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Contact [7]
3 years ago
7

Suppose that L1 : V → W and L2 : W → Z arelinear transformations and E, F, and G are orderedbases for V, W, and Z, respectively.

Show that, if Arepresents L1 relative to E and F and B representsL2 relative to F and G, then the matrix C = BA representsL2 ◦ L1: V → Z relative to E and G. Hint:Show that BA[v]E = [(L2 ◦ L1)(v)]G for all v ∈ V.

Mathematics
1 answer:
daser333 [38]3 years ago
3 0

Answer:

a) v ∈ ker(<em>L</em>) if only if [V]_{E} ∈ <em>N</em>(<em>A</em>)

b) w ∈ <em>L</em>(<em>v</em>) if and only if [W]_{F} is in the column space of <em>A</em>

<em />

<em>See attached</em>

Step-by-step explanation:

See attached the proof Considering the vector spaces <em>V</em> and <em>W</em> with other bases <em>E</em> and <em>F</em> respectively.

Let <em>L</em> be the Linear transformation form <em>V</em> and <em>W</em> and A is the matrix representing <em>L</em> relative to<em> E</em> and <em>F</em>

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Which of the given expressions results in 0 when evaluated at x = 5?
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Answer:

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Step-by-step explanation:

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3 years ago
A loan of $36,000 is made at 6.75% interest, compounded annually. After how many years will the amount due reach $57,000 or more
irakobra [83]

Answer:

After 7.04 years the amount will reach $57,000 or more

Step-by-step explanation:

The rule of the compound interest is A=P(1+\frac{r}{n})^{nt} , where

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  • P is the initial value
  • r is the rate in decimal
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∵ A loan of $36,000 is made at 6.75% interest, compounded annually

∴ P = 36,000

∴ r = 6.75% = 6.75 ÷ 100 = 0.0675

∴ n = 1 ⇒ compounded annually

∵ The amount after t years will reach $57,000 or more

∴ A = 57,000

→ To find t substitute these values in the rule above

∵ 57,000 = 36,000 (1+\frac{0.0675}{1})^{(1)(t)}

∴ 57,000 = 36,000 (1.0675)^{t}

→ Divide both sides by 36,000

∵ \frac{19}{12} =  (1.0675)^{t}

→ Insert ㏒ in both sides

∴ ㏒( \frac{19}{12} ) = ㏒ (1.0675)^{t}

→ Remember ㏒a^{n} = n ㏒(a)

∵ ㏒( \frac{19}{12} ) = <em>t</em> ㏒(1.0675)

→ Divide both sides by ㏒(1.0675)

∴ 7.035151337 = <em>t</em>

<em>∴ </em><em>t </em>≅ 7.04

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Match each complex number with its equivalent expression i^157 i^315 i^102 i^76
ryzh [129]

Answers:

i^{157} = i\\\\i^{315} = -i\\\\i^{102} = -1\\\\i^{76} = 1\\\\

=====================================================

Explanation:

By definition, i = \sqrt{-1}

Squaring both sides gets us i^2 = -1

Then multiply both sides by i to get i^3 = -i

Repeat the last step and you should get i^4 = -i^2 = -(-1) = 1

---------------

Notice we have this pattern going on:

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Once we reach i^4, we start the process over again.

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This means we'll divide the exponent over 4 and look at the remainder. We ignore the quotient completely.

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i^{157} = i^1 = i

---------------

Similarly,

315/4 = 78 remainder 3

So i^{315} = i^3 = -i

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102/4 = 25 remainder 2

i^{102} = i^2 = -1

----------------

76/4 = 19 remainder 0

i^{76} = i^0 = 1

8 0
2 years ago
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