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3 years ago

A cereal box has a length of 4 inches, a width of 5 inches,

Mathematics

2 answers:

Vanyuwa [196]3 years ago

7 0

Answer:

160

Step-by-step explanation:

base x width x height = volume

4x5x8=160 in^3

That's all there is to volume of a box

Hope that helps :)

zloy xaker [14]3 years ago

6 0

Answer:

160 yet it's supposa be in Cuboid e.g. 160cm3

Step-by-step explanation:

Volume=L×H×W

= 4×5×8

=160

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Division of fraction 3/5 divided by 10/11

Tema [17]

3/5 ÷ 10/11
multiply by the inverse

3/5 x 11/10 = 33/50

3 0

3 years ago

Read 2 more answers

A circular swimming pool has a diameter of the 26 feet.

kherson [118]

Hey there! :)

The formula for finding area of circles is: π · r²

<em>Remember that radius is half the diameter. To find radius when the diameter is given, divide by 2</em>

d = 26

26 ÷ 2 = 13
_____________________________________________________________

π · r²

r = 13
13² = 13 × 13 = 169

169 × 3.14 = 530.66 = 530.7

The area of the swimming pool is 530.7 ft²

Hope this helps :)

7 0

3 years ago

Read 2 more answers

Suppose that the amount of algae in a pond doubles every 4 hours. If the pond initially contains 90 pounds of algae, how much al

UNO [17]

After 12 hours the pound would have 720 pounds of algae.

4 0

3 years ago

How to prove this???

swat32

$\cos^3 2A + 3 \cos 2A \\
\Rightarrow \cos 2A (\cos^2 2A + 3) \\
\Rightarrow (\cos^2 A - \sin^2 A) (\cos^2 2A + 3) \\
\Rightarrow (\cos^2 A - \sin^2 A) (1 - \sin^2 2A + 3) \\
\Rightarrow (\cos^2 A - \sin^2 A) (4 - \sin^2 2A) \\
\Rightarrow (\cos^2 A - \sin^2 A) (4 - (2\sin A \cos A)(2\sin A \cos A)) \\
\Rightarrow (\cos^2 A - \sin^2 A) (4 - 4\sin^2 A \cos^2 A) \\ 
\Rightarrow 4(\cos^2 A - \sin^2 A) (1 - \sin^2 A \cos^2 A) 
$

go to right side now

$4( \cos^6 A - \sin^6 A)\\
\Rightarrow 4( \cos^3 A - \sin^3 A)(\cos^3 A + \sin^3 A)$

use $x^3 - y^3 = (x-y)(x^2 + xy + y^2)$ and $x^3 + y^3 = x^2 - xy + y^2$

$4( \cos^6 A - \sin^6 A)\\ \Rightarrow 4( \cos^3 A - \sin^3 A)(\cos^3 A + \sin^3 A) \\
\Rightarrow 4(\cos A - \sin A)(\cos^2 A + \cos A \sin A + \sin^2 A) \\
~\quad \quad\cdot ( \cos A + \sin A)(\cos^2 A - \cos A \sin A + \cos^2 A)$

so $\sin^2 A + \cos^2 A = 1$

$4( \cos^6 A - \sin^6 A)\\ \Rightarrow 4(\cos A - \sin A)(\cos^2 A + \cos A \sin A + \sin^2 A) \\ ~\quad \quad\cdot ( \cos A + \sin A)(\cos^2 A - \cos A \sin A + \cos^2 A) \\ \Rightarrow 4(\cos^2 A - \sin^2 A)(1 + \cos A \sin A )(1- \cos A \sin A ) \\ \Rightarrow 4(\cos^2 A - \sin^2 A)(1 - \cos^2 A \sin^2 A )\\ \Rightarrow 4(\cos^2 A - \sin^2 A)(1 - \sin^2 A \cos^2 A ) \\
 \Rightarrow Left hand side$

4 0

3 years ago

In the expansion (ax+by)^7, the coefficients of the first two terms are 128 and -224, respectively. Find the values of a and b

madam [21]

Answer:

a = 2, b = 3.5

Step-by-step explanation:

Expanding $(ax+by)^7$ using Binomial expansion, we have that:

$(ax+by)^7$ =

$(ax)^7(by)^0 + (ax)^6(by)^1 + (ax)^5(by)^2 + (ax)^4(by)^3 + (ax)^3(by)^4 + (ax)^2(by)^5 + (ax)^1(by)^6 + (ax)^0(by)^7$

$= (a)^7(x)^7+ (a)^6(x)^6(b)(y) + (a)^5(x)^5(b)^2(y)^2 + (a)^4(x)^4(b)^3(y)^3 + (a)^3(x)^3(b)^4(y)^4 + (a)^2(x)^2(b)^5(y)^5 + (a)(x)(b)^6(y)^6 + (b)^7(y)^7\\\\\\= (a)^7(x)^7+ (a)^6(b)(x)^6(y) + (a)^5(b)^2(x)^5(y)^2 + (a)^4(b)^3(x)^4(y)^3 + (a)^3(b)^4(x)^3(y)^4 + (a)^2(b)^5(x)^2(y)^5 + (a)(b)^6(x)(y)^6 + (b)^7(y)^7$