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2 years ago

F(x) = (x - 2)^2 + 2 find the range of the function

Mathematics

1 answer:

a_sh-v [17]2 years ago

3 0

Vertex = (2, 2)
Range = {f(x): f(x) >= 2}

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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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3 years ago

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9514 1404 393

Answer:

m = (5x -6)/(43y -4)

Step-by-step explanation:

Subtract 4m+6 to separate m-terms from other terms.

5x +4m -(4m +6) = 43my +6 -(4m +6)

5x -6 = 43my -4m

5x -6 = m(43y -4) . . . . factor out m

(5x -6)/(43y -4) = m . . . . divide by the coefficient of m

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Answer:

Step-by-step explanation:

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