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Nezavi [6.7K]
3 years ago
15

Let T: R^3 --> R^3 be a linear transformation. Let {v1, v2, v3} be a set of linearly dependent vectors in R^3. Show that the

set {T(v1), T(v2), T(v3)} is also linearly dependent.

Mathematics
1 answer:
weqwewe [10]3 years ago
4 0

Answer:

Check the explanation

Step-by-step explanation:

Kindly check the attached image below to see the step by step explanation to the question above.

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I NEED HELPPPP ASAPPP PLEASE HELP MEEEE I WILL GIVE BRAINLY AND 20 POINTS!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!
Morgarella [4.7K]

Answer:

c

Step-by-step explanation:

this is 9n>27. THIS is truuuuuuuuuuuueeeeeee

4 0
3 years ago
What value of c makes x2 − 12x + c a perfect square trinomial? −36 −24 24 36
rewona [7]
Take half of the middle coefficient.
Then square it.

-12/2 = -6

(-6)^2 = 36

c = 36
8 0
4 years ago
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Multiply the expression <br> 2/3 (6x + 15y)<br> and simplify completely.
12345 [234]

Answer: 4x+10y

you can't simplify 4x+10y

6 0
3 years ago
Which definite integral approximation formula is this: the integral from a to b of f(x)dx ≈ (b-a)/n * [<img src="https://tex.z-d
Stella [2.4K]

The answer is most likely A.

The integration interval [<em>a</em>, <em>b</em>] is split up into <em>n</em> subintervals of equal length (so each subinterval has width (<em>b</em> - <em>a</em>)/<em>n</em>, same as the coefficient of the sum of <em>y</em> terms) and approximated by the area of <em>n</em> rectangles with base (<em>b</em> - <em>a</em>)/<em>n</em> and height <em>y</em>.

<em>n</em> subintervals require <em>n</em> + 1 points, with

<em>x</em>₀ = <em>a</em>

<em>x</em>₁ = <em>a</em> + (<em>b</em> - <em>a</em>)/<em>n</em>

<em>x</em>₂ = <em>a</em> + 2(<em>b</em> - <em>a</em>)/<em>n</em>

and so on up to the last point <em>x</em> = <em>b</em>. The right endpoints are <em>x</em>₁, <em>x</em>₂, … etc. and the height of each rectangle are the corresponding <em>y </em>'s at these endpoints. Then you get the formula as given in the photo.

• "Average rate of change" isn't really relevant here. The AROC of a function <em>G(x)</em> continuous* over an interval [<em>a</em>, <em>b</em>] is equal to the slope of the secant line through <em>x</em> = <em>a</em> and <em>x</em> = <em>b</em>, i.e. the value of the difference quotient

(<em>G(b)</em> - <em>G(a)</em> ) / (<em>b</em> - <em>a</em>)

If <em>G(x)</em> happens to be the antiderivative of a function <em>g(x)</em>, then this is the same as the average value of <em>g(x)</em> on the same interval,

g_{\rm ave}=\dfrac{G(b)-G(a)}{b-a}=\dfrac1{b-a}\displaystyle\int_a^b g(x)\,\mathrm dx

(* I'm actually not totally sure that continuity is necessary for the AROC to exist; I've asked this question before without getting a particularly satisfying answer.)

• "Trapezoidal rule" doesn't apply here. Split up [<em>a</em>, <em>b</em>] into <em>n</em> subintervals of equal width (<em>b</em> - <em>a</em>)/<em>n</em>. Over the first subinterval, the area of a trapezoid with "bases" <em>y</em>₀ and <em>y</em>₁ and "height" (<em>b</em> - <em>a</em>)/<em>n</em> is

(<em>y</em>₀ + <em>y</em>₁) (<em>b</em> - <em>a</em>)/<em>n</em>

but <em>y</em>₀ is clearly missing in the sum, and also the next term in the sum would be

(<em>y</em>₁ + <em>y</em>₂) (<em>b</em> - <em>a</em>)/<em>n</em>

the sum of these two areas would reduce to

(<em>b</em> - <em>a</em>)/<em>n</em> = (<em>y</em>₀ + <u>2</u> <em>y</em>₁ + <em>y</em>₂)

which would mean all the terms in-between would need to be doubled as well to get

\displaystyle\int_a^b f(x)\,\mathrm dx\approx\frac{b-a}n\left(y_0+2y_1+2y_2+\cdots+2y_{n-1}+y_n\right)

7 0
3 years ago
Solve for X. show your work. (x+2)(x+8)=0
SOVA2 [1]
The answer to this question is X 1 = -8 X 2= -2
7 0
3 years ago
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