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2 years ago

3 + root 3 *2 +root 2

Mathematics

1 answer:

icang [17]2 years ago

4 0

Answer:

6 + 3 $\sqrt{2}$ + 2 $\sqrt{3}$ + $\sqrt{6}$

Step-by-step explanation:

Using the rule of radicals

$\sqrt{a}$ × $\sqrt{b}$ ⇔ $\sqrt{ab}$

Given

(3 + $\sqrt{3}$ )(2 + $\sqrt{2}$ )

Each term in the second factor is multiplied by each term in the first factor, that is

3(2 + $\sqrt{2}$ ) + $\sqrt{3}$ (2 + $\sqrt{2}$ ) ← distribute both parenthesis

= 6 + 3 $\sqrt{2}$ + 2 $\sqrt{3}$ + $\sqrt{6}$

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Find x using completing the square method x^2+16n+28 = 0

Nezavi [6.7K]

Answer:

x=√−16n−28 or x=−√−16n−28

Step-by-step explanation:

Step 1: Add -16n to both sides.

x2+16n+28+−16n=0+−16n

x2+28=−16n

Step 2: Add -28 to both sides.

x2+28+−28=−16n+−28

x2=−16n−28

Step 3: Take square root.

x=√−16n−28 or x=−√−16n−28

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2 years ago

Select the correct answer Consider functions p and q. p(x) = log₂ (x - 1) g(x) = 2^x - 1 Which statement is true about these fun

goldenfox [79]

The x-intercept of p(x) is x = 2, the x-intercept of g(x) is x = 0, then the correct option is B.

<h3>What can we say about the x-intercepts of the given functions?</h3>

For a function f(x), the x-intercept is the value of x such that:

f(x) = 0.

Here we have:

p(x) = log₂(x - 1)

Remember that:

logₙ(1) = 0

For any base n, then the x-intercept of p(x) is x = 2, because:

p(2) = log₂(2 - 1) = log₂(1) = 0.

The other function is:

g(x) = 2ˣ - 1

Remember that any number to the power of zero is equal to 1, then:

g(0) = 2⁰ - 1 = 1 - 1 = 0

The x-intercept of p(x) is x = 2, the x-intercept of g(x) is x = 0, then the correct option is B.

If you want to learn more about x-intercepts:

brainly.com/question/3951754

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1 year ago

Implicit differentiation Please help

Anvisha [2.4K]

Answer:

$y''(-1) =8$

General Formulas and Concepts:

Pre-Algebra

Order of Operations: BPEMDAS

Brackets
Parenthesis
Exponents
Multiplication
Division
Addition
Subtraction

Left to Right

Equality Properties

Algebra I

Factoring

Calculus

Implicit Differentiation

The derivative of a constant is equal to 0

Basic Power Rule:

f(x) = cxⁿ
f’(x) = c·nxⁿ⁻¹

Product Rule: $\frac{d}{dx} [f(x)g(x)]=f'(x)g(x) + g'(x)f(x)$

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

Quotient Rule: $\frac{d}{dx} [\frac{f(x)}{g(x)} ]=\frac{g(x)f'(x)-g'(x)f(x)}{g^2(x)}$

Step-by-step explanation:

Step 1: Define

-xy - 2y = -4

Rate of change of the tangent line at point (-1, 4)

Step 2: Differentiate Pt. 1

Find 1st Derivative

Implicit Differentiation [Product Rule/Basic Power Rule]: $-y - xy' - 2y' = 0$
[Algebra] Isolate y' terms: $-xy' - 2y' = y$
[Algebra] Factor y': $y'(-x - 2) = y$
[Algebra] Isolate y': $y' = \frac{y}{-x-2}$
[Algebra] Rewrite: $y' = \frac{-y}{x+2}$

Step 3: Find y

Define equation: $-xy - 2y = -4$
Factor y: $y(-x - 2) = -4$
Isolate y: $y = \frac{-4}{-x-2}$
Simplify: $y = \frac{4}{x+2}$

Step 4: Rewrite 1st Derivative

[Algebra] Substitute in y: $y' = \frac{-\frac{4}{x+2} }{x+2}$
[Algebra] Simplify: $y' = \frac{-4}{(x+2)^2}$

Step 5: Differentiate Pt. 2

Find 2nd Derivative

Differentiate [Quotient Rule/Basic Power Rule]: $y'' = \frac{0(x+2)^2 - 8 \cdot 2(x + 2) \cdot 1}{[(x + 2)^2]^2}$
[Derivative] Simplify: $y'' = \frac{8}{(x+2)^3}$

Step 6: Find Slope at Given Point

[Algebra] Substitute in x: $y''(-1) = \frac{8}{(-1+2)^3}$
[Algebra] Evaluate: $y''(-1) =8$

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3 + root 3 *2 +root 2​

3 + root 3 *2 +root 2