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3 years ago

5

Solve log2(x – 1) = log12(x – 1) by graphing.

2 answers:

e-lub [12.9K]3 years ago

5 0

$\log_2 (x-1) = \log_{12} (x-1)~\text{ only if }~(x-1 )=1\\\\
\Longrightarrow ~~\boxed{x =2}$

Leto [7]3 years ago

5 0

Answer:

By graphing ; $\log_2 (x-1) = \log_{12} (x-1)$ only for x =2

Step-by-step explanation:

Solve: $\log_2 (x-1) = \log_{12} (x-1)$

let $y_1=\log_2 (x-1)$ and $y_2=\log_{12} (x-1)$

To find the x for which $y_1= y_2$

A graph of these $y_1=\log_2 (x-1)$ and $y_2=\log_{12} (x-1)$ shows us that the graph intersect.

This implies that there is a single (x, y) value that satisfies both equations.

i.,e (2, 0)

Therefore, $\log_2 (x-1) = \log_{12} (x-1)$ only when x =2

You can see the graph as shown below:

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A customer bought a car and paid $1,080 in sales tax.The rate is 6%?

Ghella [55]

The price of car is $18000

Step-by-step explanation:

Given

Sales tax on car = $1080

Rate = 6%

Let x be the price of the car

We know that as the rate of sale tax is 6%, 6% of x is 1080

Using mathematical notation

0.06x = 1080

Dividing both sides by 0.06

$\frac{0.06x}{0.06} = \frac{1080}{0.06}\\\\x = 18000$

Hence,

The price of car is $18000

Keywords: Rate of tax, Percentage

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7 0

3 years ago

What is 1/3 of 18 kg?

weqwewe [10]

18*(1/3) is the same as 18/3.
The answer is 6kg.

8 0

4 years ago

Read 2 more answers

Do you agree with either one of them? Why or why not?

Annette [7]

Answer: Neither one is correct

Step-by-step explanation:

We don't know what we are actually comparing to: is it inches:feet or feet:inches? With the info provided, we can't decide which one is correct.

Outcomes:

1. Inches:feet constant of proportionality is 12,

2.feet:inches constant of proportionality is 1/12

8 0

1 year ago

Read 2 more answers

Im giving 100 Points to anyone who could solve this for me please.

myrzilka [38]

Answer:

AB = 27.2

BC = 33.5

AC = 50.4

∠A = 38°

∠ABC = 112°

Step-by-step explanation:

<u>Trigonometric ratios</u>

$\sf \sin(\theta)=\dfrac{O}{H}\quad\cos(\theta)=\dfrac{A}{H}\quad\tan(\theta)=\dfrac{O}{A}$

where:

$\theta$ is the angle
O is the side opposite the angle
A is the side adjacent the angle
H is the hypotenuse (the side opposite the right angle)

$\implies \sf \cos(30^{\circ})=\dfrac{29}{BC}$

$\implies \sf BC=\dfrac{29}{\cos(30^{\circ})}$

$\implies \sf BC=\dfrac{58\sqrt{3}}{3}=33.5\:(nearest\:tenth)$

$\implies \sf \tan(30^{\circ})=\dfrac{BD}{29}$

$\implies \sf BD=29\tan(30^{\circ})$

$\implies \sf BD=\dfrac{29\sqrt{3}}{3}$

$\implies \sf \sin(38^{\circ})=\dfrac{BD}{AB}$

$\implies \sf AB=\dfrac{\dfrac{29\sqrt{3}}{3}}{\sin(38^{\circ})}$

$\implies \sf AB=27.2\:(nearest\:tenth)$

$\implies \sf \tan(38^{\circ})=\dfrac{BD}{AD}$

$\implies \sf AD=\dfrac{\dfrac{29\sqrt{3}}{3}}{\tan(38^{\circ})}$

$\implies \sf AD=21.4\:(nearest\:tenth)$

$\implies \sf AC=AD+DC=21.4+29=50.4$

The interior angles of a triangle sum to 180°

⇒ ∠A + 52° + 90° = 180°

⇒ ∠A = 180° - 90° - 52°

⇒ ∠A = 38°

⇒ ∠ABC + 38° + 30° = 180°

⇒ ∠ABC = 180° - 38° - 30°

⇒ ∠ABC = 112°

**I have checked the measures using a graphing programme - see attached**

8 0

2 years ago

Find the critical points of the surface f(x, y) = x3 - 6xy + y3 and determine their nature.

Vedmedyk [2.9K]

Compute the gradient of $f$ .

$\nabla f(x,y) = \left\langle 3x^2 - 6y, -6x + 3y^2\right\rangle$

Set this equal to the zero vector and solve for the critical points.

$3x^2-6y = 0 \implies x^2 = 2y$

$-6x+3y^2=0 \implies y^2 = 2x \implies y = \pm\sqrt{2x}$

$\implies x^2 = \pm2\sqrt{2x}$

$\implies x^4 = 8x$

$\implies x^4 - 8x = 0$

$\implies x (x-2) (x^2 + 2x + 4) = 0$

$\implies x = 0 \text{ or } x-2 = 0 \text{ or } x^2 + 2x + 4 = 0$

$\implies x = 0 \text{ or } x = 2 \text{ or } (x+1)^2 + 3 = 0$

The last case has no real solution, so we can ignore it.

Now,

$x=0 \implies 0^2 = 2y \implies y=0$

$x=2 \implies 2^2 = 2y \implies y=2$

so we have two critical points (0, 0) and (2, 2).

Compute the Hessian matrix (i.e. Jacobian of the gradient).

$H(x,y) = \begin{bmatrix} 6x & -6 \\ -6 & 6y \end{bmatrix}$

Check the sign of the determinant of the Hessian at each of the critical points.

$\det H(0,0) = \begin{vmatrix} 0 & -6 \\ -6 & 0 \end{vmatrix} = -36 < 0$

which indicates a saddle point at (0, 0);

$\det H(2,2) = \begin{vmatrix} 12 & -6 \\ -6 & 12 \end{vmatrix} = 108 > 0$

We also have $f_{xx}(2,2) = 12 > 0$ , which together indicate a local minimum at (2, 2).

3 0

2 years ago