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3 years ago

Solve ln(6x-1) = -4 for x

Mathematics

2 answers:

Reika [66]3 years ago

5 0

Answer: Isolate the variable by dividing each side by factors that don't contain the variable.

Exact Form:

−

Decimal Form:

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0.5

Step-by-step explanation:

mars1129 [50]3 years ago

4 0

Answer:

answer,:d=-95

Step-by-step explanation:

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6x^{2}-x+4=0

6x2−x+4=0

a={\color{#c92786}{6}}

a=6

−

b={\color{#e8710a}{-1}}

b=−1

c={\color{#129eaf}{4}}

c=4

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(

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)

(

−

)

−

⋅

√

⋅

x=\frac{-({\color{#e8710a}{-1}}) \pm \sqrt{({\color{#e8710a}{-1}})^{2}-4 \cdot {\color{#c92786}{6}} \cdot {\color{#129eaf}{4}}}}{2 \cdot {\color{#c92786}{6}}}

x=2⋅6−(−1)±(−1)2−4⋅6⋅4

Simplify

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All boxes with a square base, an open top, and a volume of 60 ft cubed have a surface area given by S(x)equalsx squared plus

Karo-lina-s [1.5K]

Answer:

The absolute minimum of the surface area function on the interval $(0,\infty)$ is $S(2\sqrt[3]{15})=12\cdot \:15^{\frac{2}{3}} \:ft^2$

The dimensions of the box with minimum surface area are: the base edge $x=2\sqrt[3]{15}\:ft$ and the height $h=\sqrt[3]{15} \:ft$

Step-by-step explanation:

We are given the surface area of a box $S(x)=x^2+\frac{240}{x}$ where x is the length of the sides of the base.

Our goal is to find the absolute minimum of the the surface area function on the interval $(0,\infty)$ and the dimensions of the box with minimum surface area.

1. To find the absolute minimum you must find the derivative of the surface area ( $S'(x)$ ) and find the critical points of the derivative ( $S'(x)=0$ ).

$\frac{d}{dx} S(x)=\frac{d}{dx}(x^2+\frac{240}{x})\\\\\frac{d}{dx} S(x)=\frac{d}{dx}\left(x^2\right)+\frac{d}{dx}\left(\frac{240}{x}\right)\\\\S'(x)=2x-\frac{240}{x^2}$

Next,

$2x-\frac{240}{x^2}=0\\2xx^2-\frac{240}{x^2}x^2=0\cdot \:x^2\\2x^3-240=0\\x^3=120$

There is a undefined solution $x=0$ and a real solution $x=2\sqrt[3]{15}$ . These point divide the number line into two intervals $(0,2\sqrt[3]{15})$ and $(2\sqrt[3]{15}, \infty)$

Evaluate S'(x) at each interval to see if it's positive or negative on that interval.

$\begin{array}{cccc}Interval&x-value&S'(x)&Verdict\\(0,2\sqrt[3]{15}) &2&-56&decreasing\\(2\sqrt[3]{15}, \infty)&6&\frac{16}{3}&increasing \end{array}$

An extremum point would be a point where f(x) is defined and f'(x) changes signs.

We can see from the table that f(x) decreases before $x=2\sqrt[3]{15}$ , increases after it, and is defined at $x=2\sqrt[3]{15}$ . So f(x) has a relative minimum point at $x=2\sqrt[3]{15}$ .

To confirm that this is the point of an absolute minimum we need to find the second derivative of the surface area and show that is positive for $x=2\sqrt[3]{15}$ .

$\frac{d}{dx} S'(x)=\frac{d}{dx}(2x-\frac{240}{x^2})\\\\S''(x) =\frac{d}{dx}\left(2x\right)-\frac{d}{dx}\left(\frac{240}{x^2}\right)\\\\S''(x) =2+\frac{480}{x^3}$

and for $x=2\sqrt[3]{15}$ we get:

$2+\frac{480}{\left(2\sqrt[3]{15}\right)^3}\\\\\frac{480}{\left(2\sqrt[3]{15}\right)^3}=2^2\\\\2+4=6>0$

Therefore S(x) has a minimum at $x=2\sqrt[3]{15}$ which is:

$S(2\sqrt[3]{15})=(2\sqrt[3]{15})^2+\frac{240}{2\sqrt[3]{15}} \\\\2^2\cdot \:15^{\frac{2}{3}}+2^3\cdot \:15^{\frac{2}{3}}\\\\4\cdot \:15^{\frac{2}{3}}+8\cdot \:15^{\frac{2}{3}}\\\\S(2\sqrt[3]{15})=12\cdot \:15^{\frac{2}{3}} \:ft^2$

2. To find the third dimension of the box with minimum surface area:

We know that the volume is 60 $ft^3$ and the volume of a box with a square base is $V=x^2h$ , we solve for h

$h=\frac{V}{x^2}$

Substituting V = 60 $ft^3$ and $x=2\sqrt[3]{15}$

$h=\frac{60}{(2\sqrt[3]{15})^2}\\\\h=\frac{60}{2^2\cdot \:15^{\frac{2}{3}}}\\\\h=\sqrt[3]{15} \:ft$

The dimension are the base edge $x=2\sqrt[3]{15}\:ft$ and the height $h=\sqrt[3]{15} \:ft$

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3 years ago

PLS HELP I WILL MARK YOU THE BRAINLEST<br> Graph y>1−3x.

viva [34]

Answer:

Step-by-step explanation: ny line can be graphed using two points. Select two

values, and plug them into the equation to find the corresponding

values.

Graph the line using the slope and the y-intercept, or the points.

Slope:

−

y-intercept:

(

)

6 0

3 years ago

Which shows the first step in the solution to the<br> equation log₂x + log₂(x - 6) = 4?

iren [92.7K]

log was used calculate big numbers before calculators

log is a re-arranged way to show a number with an exponent

example

log₂ 16 = 4 means 2^4 = 16

logx(Z) = y means x^y=Z

log(x-6)/log(2) + log(x)/log(2) = 4

(log(x-6)+ log(x))/log(2) = 4

(log(x-6)+ log(x)) = 4log(2)

(log(x-6)x) = log(16)

x=8

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2 years ago

You are scheduled to work 34 hours per week at your new job, where you earn $13.1 per

Lyrx [107]

13.1 an hour, 34 hours a week
13.1 x 34 = 445.4 a week
to find out how many weeks, divide 4300 and 445.4 which is 9.65 weeks, which is technically 10 weeks

it would take 10 weeks to earn 4300

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1 year ago

PLEASE LED ME THE ANSWER<br> ILL GIVE YOU 20 POINTS

krok68 [10]

Answer:

DEC+DEF=180

DEC=180-116

DEC=64°

in triangle DCE

angle D+angle C+angle E=180

7y+6+4y+64=180

11y+70=180

11y=180-70

11y=110

y=110/11

y=10°

angle C=4y

=4(10)

=40°

3 0

3 years ago