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2 years ago

A rectangular prism is 4 cm long, 3 cm wide and 8 cm tall. What is the volume of the prism?

Mathematics

2 answers:

Andrei [34K]2 years ago

8 0

Answer:

96 cm³

Step-by-step explanation:

v = 4 x 3 x 8 = 96 cm³

SpyIntel [72]2 years ago

3 0

Answer:

The volume of the prism is 96cm^3

Step-by-step explanation:

You multpily 4*3*8=96

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Please help me! i'm stuck on it

den301095 [7]

Answer:

x = 15.4

Step-by-step explanation:

Because this is a right triangle, you can use the pythagorean theorem to find the length of the hypotenuse. the theorem is a^2 + b^2 = c^2

9^2 + 12.5^2 = c^2

solving this will give you 15.4

8 0

3 years ago

Can someone answer these questions ? thank you

Naya [18.7K]

Answer:

Step-by-step explanation:

12a)To rationalize the denominator, multiply the denominator and numerator by √5.

$\frac{15}{\sqrt{5}}=\frac{15*\sqrt{5}}{\sqrt{5}*\sqrt{5}}\\\\=\frac{15\sqrt{5}}{5}\\\\=3\sqrt{5}$

b) (a+b)² = a² + 2ab + b²

(1 +√3)² = 1² + 2*1*√3 + (√3)²

= 1 + 2√3 + 3

= 4 + 2√3

a = 4 ; b =2

13) (a + b)(a - b) = a² - b²

$\frac{(6-\sqrt{5})(6+\sqrt{5})}{\sqrt{31}}=\frac{6^{2}-(\sqrt{5})^{2}}{\sqrt{31}}\\\\ =\frac{36-5}{\sqrt{31}}\\\\=\frac{31}{\sqrt{31}}\\\\=\frac{31*\sqrt{31}}{\sqrt{31}*\sqrt{31}}\\\\=\frac{31\sqrt{31}}{31}\\\\=\sqrt{31}$

5 0

3 years ago

a company makes carboard cylindrical boxes with a diameter of 10 inches and a height of 8 inches.how many square inches of carbo

NeTakaya

Answer:

408.2sq inches

Step-by-step explanation:

Area of the cylindrical box = 2πr(r+h)

r is the radius = diameter/2

r = 10/2 = 5in

h is the height = 8in

Substitute

Area of the cylindrical box = 2(3.14)(5)(5+8)

Area of the cylindrical box = 2 * 3.14 * 5 * 13

Area of the cylindrical box = 408.2sq inches

8 0

3 years ago

Write an algebraic expression for 10 less than the difference of 2 times a number and 7

notka56 [123]

Answer:

(2n - 7) - 10

Step-by-step explanation:

(2n - 7) represents "difference of 2 times a number and 7"

- 10 represents "10 less"

6 0

3 years ago

Read 2 more answers

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