago
Explain how you can find the area of a polygon
anyanavicka [17]

Answer:

put a ruler in it and there it is

7 0
3 years ago
I need help quick please
qwelly [4]
A=90
B=53
The reason I am saying this is because you can see the 90 degree angle in A
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