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geniusboy [140]
2 years ago
5

WILL MARK BRAINLIEST TO THE BEST ANSWERS. The interior angles of a pentagon measure x°, (2x)°, (2.5x)° , (x + 10)° , and (2x + 2

0)°. Determine the value of x. PLEASE SHOW ALL SOLUTIONS NESSICARY. THANK YOU.......
Mathematics
2 answers:
Llana [10]2 years ago
8 0

Answer:

  • x = 60°

Step-by-step explanation:

<u>Sum of interior angles of a regular polygon is: </u>

  • 180°(n-2)

<u>Pentagon has 5 sides, so sum of angles is: </u>

  • 180°(5-2) = 180°*3 = 540°

<u>For the given pentagon we have sum of angle measures as below, and solving for x:</u>

  • x + 2x + 2.5x + (x +10) + (2x + 20) = 540°
  • 8.5x + 30° = 540°
  • 8.5x = 510°
  • x = 510°/8.5
  • x = 60°
nalin [4]2 years ago
6 0

Answer:

x=60

Step-by-step explanation:

The sum of interior angles of a regular Pentagon is 540°

x+2x+2.5x+x+10+2x+20=540

8.5x=540-30

8.5x=510

x=510/8.5=60

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Answer:

x+(x+1)=471

Step-by-step explanation:

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A race car travels 100 feet in .5 seconds. At this rate of speed, how many feeet will the race car travel in a minute?
soldier1979 [14.2K]

Answer:

Step-by-step explanation:

\frac{100ft}{0.5s} * 60s

\frac{100 * 60}{0.5}

\frac{6000}{0.5}

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3 years ago
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Consider the function f(x)=xln(x). Let Tn be the nth degree Taylor approximation of f(2) about x=1. Find: T1, T2, T3. find |R3|
Fynjy0 [20]

Answer:

R3 <= 0.083

Step-by-step explanation:

f(x)=xlnx,

The derivatives are as follows:

f'(x)=1+lnx,

f"(x)=1/x,

f"'(x)=-1/x²

f^(4)(x)=2/x³

Simialrly;

f(1) = 0,

f'(1) = 1,

f"(1) = 1,

f"'(1) = -1,

f^(4)(1) = 2

As such;

T1 = f(1) + f'(1)(x-1)

T1 = 0+1(x-1)

T1 = x - 1

T2 = f(1)+f'(1)(x-1)+f"(1)/2(x-1)^2

T2 = 0+1(x-1)+1(x-1)^2

T2 = x-1+(x²-2x+1)/2

T2 = x²/2 - 1/2

T3 = f(1)+f'(1)(x-1)+f"(1)/2(x-1)^2+f"'(1)/6(x-1)^3

T3 = 0+1(x-1)+1/2(x-1)^2-1/6(x-1)^3

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Thus, T1(2) = 2 - 1

T1(2) = 1

T2 (2) = 2²/2 - 1/2

T2 (2) = 3/2

T2 (2) = 1.5

T3(2) = 1/6 (-2^3 + 6 *2^2 - 3 *2 - 2)

T3(2) = 4/3

T3(2) = 1.333

Since;

f(2) = 2 × ln(2)

f(2) = 2×0.693147 =

f(2) = 1.386294

Since;

f(2) >T3; it is significant to posit that T3 is an underestimate of f(2).

Then; we have, R3 <= | f^(4)(c)/(4!)(x-1)^4 |,

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f^(4)(x)=2/x^3, we have, |f^(4)(c)| <= 2

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R3 <= |2/(4!)(2-1)^4|

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that simply means that the point (0, 2)  is on the false area, so that's the area we do NOT shade, so <u>we shade the other side</u>.

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check the picture below.

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