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kati45 [8]
2 years ago
7

What is the area of the real object that the scale drawing models?

Mathematics
1 answer:
timurjin [86]2 years ago
3 0

Answer:

c: 120

Step-by-step explanation:

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Mrs. Cazares went shopping for candy for her students. She bought 6 bags of candy. She had a coupon for $6 off her purchase. If
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Answer:

7

Step-by-step explanation:

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Imeldas jigsaw puzzle has 500 pieces. She estimates that she put togethe approximately 95 of the pieces.
tekilochka [14]

Answer:

19% done

81% left

Step-by-step explanation:

95/500=19

100-19=81

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A rectangle has a length of 6.5m and a width of 7.3m what is the area and the perimeter
Bad White [126]

Step-by-step explanation:

<h3><em><u>Given</u></em><em><u>:</u></em></h3>

Length of the rectangle = 6.5 m

Width of the rectangule = 7.3 m

<h3><em><u>Then</u></em><em><u>:</u></em></h3>

<u>First</u><u> </u><u>case</u><u>,</u>

Area of the rectangle

= length × width

= 6.5 m × 7.3 m

= <em><u>47.45</u></em><em><u> </u></em><em><u>s</u></em><em><u>q</u></em><em><u>.</u></em><em><u>m</u></em><em><u> </u></em><em><u>(</u></em><em><u>Ans</u></em><em><u>)</u></em><em><u>(</u></em><em><u>i</u></em><em><u>)</u></em>

<u>Second</u><u> </u><u>case</u><u>,</u>

Perimeter of the rectangle

= 2(length + width)

= 2(6.5 + 7.3)m

= 2 × 13.8 m

= <em><u>27</u></em><em><u>.</u></em><em><u>6</u></em><em><u> </u></em><em><u>m</u></em><em><u> </u></em><em><u>(</u></em><em><u>Ans</u></em><em><u>)</u></em><em><u>(</u></em><em><u>ii</u></em><em><u>)</u></em>

5 0
2 years ago
Read 2 more answers
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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