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loris [4]
3 years ago
15

Which table represents a linear function?

Mathematics
1 answer:
kykrilka [37]3 years ago
7 0

Answer:

Option 3 (C)

Step-by-step explanation:

It is the only one that changes the same amount every time ( times 2 )

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Solve the system of linear equations 2x + 3y =16.9<br> 5x =y + 7.4
andrew11 [14]

Answer:

y = 23.4

Step-by-step explanation:

yeahhhh

5 0
3 years ago
6. Solve the problem.
Assoli18 [71]

Answer: 32

Step-by-step explanation:

we have in this case  right triangle with a leg equal to 24 and a hypotenuse equal to 40

c^2=a^2+b^2

40^2=24^2+b^2

b^2=1600-576=1024

b=sqrt1024=32

4 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
Given the function rule f(x) = x^2 - 5x + 1, what is the output of f(-3)?
mojhsa [17]
 f(x) = x²<span> - 5x + 1

f(-3) = (-3)</span>² - 5(-3) + 1
<span>
f(-3) = 25


</span>--------------------------------------------------<span>
Answer: f(-3) = 25 (Answer C)
</span>--------------------------------------------------
4 0
4 years ago
Find the sum and product of the roots.
xz_007 [3.2K]

Answer:

sum= -3 product=-4

Step-by-step explanation:

2x^2-2x+8x-8

2x(x-1)+8(x-1)

(2x+8)(x-1)

x=-4 x=1

8 0
4 years ago
Read 2 more answers
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