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3 years ago

|6m−2|=0 please help

Mathematics

2 answers:

Deffense [45]3 years ago

6 0

Answer:

m = 1/3

Step-by-step explanation:

stellarik [79]3 years ago

3 0

Answer: m = 1/3

Step-by-step explanation: Lets solve the equation |6m - 2| = 0.

|6m - 2| = 0

Find absolute values of 6m - 2

-6m + 2 = 0 (Now solve both sides of the equation.)

-6m + 2 - 2 = 0 - 2

-6m = -2

-6m/-6 = -2/-6

m = 1/3

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Which are the solutions of x2 = 19x + 1?

Finger [1]

Answer:

$(\frac{19-\sqrt{365}} {2},\frac{19+\sqrt{365}} {2})$

Step-by-step explanation:

we have

$x^2=19x+1$

we know that

The formula to solve a quadratic equation of the form

$ax^{2} +bx+c=0$

is equal to

$x=\frac{-b\pm\sqrt{b^{2}-4ac}} {2a}$

in this problem we have

$-x^{2}-19x-1=0$

$a=1\\b=-19\\c=-1$

substitute in the formula

$x=\frac{-(-19)\pm\sqrt{-19^{2}-4(1)(-1)}} {2(1)}$

$x=\frac{19\pm\sqrt{365}} {2}$

$x=\frac{19+\sqrt{365}} {2}$

$x=\frac{19-\sqrt{365}} {2}$

$(\frac{19-\sqrt{365}} {2},\frac{19+\sqrt{365}} {2})$

therefore

StartFraction 19 minus StartRoot 365 EndRoot Over 2 EndFraction comma StartFraction 19 + StartRoot 365 EndRoot Over 2 EndFraction

6 0

3 years ago

Read 2 more answers

The velocity v of blood that flows in a blood vessel with radius r and length ℓ at a distance r from the central axis is v(r) =

pogonyaev

$v(r)=4P\eta\ell(R^2-r^2)$

(assuming $R$ is the radius of the vessel, and that $0\le r\le R$ )

The average velocity of the blood is given by

$\displaystyle\frac1{R-0}\int_{r=0}^{r=R}v(r)\,\mathrm dr$
$=\displaystyle\frac{4P\eta\ell}R\int_{r=0}^{r=R}(R^2-r^2)\,\mathrm dr$
$=\dfrac{4P\eta\ell}R\left(\dfrac23R^3\right)$
$=\dfrac{8P\eta\ell}3R^2$