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Papessa [141]
3 years ago
9

What is the length of the diagonal?

Mathematics
1 answer:
JulijaS [17]3 years ago
6 0

Answer:

\sqrt{865} feet

Step-by-step explanation:

Using pythagoras theorem,

length \:  =  \sqrt{17 {}^{2}  + 24 {}^{2} }  \\  =  \sqrt{865} ft

You might be interested in
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
5) Consider the diagram. What is QS? <br> 2 units<br> 5 units<br> 17 units<br> 33 units
Shkiper50 [21]

Answer:

17 units

Step-by-step explanation:

8 0
2 years ago
Read 2 more answers
Simplify,write equivalent expressions. 1/6(15a+20b)
lutik1710 [3]
1/6* (15a+20b)
= 1/6* 15a + 1/6* 20b (Distributive property)
= 2.5a + 10/3*b

The final answer is 2.5a+ 10/3b

Hope this helps~
6 0
3 years ago
What is the area of the irregular figure below?
expeople1 [14]

Answer:

360 centimeters squared

Step-by-step explanation:

7 0
2 years ago
Read 2 more answers
In square ABCD, point M is the midpoint of side AB and point N is the midpoint of side BC. What is the ratio of the area of tria
Phoenix [80]

Answer:

Ratio of triangle AMN to square ABCD = 1/8

Step-by-step explanation:

area AMN = Area ABN - Area MNB

Where AB = BC = x

Area AMN = [ ½ *( x * ½x )] - [ ½( ½x * ½x)] = x²/8

Area of ABCD = x* x = x²

So therefore:

The ratio of area of triangle AMN to area of square ABCD would be

= (x²/8) / x²

=1/8

6 0
3 years ago
Read 2 more answers
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