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3 years ago

Compare the values of the two expressions using -3(-2) + 3 __3(-4)+9

Mathematics

1 answer:

schepotkina [342]3 years ago

3 0

Answer:

Step-by-step explanation:

expression 1: -3(-2) + 3 = 9

expression 2: 3(-4) + 9 = 3

9 > 3

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Answer:

$\displaystyle \lim_{h \to 0} \frac{\sqrt{3} \Delta (\frac{\pi}{3})}{h^2} = \boxed{ 144 \sqrt{3} }$

General Formulas and Concepts:
Pre-Calculus

2x2 Matrix Determinant:
$\displaystyle \left| \begin{array}{ccc} a & b \\ c & d \end{array} \right| = ad - bc$

3x3 Matrix Determinant:
$\displaystyle \left| \begin{array}{ccc} a & b & c \\ d & e & f \\ g & h & i \end{array} \right| = a \left| \begin{array}{ccc} e & f \\ h & i \end{array} \right| - b \left| \begin{array}{ccc} d & f \\ g & i \end{array} \right| + c \left| \begin{array}{ccc} d & e \\ g & h \end{array} \right|$

Calculus

Limits

Limit Rule [Variable Direct Substitution]:
$\displaystyle \lim_{x \to c} x = c$

Limit Property [Multiplied Constant]:
$\displaystyle \lim_{x \to c} bf(x) = b \lim_{x \to c} f(x)$

Special Limit Rule [L’Hopital’s Rule]:
$\displaystyle \lim_{x \to c} \frac{f(x)}{g(x)} = \lim_{x \to c} \frac{f'(x)}{g'(x)}$

Derivatives

Derivatives
Derivative Notation

Derivative Property [Addition/Subtraction]:
$\displaystyle (u + v)' = u' + v'$

Derivative Rule [Chain Rule]:
$\displaystyle [u(v)]' = u'(v)v'$

Step-by-step explanation:

*Note:

I will not be able to fit in all the derivative work and will assume you can take derivatives with ease.

Step 1: Define

Identify given.

$\displaystyle \Delta (x) = \left| \begin{array}{ccc} \tan x & \tan (x + h) & \tan (x + 2h) \\ \tan (x + 2h) & \tan x & \tan (x + h) \\ \tan (x + h) & \tan (x + 2h) & \tan x \end{array} \right|$

$\displaystyle \lim_{h \to 0} \frac{\sqrt{3} \Delta (\frac{\pi}{3})}{h^2}$

Step 2: Find Limit Pt. 1

[Function] Simplify [3x3 and 2x2 Matrix Determinant]:
$\displaystyle \Delta (x) = \tan^3 (2h + x) + \tan^3 (h + x) + \tan^3 x - 3 \tan x \tan (h + x) \tan (2h + x)$
[Function] Substitute in x:
$\displaystyle \Delta \bigg( \frac{\pi}{3} \bigg) = \tan^3 \bigg( 2h+ \frac{\pi}{3} \bigg) + \tan^3 \bigg( h + \frac{\pi}{3} \bigg) + 3\sqrt{3} - 3\sqrt{3} \tan \bigg( h + \frac{\pi}{3} \bigg) \tan \bigg( 2h+ \frac{\pi}{3} \bigg)$

Step 3: Find Limit Pt. 2

[Limit] Rewrite [Limit Property - Multiplied Constant]:
$\displaystyle \lim_{h \to 0} \frac{\sqrt{3} \Delta (\frac{\pi}{3})}{h^2} = \sqrt{3} \lim_{h \to 0} \frac{\Delta (\frac{\pi}{3})}{h^2}$
[Limit] Apply Limit Rule [Variable Direct Substitution]:
$\displaystyle \lim_{h \to 0} \frac{\sqrt{3} \Delta (\frac{\pi}{3})}{h^2} = \sqrt{3} \bigg( \frac{0}{0} \bigg)$

Since we have an indeterminant form, we will have to use L'Hopital's Rule. We can differentiate using basic differentiation techniques listed above under "Calculus":

$\displaystyle \frac{d \Delta (\frac{\pi}{3})}{dh} = -3\sqrt{3} \bigg[ \tan^2 \bigg( h + \frac{\pi}{3} \bigg) + 1 \bigg] \tan \bigg( 2h + \frac{\pi}{3} \bigg) + tan^2 \bigg( h + \frac{\pi}{3} \bigg) \bigg[ 3 \tan^2 \bigg( h + \frac{\pi}{3} + 3 \bigg] - 3\sqrt{3} \tan \bigg( h + \frac{\pi}{3} \bigg) \bigg[ 2 \tan^2 \bigg( 2h + \frac{\pi}{3} \bigg) + 2 \bigg] + \tan^2 \bigg( 2h + \frac{\pi}{3} \bigg) \bigg[ 6 \tan^2 \bigg( 2h + \frac{\pi}{3} \bigg) + 6 \bigg]$

$\displaystyle \frac{d}{dh} h^2 = 2h$

Using L'Hopital's Rule, we can substitute the derivatives and evaluate again. When we do so, we should get another indeterminant form. We will need to use L'Hopital's Rule again:

$\displaystyle \frac{d^2 \Delta (\frac{\pi}{3})}{dh^2} = \tan \bigg( h + \frac{\pi}{3} \bigg) \bigg[ 2 \tan^2 \bigg( h + \frac{\pi}{3} \bigg) + 2 \bigg] \bigg[ \tan^2 \bigg( h + \frac{\pi}{3} \bigg) + 1 \bigg] - 2\sqrt{3} \bigg[ \tan^2 \bigg( h + \frac{\pi}{3} \bigg) + 1 \bigg] \bigg[ \tan^2 \bigg( 2h + \frac{\pi}{3} \bigg) + 1 \bigg] - \sqrt{3} \bigg[ \tan^2 \bigg( h + \frac{\pi}{3} \bigg) + 1 \bigg] \bigg[ 2 \tan^2 \bigg( 2h + \frac{\pi}{3} \bigg) + 2 \bigg]$

$\displaystyle + \tan^3 \bigg( h + \frac{\pi}{3} \bigg) \bigg[ 2 \tan^2 \bigg( h + \frac{\pi}{3} \bigg) + 2 \bigg] - \sqrt{3} \tan \bigg( h + \frac{\pi}{3} \bigg) \tan \bigg( 2h + \frac{\pi}{3} \bigg) \bigg[ 2 \tan^2 \bigg( h + \frac{\pi}{3} \bigg) + 2 \bigg] + \tan \bigg( 2h + \frac{\pi}{3} \bigg) \bigg[ 2 \tan^2 \bigg( 2h + \frac{\pi}{3} \bigg) + 2 \bigg] \bigg[ 4 \tan^2 \bigg( 2h + \frac{\pi}{3} \bigg) + 4 \bigg]$

$\displaystyle - 2\sqrt{3} \tan \bigg( h + \frac{\pi}{3} \bigg) \tan \bigg( 2h + \frac{\pi}{3} \bigg) \bigg[ 4 \tan^2 \bigg( 2h + \frac{\pi}{3} \bigg) + 4 \bigg] + 2 \tan^3 \bigg( 2h + \frac{\pi}{3} \bigg) \bigg[ 4 \tan^2 \bigg( 2h + \frac{\pi}{3} \bigg) + 4 \bigg]$

$\displaystyle \frac{d^2}{dh^2} h^2 = 2$

Substituting in the 2nd derivative found via L'Hopital's Rule should now give us a numerical value when evaluating the limit using limit rules and the unit circle:

$\displaystyle \lim_{h \to 0} \frac{\sqrt{3} \Delta (\frac{\pi}{3})}{h^2} = \boxed{ 144 \sqrt{3} }$

∴ we have evaluated the given limit.

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Learn more about limits: brainly.com/question/27438198

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3 0

2 years ago

Which table shows ordered pairs that satisfy the function y=4x-2

s2008m [1.1K]

<h3>Answer: Choice D</h3>

==========================================================

Explanation:

If you plug in x = -2, then,

y = 4x-2

y = 4(-2)-2

y = -8-2

y = -10

This shows that the answer must be choice D, as this is the only answer that has x = -2 lead to y = -10. Note the first column of table D.

We could stop here if you wanted.

----------------------

If we kept going and plugged in x = 0, then,

y = 4x-2

y = 4(0)-2

y = 0-2

y = -2

We have x = 0 lead to y = -2. This matches with the second column of table D.

-----------------------

Lastly, if we plugged in x = 2, then we get,

y = 4x-2

y = 4(2)-2

y = 8-2

y = 6

So the third column of table D is also confirmed.

3 0

3 years ago

For what value of a is the volume of the tetrahedron formed by the coordinate planes and the plane (x/a) (y/10) (z/6)

kotykmax [81]

This question is incomplete, the complete question is;

For what value of a is the volume of the tetrahedron formed by the coordinate planes and the plane (x/a) + (y/10) + (z/6) = 1 equal to 10?

Answer: the value of a is 1

Step-by-step explanation:

Given that;

Volume of tetrahedron bounded by plane (x/a) + (y/10) + (z/6) = 1

and coordinate plane is; V = 1/6|abc|

(x/a) + (y/10) + (z/6) = 1

volume = 10

10 = 1/6 | a × 10 × 6 |

60 = a × 10 × 6

60 = 60a

a = 60 / 60

a = 1

Therefore the value of a is 1

3 0

3 years ago

Compare the values of the two expressions using -3(-2) + 3 __3(-4)+9​

Compare the values of the two expressions using -3(-2) + 3 __3(-4)+9