All categories

2 years ago

Need help (pic included)

Mathematics

1 answer:

ValentinkaMS [17]2 years ago

4 0

Answer:

(-4,-2)

x=-4

y=-2

Step-by-step explanation:

-3x-4y=20

3(x-10y=16)=3x-30y=48

The reason I multiplied 3 to the second equation is for when we add the equations together the x will cancel out.

-3x-4y=20

+ <u>3x-30y=48</u>

-34y=68

Divide -34 from both sides.

y=-2

To find x you need to plug in -2 for y into one of the equations.

x-10y=16

x-10(-2)=16

Remember a negative times a negative equals a positive.

x+20=16

Subtract 20 from both sides.

x=-4

Hope this helps!

If not, I am sorry.

You might be interested in

Geometry math question no Guessing and Please show work thank you

Fiesta28 [93]

Theorem

In a triangle, the measure of an exterior angle equals the sum of the measures of its two remote interior angles.

x + y = z

4n - 18 + n + 8 = 133 - 6n

5n - 10 = 133 - 6n

11n = 143

n = 13

z = 133 - 6n = 133 - 6(13) = 133 - 78 = 55

Answer: C. 55

3 0

3 years ago

Read 2 more answers

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

Setler79 [48]

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:

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

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

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

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