Mathematics: 5^(-x)+7=2x+4 This was on plato

5^(-x)+7=2x+4 This was on plato - id57247738496, aroland1990x

Ответ:

avelyp

02.02.2022

Below

I hope its not too complicated

$x=\frac{\text{W}_0\left(\frac{\ln \left(5\right)}{2e^{\frac{3\ln \left(5\right)}{2}}}\right)}{\ln \left(5\right)}+\frac{3}{2}$

Step-by-step explanation:

$5^{\left(-x\right)}+7=2x+4\\\\\mathrm{Prepare}\:5^{\left(-x\right)}+7=2x+4\:\mathrm{for\:Lambert\:form}:\quad 1=\left(2x-3\right)e^{\ln \left(5\right)x}\\\\\mathrm{Rewrite\:the\:equation\:with\:}\\\left(x-\frac{3}{2}\right)\ln \left(5\right)=u\mathrm{\:and\:}x=\frac{u}{\ln \left(5\right)}+\frac{3}{2}\\\\1=\left(2\left(\frac{u}{\ln \left(5\right)}+\frac{3}{2}\right)-3\right)e^{\ln \left(5\right)\left(\frac{u}{\ln \left(5\right)}+\frac{3}{2}\right)}$

$Simplify\\\\\mathrm{Rewrite}\:1=\frac{2e^{u+\frac{3}{2}\ln \left(5\right)}u}{\ln \left(5\right)}\:\\\\\mathrm{in\:Lambert\:form}:\quad \frac{e^{\frac{2u+3\ln \left(5\right)}{2}}u}{e^{\frac{3\ln \left(5\right)}{2}}}=\frac{\ln \left(5\right)}{2e^{\frac{3\ln \left(5\right)}{2}}}$

$\mathrm{Solve\:}\:\frac{e^{\frac{2u+3\ln \left(5\right)}{2}}u}{e^{\frac{3\ln \left(5\right)}{2}}}=\frac{\ln \left(5\right)}{2e^{\frac{3\ln \left(5\right)}{2}}}:\quad u=\text{W}_0\left(\frac{\ln \left(5\right)}{2e^{\frac{3\ln \left(5\right)}{2}}}\right)\\\\\mathrm{Substitute\:back}\:u=\left(x-\frac{3}{2}\right)\ln \left(5\right),\:\mathrm{solve\:for}\:x$

$\mathrm{Solve\:}\:\left(x-\frac{3}{2}\right)\ln \left(5\right)=\text{W}_0\left(\frac{\ln \left(5\right)}{2e^{\frac{3\ln \left(5\right)}{2}}}\right):\\\quad x=\frac{\text{W}_0\left(\frac{\ln \left(5\right)}{2e^{\frac{3\ln \left(5\right)}{2}}}\right)}{\ln \left(5\right)}+\frac{3}{2}$

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Ответ:

harringtonrob16

04.10.2022

Mathematics

Step-by-step explanation:

The center of the circle is the midpoint of the two end points of the diameter.

Formula

Center = (x2 + x1)/2 , (y2 + y1)/2

Givens

x2 = 4

x1 = - 10

y2 = 6

y1 = - 2

Solution

Center = (4 - 10)/2, (6 - 2)/2

Center = -6/2 , 4/2

Center = - 3 , 2

So far what you have is

(x+3)^2 + (y - 2)^2 = r^2

Now you have to find the radius.

You can use either of the endpoints to find the radius.

find the distance from (4,6) to (-3,2)

r^2 = ( (x2 - x1)^2 + (y2 - y1)^2 )

x2 = 4

x1 = -3

y2 = 6

y1 = 2

r^2 = ( (4 - -3)^2 + (6 - 2)^2 )

r^2 = ( (7)^2 + 4^2)

r^2 = ( 49 + 16)

r^2 = 65

Ultimate formula is

(x+3)^2 + (y - 2)^2 = 65

The radius is √65 = 8.06

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