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Leviafan [203]
1 year ago
9

On the grid draw the graph of y= 1- 3x for values of x from -2 to 3

Mathematics
1 answer:
tino4ka555 [31]1 year ago
5 0

Answer:

see solution

Step-by-step explanation:

equation: y = 1 - 3x

when x is -2:
y = 1 -3 * -2  = y = 1+6 = y = 7
coordinates = (x,y) = (-2,7)

when x is -1:
y = 1-3*-1 = y = 1+3 = y = 4
coordinates = (x,y) = (-1,4)

when x is 0:
y = 1 - 3 * 0 = y = 1-0 = y =1
coordinates = (x,y) = (0,1)

when x is 1
y = 1-3*1 = y = 1-3 = y = -2
coordinates = (x,y) = (1,-2)

When x is 2
y = 1-3*2 = y = 1-6 = y = -5
coordinates = (x,y) = (2,-5)

when x is 3
y = 1-3*3 = y = 1-9 = y = -8
coordinates = (x,y) = (3,-8)

Plot these points on the graph and join up the dots to see the line.

bye

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Answer:

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)}

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\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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