Using a = 5, b = 12 and c = 13, is triangle ABC a right angle triangle? Yes or No

Help with num 1 please.

KengaRu [80]

Answer:

(i) $\displaystyle y' = (6x - 1)ln(2x + 1) + \frac{2x(3x - 1)}{2x + 1}$

(ii) $\displaystyle y' = \frac{2x}{ln(x)} - \frac{x^2 + 2}{x(lnx)^2}$

(iii) $\displaystyle y' = \frac{e^x[xln(2x) + 1]}{x}$

General Formulas and Concepts:

Calculus

Differentiation

Derivatives
Derivative Notation

Derivative Property [Multiplied Constant]: $\displaystyle \frac{d}{dx} [cf(x)] = c \cdot f'(x)$

Derivative Property [Addition/Subtraction]: $\displaystyle \frac{d}{dx}[f(x) + g(x)] = \frac{d}{dx}[f(x)] + \frac{d}{dx}[g(x)]$

Basic Power Rule:

f(x) = cxⁿ
f’(x) = c·nxⁿ⁻¹

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

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

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

Exponential Differentiation

Logarithmic Differentiation

Step-by-step explanation:

(i)

Step 1: Define

Identify

$\displaystyle y = (3x^2 - x)ln(2x + 1)$

Step 2: Differentiate

Product Rule: $\displaystyle y' = (3x^2 - x)'ln(2x + 1) + (3x^2 - x)[ln(2x + 1)]'$
Basic Power Rule/Logarithmic Differentiation [Chain Rule]: $\displaystyle y' = (6x - 1)ln(2x + 1) + (3x^2 - x)\frac{1}{2x + 1}(2x + 1)'$
Basic Power Rule: $\displaystyle y' = (6x - 1)ln(2x + 1) + (3x^2 - x)\frac{2}{2x + 1}$
Simplify [Factor]: $\displaystyle y' = (6x - 1)ln(2x + 1) + \frac{2x(3x - 1)}{2x + 1}$

(ii)

Step 1: Define

Identify

$\displaystyle y = \frac{x^2 + 2}{lnx}$

Step 2: Differentiate

Quotient Rule: $\displaystyle y' = \frac{(x^2 + 2)'lnx - (x^2 + 2)(lnx)'}{(lnx)^2}$
Basic Power Rule/Logarithmic Differentiation: $\displaystyle y' = \frac{2xlnx - (x^2 + 2)\frac{1}{x}}{(lnx)^2}$
Rewrite: $\displaystyle y' = \frac{2xlnx}{(lnx)^2} - \frac{(x^2 + 2)\frac{1}{x}}{(lnx)^2}$
Simplify: $\displaystyle y' = \frac{2x}{ln(x)} - \frac{x^2 + 2}{x(lnx)^2}$

(iii)

Step 1: Define

Identify

$\displaystyle y = e^xln(2x)$

Step 2: Differentiate

Product Rule: $\displaystyle y' = (e^x)'ln(2x) + e^x[ln(2x)]'$
Exponential Differentiation/Logarithmic Differentiation [Chain Rule]: $\displaystyle y' = e^xln(2x) + e^x(\frac{1}{2x})(2x)'$
Basic Power Rule: $\displaystyle y' = e^xln(2x) + e^x(\frac{1}{2x})2$
Simplify: $\displaystyle y' = e^xln(2x) + \frac{e^x}{x}$
Rewrite: $\displaystyle y' = \frac{xe^xln(2x) + e^x}{x}$
Factor: $\displaystyle y' = \frac{e^x[xln(2x) + 1]}{x}$

Topic: AP Calculus AB/BC (Calculus I/I + II)

Unit: Differentiation

Book: College Calculus 10e

6 0

3 years ago

What is the blank answer?

Taya2010 [7]

Answer:

you.. need to provide a picture

Step-by-step explanation:

7 0

3 years ago

I need a lot of help from the smartest people now right now

Dominik [7]

Answer:

K) I, II, and III

Step-by-step explanation:

Given the quadratic equation in standard form, h = -at² + bt + c, where h is the height or the projectile of a baseball that changes over time, t. In the given quadratic equation, c represents the constant term. Altering the constant term, c, affects the h-intercept, the maximum value of h, and the t-intercept of the quadratic equation.

<h2>I. The h-intercept</h2>

The h-intercept is the value of the height, h, when t = 0. This means that setting t = 0 will leave you with the value of the constant term. In other words:

Set t = 0:

h = -at² + bt + c

h = -a(0)² + b(0) + c

h = -a(0) + 0 + c

h = 0 + c

h = c

Therefore, the value of the h-intercept is the value of c.

Hence, altering the value of c will also change the value of the h-intercept.

<h2>II. The maximum value of h</h2>

The maximum value of h occurs at the vertex, (t, h ). Changing the value of c affects the equation, especially the maximum value of h. To find the value of the t-coordinate of the vertex, use the following formula:

t = -b/2a

The value of the t-coordinate will then be substituted into the equation to find its corresponding h-coordinate. Thus, changing the value of c affects the corresponding h-coordinate of the vertex because you'll have to add the constant term into the rest of the terms within the equation. Therefore, altering the value of c affects the maximum value of h.

<h2>III. The t-intercept</h2>

The t-intercept is the point on the graph where it crosses the t-axis, and is also the value of t when h = 0. The t-intercept is the zero or the solution to the given equation. To find the t-intercept, set h = 0, and solve for the value of t. Solving for the value of t includes the addition of the constant term, c, with the rest of the terms in the equation. Therefore, altering the value of c also affects the t-intercept.

Therefore, the correct answer is Option K: I, II, and III.

5 0

2 years ago

How can you find 50% of any number? Select all that apply.

Art [367]

Answer:

multiply it by 0.5

Step-by-step explanation:

8 0

3 years ago

Read 2 more answers

Q13 Please help me solve this….

KonstantinChe [14]

Answer:

Choice C

$\frac{\sqrt{186}+\sqrt{15}}{18}$

Step-by-step explanation:

The quadrant in which an angle lies determines the signs of the trigonometric functions sin, cos and tan

If an angle Θ lies in quadrant IV, cos(Θ) is positive and both sin(Θ) and tan(Θ) are negative

Two of the trigonometric identities we can use are

1. $\sin^2(\theta) + cos^2(\theta) = 1$ and

2. $\cos(A-B) = \cos A\cos B - \sin A\sin B$

Using identity 1, we can solve for cos(s) and cos(t)

$sin(s)=-\frac{\sqrt{3}}{3}, sin^{2}(s)=\frac{3}{9}=\frac{1}{3}\\cos^{2}(s)=1-sin^{2}(s)=1-\frac{1}{3}=\frac{2}{3}; cos(s)=\pm\sqrt{\frac{2}{3}}\\\\$

$sin(t)=-\frac{\sqrt{5}}{6}, sin^{2}(t)=\frac{5}{36}cos^{2}(t)=1-sin^{2}(t)=1-\frac{5}{36}=\frac{31}{36};cos(t)=\pm\frac{\sqrt{31}}{6}$

Since both angles lie in quadrant IV, both cos(s) and cos(t) must be positive so we only consider the positive signs of both values

Using identity 2, we can solve for cos(s-t)

$cos(s-t)=cos(s)cos(t)+sin(s)sin(t)=\frac{\sqrt{2}}{\sqrt{3}}.\frac{\sqrt{31}}{6}+(-\frac{\sqrt{3}}{3})(-\frac{\sqrt{5}}{6})$

Multiplying numerator and denominator of the first term by $\sqrt{3}$ gives us the final expression as

$\frac{\sqrt{3}}{\sqrt{3}}\frac{\sqrt{2}}{\sqrt{3}}.\frac{\sqrt{31}}{6}+(-\frac{\sqrt{3}}{3})(-\frac{\sqrt{5}}{6})=\frac{\sqrt{186}}{18}+\frac{\sqrt{15}}{18}= \frac{\sqrt{186}+\sqrt{15}}{18}$

4 0

2 years ago

Using a = 5, b = 12 and c = 13, is triangle ABC a right angle triangle? Yes or No​

Using a = 5, b = 12 and c = 13, is triangle ABC a right angle triangle? Yes or No