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ICE Princess25 [194]
3 years ago
9

Please help me on this! Thank you!

Mathematics
1 answer:
Fantom [35]3 years ago
5 0

Answer:

The area of the garden is 17 yd^2

Step-by-step explanation:

Ok, let's divide the garden into 2 parts. The small part to the left of the patio let's call it area A. This area has 1 yd of width and 2 yd of length as it has the same length as the patio.

Therefore area A has an area of 1*2=2 yd

Let's call the remaining area area B.

We can see that its length is 3 yd and its width is 5 yd so:

3*5=15

Summing both areas to obtain the area of the garden we get

15+2=17

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Hilary has 108 building blocks in the red bucket, which is 2 times as many blocks
Setler79 [48]

Step-by-step explanation:

red one has 108

then green one has 108/2

then m=54

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3 years ago
Please hurry find the measure of P
Inessa [10]

Answer:

m∠P  = 65°

Step-by-step explanation:

An Isosceles triangle has two sides of equal length and two interior angles of equal measure.

Sum of interior angles of a triangle = 180°

If QO ≅ PQ and the enclosed m∠Q = 50° then m∠P ≅ m∠O

⇒ m∠Q + m∠P + m∠O = 180

⇒ 50 + m∠P + m∠O = 180

⇒ m∠P + m∠O = 130

As m∠P ≅ m∠O:

⇒ m∠P  = 130 ÷ 2 = 65°

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If f(x)=|x+2| is changed to g(x)=3f(x)+4, how is the graph of the function transformed?
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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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