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3 years ago

Logbase3 (x+2)= log base3 (2x^2-1) steps 1 through 6

1 answer:

5 0

$log_3(x+2)=log_3(2x^2-1)\\3^{log_3(x+2)}=3^{log_3(2x^2-1)}\\x+2=2x^2-1\\-x -2 -x -2\\0=2x^2-x-3\\2*-3=-6\\Test factors -3 and 2:\\-3*2=-6, -3+2=-1\\(x-3)(x+2)=0\\Use slide and divide with 2:\\(x-3/2)(x+2/2)=0\\=>\\(2x-3)(x+1)=0\\Set both equal to 0:\\x+1=0\\x=-1\\2x-3=0\\2x=3\\/2 /2\\x=3/2\\$

Answer:

x=-1, 3/2

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Simplify....... -9-6c-j

marin [14]

Answer: −6c−j−9

Step-by-step explanation:

3 0

3 years ago

Find all the solutions for the equation:

Contact [7]

$2y^2\,\mathrm dx-(x+y)^2\,\mathrm dy=0$

Divide both sides by $x^2\,\mathrm dx$ to get

$2\left(\dfrac yx\right)^2-\left(1+\dfrac yx\right)^2\dfrac{\mathrm dy}{\mathrm dx}=0$

$\dfrac{\mathrm dy}{\mathrm dx}=\dfrac{2\left(\frac yx\right)^2}{\left(1+\frac yx\right)^2}$

Substitute $v(x)=\dfrac{y(x)}x$ , so that $\dfrac{\mathrm dv(x)}{\mathrm dx}=\dfrac{x\frac{\mathrm dy(x)}{\mathrm dx}-y(x)}{x^2}$ . Then

$x\dfrac{\mathrm dv}{\mathrm dx}+v=\dfrac{2v^2}{(1+v)^2}$

$x\dfrac{\mathrm dv}{\mathrm dx}=\dfrac{2v^2-v(1+v)^2}{(1+v)^2}$

$x\dfrac{\mathrm dv}{\mathrm dx}=-\dfrac{v(1+v^2)}{(1+v)^2}$

The remaining ODE is separable. Separating the variables gives

$\dfrac{(1+v)^2}{v(1+v^2)}\,\mathrm dv=-\dfrac{\mathrm dx}x$

Integrate both sides. On the left, split up the integrand into partial fractions.

$\dfrac{(1+v)^2}{v(1+v^2)}=\dfrac{v^2+2v+1}{v(v^2+1)}=\dfrac av+\dfrac{bv+c}{v^2+1}$

$\implies v^2+2v+1=a(v^2+1)+(bv+c)v$

$\implies v^2+2v+1=(a+b)v^2+cv+a$

$\implies a=1,b=0,c=2$

Then

$\displaystyle\int\frac{(1+v)^2}{v(1+v^2)}\,\mathrm dv=\int\left(\frac1v+\frac2{v^2+1}\right)\,\mathrm dv=\ln|v|+2\tan^{-1}v$

On the right, we have

$\displaystyle-\int\frac{\mathrm dx}x=-\ln|x|+C$

Solving for $v(x)$ explicitly is unlikely to succeed, so we leave the solution in implicit form,

$\ln|v(x)|+2\tan^{-1}v(x)=-\ln|x|+C$

and finally solve in terms of $y(x)$ by replacing $v(x)=\dfrac{y(x)}x$ :

$\ln\left|\frac{y(x)}x\right|+2\tan^{-1}\dfrac{y(x)}x=-\ln|x|+C$

$\ln|y(x)|-\ln|x|+2\tan^{-1}\dfrac{y(x)}x=-\ln|x|+C$

$\boxed{\ln|y(x)|+2\tan^{-1}\dfrac{y(x)}x=C}$

7 0

3 years ago

Tim and Cynthia leave Cynthia's house at the same time. Tim drives north and Cynthia drives west. Tim's average speed is 10 mph

RSB [31]

Answer:

54.2 mph

Step-by-step explanation:

Since the drivers leave at the same time and travel in directions 90° from each other, their separation speed can be found using the Pythagorean theorem. Let c represent Cynthia's average speed. Then (c-10) will represent Tim's average speed. Their combined separation speed is 70 miles per hour, so we can write ...

70² = c² +(c -10)²

4900 = 2c² -20c +100 . . . . eliminate parentheses

c² -10c -2400 = 0 . . . . . . . . divide by 2; put in standard form

(c -5)² -2425 = 0 . . . . . . . . . rearrange to vertex form

c = 5 ±5√97 . . . . . . . . . . . . solve for c, simplify radical

c ≈ 54.244 ≈ 54.2 . . . . . . . use the positive solution

Cynthia's average speed was 54.2 miles per hour.

6 0

3 years ago

The volume of this prism is 4375cm3. The area of the cross-section is 125cm2. Work out x

fomenos

Answer:

x is 35 cm

Step-by-step explanation:

To find the value of x, we will divide the volume by the cross-section

Mathematically, that will be;

4375/125 = 35 cm

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3 years ago

Determine the model that best describes the data and then write an equation for the function that models the data.

zhannawk [14.2K]

Answer:

s = 4·3^t

Step-by-step explanation:

After the first term, each term in the sequence is 3 times the previous one. This is a description of a geometric sequence. Here, the first term is 4, so the general term is ...

an = 4·3^n . . . . for n counting from 0

We can name the function "s" for "speed" and use "t" for the time in seconds. Then the equation is ...

s = 4·3^t

4 0

4 years ago