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barxatty [35]
3 years ago
13

Can somebody help me with the Math's questions on Pythagoras Theorem? Thanks

Mathematics
1 answer:
Svetlanka [38]3 years ago
8 0

Answer:

Part a) AB=17\ m

Part b) AC=20.8\ m

Step-by-step explanation:

see the attached figure with letters to better understand the problem

Part a) Find the length AB

Applying the Pythagoras Theorem in the right triangle ABH

AB^2=AH^2+HB^2

where

AB is the hypotenuse (the greater side)

AH and HB are the legs of the right triangle

we have

AH=FG=8\ m

HB=GB-ED=26-11=15\ m

substitute the values

AB^2=8^2+15^2

AB^2=289

AB=17\ m

Part a) Find the length AC

Applying the Pythagoras Theorem in the right triangle ABC

AC^2=AB^2+BC^2

where

AC is the hypotenuse (the greater side)

AB and BC are the legs of the right triangle

we have

BC=12\ m

AB=17\ m

substitute

AC^2=17^2+12^2

AC^2=433

AC=20.8\ m

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How do you do both parts?
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Step-by-step explanation:

f(x) = 1 / (5 + x)

Divide top and bottom by 5.

f(x) = ⅕ / (1 + ⅕x)

f(x) = ⅕ / (1 − (-⅕x))

Write as a geometric series.

f(x) = ∑ₙ₌₀°° ⅕ (-⅕x)ⁿ

f(x) = ∑ₙ₌₀°° ⅕ (-1)ⁿ (⅕)ⁿ xⁿ

f(x) = ∑ₙ₌₀°° (-1)ⁿ (⅕)ⁿ⁺¹ xⁿ

Use ratio test:

lim(n→∞)│aₙ₊₁ / aₙ│< 1

lim(n→∞)│[(-1)ⁿ⁺¹ (⅕)ⁿ⁺² xⁿ⁺¹] / [(-1)ⁿ (⅕)ⁿ⁺¹ xⁿ]│< 1

lim(n→∞)│-1 (⅕) x│< 1

│x/5│< 1

│x│< 5

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If x = -5, ∑ₙ₌₀°° ⅕ (-1)ⁿ (⅕)ⁿ (-5)ⁿ = ∑ₙ₌₀°° ⅕ (1)ⁿ, which diverges.

If x = 5, ∑ₙ₌₀°° ⅕ (-1)ⁿ (⅕)ⁿ (5)ⁿ = ∑ₙ₌₀°° ⅕ (-1)ⁿ, which diverges.

The interval of convergence is therefore (-5, 5).

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Answer:

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