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Iteru [2.4K]
2 years ago
10

PLEASE SHOW WORK

Mathematics
1 answer:
Dmitrij [34]2 years ago
6 0

Answer:

Below in bold

Step.-by-step explanation:

It consists of 5 isosceles triangles of equal sides 4 cm and vertex angles 360 / 5

= 72 degrees.

Area of 1 triangle = 1/2 * 4^2 *sin 72

Area of the whole pentagon = 5 * 1/2 *4^2 * sin 72

= 38.04 cm^2.

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Peter's glasses each hold 8 fluid ounces how many glasses of juice can Peter pour from a bottle that holds 3 quarts
Artyom0805 [142]

Answer:

12 glasses can be hold

Step-by-step explanation:

Find the Amount of quarts equal to fluid ounces

After you find it divide the number to 8

Brainliest is needed plz

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3 years ago
Right Triangle and Trigonometry , Please help!!! Will give Brainliest!!!
Harman [31]
That would be true because the adjacent side to B is 4 and the hypotenuse is 5 and the formula for cos b is adjacent/hypotenuse

Hope this helps!
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3 years ago
What is 0.512 as a percentage
Sonja [21]

Answer:

51.2%

Step-by-step explanation:

7 0
3 years ago
Read 2 more answers
A city has a population of 230,000 people. Suppose that each year the population grows by 9%. What will the population be after
zloy xaker [14]

<u>Answer:</u>

<h2><u>385,733</u></h2>

Step-by-step explanation:

If it grows by 9% each year, you can simply do the current population times 109%. Because its an additional 9%.

230,000*109% = 250,700

After the first year, there is a population of 250,700 people.

250,700*109% = 273,263

After the second year, there is a population of 273,263 people.

273,263*109% = 297,856.67

After the third year, there is a population of 297,856.67 people.

297,856.67*109% = 324,663.77

After the fourth year, there is a population of 324,663.77 people.

324,663.77*109% = 353,883.509

After the fifth year, there is a population of 353,883.509 people.

353,883.509*109% = 385,733.025

After the sixth year, there is a population of 385,733.025 people

Round it up to 385,733 people.

7 0
3 years ago
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

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

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 |,

Since;

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

Finally;

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

R3 <= | 2 / 24× 1 |

R3 <= 1/12

R3 <= 0.083

5 0
3 years ago
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