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dimulka [17.4K]
3 years ago
14

What is the value of ?A. 7.5 B. 17.5 C. 60.5 D. 76.5

Mathematics
2 answers:
REY [17]3 years ago
8 0

Answer:

76.5

Step-by-step explanation:

Recall that for any positive integer n

n! = 1*2*3*4*...*n

For example

5! = 1*2*3*4*5 = 120

so

\sum_{n=2}^6\frac{(n-1)!}{2}=\frac{(2-1)!}{2}+\frac{(3-1)!}{2}+\frac{(4-1)!}{2}+\frac{(5-1)!}{2}+\frac{(6-1)!}{2}=\\\\=\frac{1!}{2}+\frac{2!}{2}+\frac{3!}{2}+\frac{4!}{2}+\frac{5!}{2}=\frac{1}{2}+\frac{1*2}{2}+\frac{1*2*3}{2}+\frac{1*2*3*4}{2}+\frac{1*2*3*4*5}{2}=\\\\=\frac{1}{2}+1+3+12+60=\boxed{76.5}

Marina CMI [18]3 years ago
7 0

Answer:

D on edge

Step-by-step explanation:

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What is the completely factored form of this polynomial?<br> 18x^3– 120x^2-42x
Vlada [557]

Answer:

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Step-by-step explanation:

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ASHA 777 [7]
Answer:  " b = 3 ⅕ ; or, write as: 3.2  .
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Given:  "A = (1/2) * b * h " ;

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2A = (b*h) ;

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Given A = 48 ;  and h = 30 ; Plug in these values; and solve for "b" ; 

→ b = (2*48) / 30 ; 

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3 years ago
99 POINT QUESTION, PLUS BRAINLIEST!!!
Elden [556K]
We draw region ABC. Lines that connect y = 0 and y = x³ are vertical so:
(i) prependicular to the axis x - disc method;
(ii) parallel to the axis y - shell method;
(iii) parallel to the line x = 18 - shell method.

Limits of integration for x are easy x₁ = 0 and x₂ = 9.
Now, we have all information, so we could calculate volume.

(i)

V=\pi\cdot\int\limits_a^bf^2(x)\, dx\qquad\implies \qquad a=0\qquad b=9\qquad f(x)=x^3


V=\pi\cdot\int\limits_0^9(x^3)^2\, dx=\pi\cdot\int\limits_0^9x^6\, dx=\pi\cdot\left[\dfrac{x^7}{7}\right]_0^9=\pi\cdot\left(\dfrac{9^7}{7}-\dfrac{0^7}{7}\right)=\dfrac{9^7}{7}\pi=\\\\\\=\boxed{\dfrac{4782969}{7}\pi}

Answer B. or D.

(ii)

V=2\pi\cdot\int\limits_a^bx\cdot f(x)\, dx


V=2\pi\cdot\int\limits_0^{9}(x\cdot x^3)\, dx=2\pi\cdot\int\limits_0^{9}x^4\, dx=&#10;2\pi\cdot\left[\dfrac{x^5}{5}\right]_0^9=2\pi\cdot\left(\dfrac{9^5}{5}-\dfrac{0^5}{5}\right)=\\\\\\=2\pi\cdot\dfrac{9^5}{5}=\boxed{\dfrac{118098}{5}\pi}

So we know that the correct answer is D.

(iii)
Line x = h

V=2\pi\cdot\int\limits_a^b(h-x)\cdot f(x)\, dx\qquad\implies\qquad h=18


V=2\pi\cdot\int\limits_0^9\big((18-x)\cdot x^3\big)\, dx=2\pi\cdot\int\limits_0^9(18x^3-x^4)\, dx=\\\\\\=2\pi\cdot\left(\int\limits_0^918x^3\, dx-\int\limits_0^9x^4\, dx\right)=2\pi\cdot\left(18\int\limits_0^9x^3\, dx-\int\limits_0^9x^4\, dx\right)=\\\\\\=2\pi\cdot\left(18\left[\dfrac{x^4}{4}\right]_0^9-\left[\dfrac{x^5}{5}\right]_0^9\right)=2\pi\cdot\Biggl(18\biggl(\dfrac{9^4}{4}-\dfrac{0^4}{4}\biggr)-\biggl(\dfrac{9^5}{5}-\dfrac{0^5}{5}\biggr)\Biggr)=\\\\\\

=2\pi\cdot\left(18\cdot\dfrac{9^4}{4}-\dfrac{9^5}{5}\right)=2\pi\cdot\dfrac{177147}{10}=\boxed{\dfrac{177147\pi}{5}}

Answer D. just as before.

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nirvana33 [79]

Answer:

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4 0
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