Find the value of the following expression:

Mathematics

2 answers:

weqwewe [10]4 years ago

4 0

Thank you, I was having a little bit of trouble with this, but then you helped.

Misha Larkins [42]4 years ago

3 0

$(2^8*3^{-5}*6^0)^{-2}*( \cfrac{3^{-2}}{2^3} )^4*2^{28}\\ \\\\ = (\cfrac{2^8}{3^5}*1)^{-2} *( \cfrac{1}{3^2*2^3} )^4*2^{28}\\\\\\=( \cfrac{3^5}{2^8} )^2* \cfrac{1^4}{3^{2*4}*2^{3*4}} *2^{28}\\\\\\= \cfrac{3^{5*2}}{2^{8*2}} * \cfrac{1}{3^8*2^{12}} *2^{28}\\\\\\=\cfrac{3^{10}}{2^{16}} * \cfrac{2^{28}}{3^8*2^{12}}\\\\\\= \cfrac{3^{10}*2^{28}}{3^8*2^{16+12}} \\\\\\=\cfrac{3^{10}*2^{28}}{3^8*2^{28}} \\\\=3^{10-8}*2^{28-28}\\\\=3^2*2^0\\\\=9*1\\\\=9$

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The area of a circle 220mm in diameter is approximately how many times greater than the area of a circle 30mm in diameter?

zmey [24]

The first circle is approximately 44 times greater than the second cirle

Radius = 110 mm
110² = 10,000 mm² = Area

Radius = 15 mm
15² = 225 mm² = Area

Hope this helps!

5 0

4 years ago

HELP!!

True [87]

Yes is 12% good job hope this helps

8 0

3 years ago

HELP ASAP photo attached

ankoles [38]

Answer:

D. undefined

General Formulas and Concepts:

Calculus

Derivatives

Derivative Notation

Derivative of a constant is 0

Basic Power Rule:

f(x) = cxⁿ
f’(x) = c·nxⁿ⁻¹

Derivative Rule [Chain Rule]: $\displaystyle \frac{d}{dx}[f(g(x))] =f'(g(x)) \cdot g'(x)$

Trig Derivative: $\displaystyle \frac{d}{dx}[sinu] = u'cosu$

Derivatives of Parametrics: $\displaystyle \frac{dy}{dx} = \frac{\frac{dy}{dt}}{\frac{dx}{dt}}$

Step-by-step explanation:

Step 1: Define

$\displaystyle \frac{dx}{dt} = 5$

$\displaystyle \frac{dy}{dt} = sin(t^2)$

Step 2: Differentiate

[x Derivative] Basic Power Rule: $\displaystyle \frac{d^2x}{dt^2} = 0$
[y Derivative] Trig Derivative [Chain Rule]: $\displaystyle \frac{d^2y}{dt^2} = cos(t^2) \cdot \frac{d}{dt}[t^2]$
[y Derivative] Basic Power Rule: $\displaystyle \frac{d^2y}{dt^2} = cos(t^2) \cdot 2t^{2 - 1}$
[y Derivative] Simplify: $\displaystyle \frac{d^2y}{dt^2} = 2tcos(t^2)$
[Derivative] Rewrite: $\displaystyle \frac{d^2y}{dx^2} = \frac{2tcos(t^2)}{0}$