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sukhopar [10]
2 years ago
14

Look at the images please help

Mathematics
1 answer:
Wittaler [7]2 years ago
6 0

Answer:

SA=672\ yd^2

The net in the attached figure

Step-by-step explanation:

we know that

The surface area of the square pyramid using a net, is equal to the area of a square plus the area of its four lateral triangular faces

so

SA=16^2+4[\frac{1}{2}(16)(13)]

SA=256+416=672\ yd^2

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I really need help with this ASAP​
Alexandra [31]

Answer:

No. of boxes = 10

Oranges per box = 56

Total no. of oranges = 56*10=560

No. of bad oranges = 560/40= 14

(i) Probability of bad orange = 14/560 = 1/40

(ii) no of oranges expected to be bad = 14 ( found above)

Hope it helps.............. :)

7 0
3 years ago
What is f(-2) of the function given below? *<br> f(x) = 3x2 - 4x + 1
slava [35]

Answer:

f(-2) = 21

Step-by-step explanation:

Step 1: Define

f(x) = 3x² - 4x + 1

f(-2) is x = -2

Step 2: Substitute and Evaluate

f(-2) = 3(-2)² - 4(-2) + 1

f(-2) = 3(4) + 8 + 1

f(-2) = 12 + 9

f(-2) = 21

5 0
3 years ago
Define the double factorial of n, denoted n!!, as follows:n!!={1⋅3⋅5⋅⋅⋅⋅(n−2)⋅n} if n is odd{2⋅4⋅6⋅⋅⋅⋅(n−2)⋅n} if n is evenand (
tekilochka [14]

Answer:

Radius of convergence of power series is \lim_{n \to \infty}\frac{a_{n}}{a_{n+1}}=\frac{1}{108}

Step-by-step explanation:

Given that:

n!! = 1⋅3⋅5⋅⋅⋅⋅(n−2)⋅n        n is odd

n!! = 2⋅4⋅6⋅⋅⋅⋅(n−2)⋅n       n is even

(-1)!! = 0!! = 1

We have to find the radius of convergence of power series:

\sum_{n=1}^{\infty}[\frac{8^{n}n!(3n+3)!(2n)!!}{2^{n}[(n+9)!]^{3}(4n+3)!!}](8x+6)^{n}\\\\\sum_{n=1}^{\infty}[\frac{8^{n}n!(3n+3)!(2n)!!}{2^{n}[(n+9)!]^{3}(4n+3)!!}]2^{n}(4x+3)^{n}\\\\\sum_{n=1}^{\infty}[\frac{8^{n}n!(3n+3)!(2n)!!}{[(n+9)!]^{3}(4n+3)!!}](x+\frac{3}{4})^{n}\\

Power series centered at x = a is:

\sum_{n=1}^{\infty}c_{n}(x-a)^{n}

\sum_{n=1}^{\infty}[\frac{8^{n}n!(3n+3)!(2n)!!}{2^{n}[(n+9)!]^{3}(4n+3)!!}](8x+6)^{n}\\\\\sum_{n=1}^{\infty}[\frac{8^{n}n!(3n+3)!(2n)!!}{2^{n}[(n+9)!]^{3}(4n+3)!!}]2^{n}(4x+3)^{n}\\\\\sum_{n=1}^{\infty}[\frac{8^{n}4^{n}n!(3n+3)!(2n)!!}{[(n+9)!]^{3}(4n+3)!!}](x+\frac{3}{4})^{n}\\

a_{n}=[\frac{8^{n}4^{n}n!(3n+3)!(2n)!!}{[(n+9)!]^{3}(4n+3)!!}]\\\\a_{n+1}=[\frac{8^{n+1}4^{n+1}n!(3(n+1)+3)!(2(n+1))!!}{[(n+1+9)!]^{3}(4(n+1)+3)!!}]\\\\a_{n+1}=[\frac{8^{n+1}4^{n+1}(n+1)!(3n+6)!(2n+2)!!}{[(n+10)!]^{3}(4n+7)!!}]

Applying the ratio test:

\frac{a_{n}}{a_{n+1}}=\frac{[\frac{32^{n}n!(3n+3)!(2n)!!}{[(n+9)!]^{3}(4n+3)!!}]}{[\frac{32^{n+1}(n+1)!(3n+6)!(2n+2)!!}{[(n+10)!]^{3}(4n+7)!!}]}

\frac{a_{n}}{a_{n+1}}=\frac{(n+10)^{3}(4n+7)(4n+5)}{32(n+1)(3n+4)(3n+5)(3n+6)+(2n+2)}

Applying n → ∞

\lim_{n \to \infty}\frac{a_{n}}{a_{n+1}}= \lim_{n \to \infty}\frac{(n+10)^{3}(4n+7)(4n+5)}{32(n+1)(3n+4)(3n+5)(3n+6)+(2n+2)}

The numerator as well denominator of \frac{a_{n}}{a_{n+1}} are polynomials of fifth degree with leading coefficients:

(1^{3})(4)(4)=16\\(32)(1)(3)(3)(3)(2)=1728\\ \lim_{n \to \infty}\frac{a_{n}}{a_{n+1}}=\frac{16}{1728}=\frac{1}{108}

4 0
2 years ago
I don’t need help with 17 and 18 only those 4 problems on top plz help me and follow the directions on the top
Dovator [93]
13. 44 + 40 = 84

14. 15 + 81 = 96

15. 13 + 52 = 65

16. 64 + 28 = 92
5 0
3 years ago
Read 2 more answers
The width of a rectangle is y feet long. The rectangle is 4 feet longer than it is wide. What is the area of the rectangle? Whic
Korolek [52]

y(y + 4) sq ft  =   is the answer

   


8 0
3 years ago
Read 2 more answers
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