Which expression is equivalent to the given expression?<br><br> 6y−12

I need help with problem 1 with a through explanation and solution please <br><br>

ki77a [65]

Explanation:

The cubic ...

f(x) = ax³ +bx² +cx +d

has derivatives ...

f'(x) = 3ax² +2bx +c

f''(x) = 6ax +2b

By definition, there will be a point of inflection where the second derivative is zero (changes sign). The second derivative is a linear equation in x, so can only have one zero. Since it is given that a≠0, we are assured that the line described by f''(x) will cross the x-axis at ...

f''(x) = 0 = 6ax +2b ⇒ x = -b/(3a)

The single point of inflection is at x = -b/(3a).

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The cubic will have a local extreme where the first derivative is zero and the second derivative is not zero. These will only occur when the discriminant of the first derivative quadratic is positive. Their location can be found by applying the quadratic formula to the first derivative.

$x=\dfrac{-2b\pm\sqrt{(2b)^2-4(3a)(c)}}{2(3a)} = \dfrac{-2b\pm\sqrt{4b^2-12ac}}{6a}\\\\x=\dfrac{-b\pm\sqrt{b^2-3ac}}{3a}\qquad\text{extreme point locations when $b^2>3ac$}$

There will be zero or two local extremes. A local extreme cannot occur at the point of inflection, which is where the formula would tell you it is when there is only one.

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Part A tells you the point of inflection is at x= -b/(3a).

Part B tells you the midpoint of the local extremes is x = -b/(3a). (This is half the sum of the x-values of the extreme points.) You will notice these are the same point.

The extreme points are located symmetrically about their midpoint, so are located symmetrically about the point of inflection.

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Additional comment

There are other interesting features of cubics with two local extremes. The points where the horizontal tangents meet the graph, together with the point of inflection, have equally-spaced x-coordinates. The point of inflection is the midpoint, both horizontally and vertically, between the local extreme points.

6 0

3 years ago

Calculus 2. Please help

Anarel [89]

Answer:

$\displaystyle \int\limits^1_0 {\frac{1}{xe^{x^2}}} \, dx = \infty$

General Formulas and Concepts:

<u>Algebra I</u>

Exponential Rule [Rewrite]: $\displaystyle b^{-m} = \frac{1}{b^m}$

<u>Calculus</u>

Limits

Right-Side Limit: $\displaystyle \lim_{x \to c^+} f(x)$

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

Derivatives

Derivative Notation

Basic Power Rule:

f(x) = cxⁿ
f’(x) = c·nxⁿ⁻¹

Integrals

Definite Integrals

Integration Constant C

Integration Rule [Fundamental Theorem of Calculus 1]: $\displaystyle \int\limits^b_a {f(x)} \, dx = F(b) - F(a)$

Integration Property [Multiplied Constant]: $\displaystyle \int {cf(x)} \, dx = c \int {f(x)} \, dx$

U-Substitution

U-Solve

Improper Integrals

Exponential Integral Function: $\displaystyle \int {\frac{e^x}{x}} \, dx = Ei(x) + C$

Step-by-step explanation:

<u>Step 1: Define</u>

<em>Identify</em>

$\displaystyle \int\limits^1_0 {\frac{1}{xe^{x^2}} \, dx$

<u>Step 2: Integrate Pt. 1</u>

[Integral] Rewrite [Exponential Rule - Rewrite]: $\displaystyle \int\limits^1_0 {\frac{1}{xe^{x^2}} \, dx = \int\limits^1_0 {\frac{e^{-x^2}}{x} \, dx$
[Integral] Rewrite [Improper Integral]: $\displaystyle \int\limits^1_0 {\frac{1}{xe^{x^2}} \, dx = \lim_{a \to 0^+} \int\limits^1_a {\frac{e^{-x^2}}{x} \, dx$

<u>Step 3: Integrate Pt. 2</u>

<em>Identify variables for u-substitution.</em>

Set: $\displaystyle u = -x^2$
Differentiate [Basic Power Rule]: $\displaystyle \frac{du}{dx} = -2x$
[Derivative] Rewrite: $\displaystyle du = -2x \ dx$

<em>Rewrite u-substitution to format u-solve.</em>

Rewrite <em>du</em>: $\displaystyle dx = \frac{-1}{2x} \ dx$

<u>Step 4: Integrate Pt. 3</u>

[Integral] Rewrite [Integration Property - Multiplied Constant]: $\displaystyle \int\limits^1_0 {\frac{1}{xe^{x^2}} \, dx = \lim_{a \to 0^+} -\int\limits^1_a {-\frac{e^{-x^2}}{x} \, dx$
[Integral] Substitute in variables: $\displaystyle \int\limits^1_0 {\frac{1}{xe^{x^2}} \, dx = \lim_{a \to 0^+} -\int\limits^1_a {\frac{e^{u}}{-2u} \, du$
[Integral] Rewrite [Integration Property - Multiplied Constant]: $\displaystyle \int\limits^1_0 {\frac{1}{xe^{x^2}} \, dx = \lim_{a \to 0^+} \frac{1}{2}\int\limits^1_a {\frac{e^{u}}{u} \, du$
[Integral] Substitute [Exponential Integral Function]: $\displaystyle \int\limits^1_0 {\frac{1}{xe^{x^2}} \, dx = \lim_{a \to 0^+} \frac{1}{2}[Ei(u)] \bigg| \limits^1_a$
Back-Substitute: $\displaystyle \int\limits^1_0 {\frac{1}{xe^{x^2}} \, dx = \lim_{a \to 0^+} \frac{1}{2}[Ei(-x^2)] \bigg| \limits^1_a$
Evaluate [Integration Rule - FTC 1]: $\displaystyle \int\limits^1_0 {\frac{1}{xe^{x^2}} \, dx = \lim_{a \to 0^+} \frac{1}{2}[Ei(-1) - Ei(a)]$
Simplify: $\displaystyle \int\limits^1_0 {\frac{1}{xe^{x^2}} \, dx = \lim_{a \to 0^+} \frac{Ei(-1) - Ei(a)}{2}$
Evaluate limit [Limit Rule - Variable Direct Substitution]: $\displaystyle \int\limits^1_0 {\frac{1}{xe^{x^2}} \, dx = \infty$

∴ $\displaystyle \int\limits^1_0 {\frac{1}{xe^{x^2}} \, dx$ diverges.

Topic: Multivariable Calculus

7 0

3 years ago

PLEASE ANSWER ASAP FOR BRAINLEST SUPER EASY QUESTION!!!!!!!!!!!!!!!!!!!!

GenaCL600 [577]

Answer:

26.8

Step-by-step explanation:

you round up 5 and make the 7 an 8

7 0

3 years ago

Read 2 more answers

(x÷(y÷z))÷((x÷y)÷z)<br>also the variables are not specified.

Bumek [7]

The answer is: z² .
__________________________
Given: <span>(x÷(y÷z))÷((x÷y)÷z) ; without any specified values for the variables;
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we shall simplify.
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We have:
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</span>(x÷(y÷z)) / ((x÷y)÷z) .
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Start with the first term; or, "numerator": (x÷(y÷z)) ;
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x ÷ (y / z) = (x / 1) * (z / y) = (x * z) / (1 *y) = [(xz) / y ]
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Then, take the second term; or "denominator":
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((x ÷ y) ÷z ) = (x / y) / z = (x / y) * (1 / z) = (x *1) / (y *z) = [x / (zy)]
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So (x÷(y÷z)) / ((x÷y)÷z) = (x÷(y÷z)) ÷ ((x÷y)÷z) =

[(xz) / y ] ÷ [x / (zy)] = [(xz) / y ] / [x / (zy)] =
[(xz) / y ] * [(zy) / x] ;
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The 2 (two) z's "cancel out" to "1" ; and
The 2 (two) y's = "cancel out" to "1" ;
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And we are left with: z * z = z² . The answer is: z² .
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4 0

4 years ago

You do a study of hypnotherapy to determine how effective it is in increasing the number of hours of sleep subjects get each nig

brilliants [131]

Answer:

Confidence interval

$8.98-2.2\frac{1.29}{\sqrt{12}}=8.16$

$8.98+2.2\frac{1.29}{\sqrt{12}}=9.80$

And the margin of error would be:

$ME=2.2\frac{1.29}{\sqrt{12}}=0.819$

Step-by-step explanation:

For this case we have the followig dataset:

DATA: 8.2; 9.1; 7.7; 8.6; 6.9; 11.2; 10.1; 9.9; 8.9; 9.2; 7.5; 10.5

We can calculate the mean and the deviation with the following formulas:

$\bar X = \frac{\sum_{i=1}^n X_i}{n}$

$s =\sqrt{\frac{\sum_{i=1}^n (X_i -\bar X)^2}{n-1}}$

And replacing we got:

$\bar X= 8.98[/tex[tex]s = 1.29$

The confidence interval for the mean is given by the following formula:

$\bar X \pm t_{\alpha/2}\frac{s}{\sqrt{n}}$ (1)

The degrees of freedom are given by:

$df=n-1=12-1=11$

The Confidence level is 0.95 or 95%, the value of significance is $\alpha=0.05$ and $\alpha/2 =0.025$ , and the critical value would be $t_{\alpha/2}=2.20$

Repplacing the info we got:

$8.98-2.2\frac{1.29}{\sqrt{12}}=8.16$

$8.98+2.2\frac{1.29}{\sqrt{12}}=9.80$ And the margin of error would be:

$ME=2.2\frac{1.29}{\sqrt{12}}=0.819$

7 0

3 years ago

Which expression is equivalent to the given expression? 6y−12