Pleaseee help !!!!!

Mathematics

1 answer:

BigorU [14]3 years ago

3 0

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1 kg = 2.2lb<br><br> 6.3lb = ___kg<br> show work

poizon [28]

$\dfrac{1kg}{xkg}=\dfrac{2.2lb}{6.3lb}\ \ \ \ |cross\ multiply\\\\2.2x=1\cdot6.3\\\\2.2x=6.3\ \ \ |divide\ both\ sides\ by\ 2.2\\\\x=\dfrac{6.3}{2.2}\\\\x=\dfrac{63}{22}\\\\\boxed{6.3lb=2\frac{19}{22}kg\approx2.86kg}$

4 0

3 years ago

Find the greatest solution for x+y when x^2+y^2 = 7, x^3+y^3=10

damaskus [11]

Answer:

Step-by-step explanation:

set

$f(x,y)=x+y\\$

constrain:

$g(x,y)=x^2+y^2 = 7\\h(x,y)=x^3+y^3=10$

Partial derivatives:

$f_{x}=1\\f_{y} =1 \\g_{x}=2x \\g_{y}=2y\\h_{x}=3x^2 \\h_{y}=3y^2$

Lagrange multiplier:

$grad(f)=a*grad(g)+b*grad(h)\\$

$\left[\begin{array}{ccc}1\\1\end{array}\right]=a\left[\begin{array}{ccc}2x\\2y\end{array}\right]+b\left[\begin{array}{ccc}3x^2\\3y^2\end{array}\right]$

4 equations:

$1=2ax+3bx^2\\1=2ay+3by^2\\x^2+y^2=7\\x^3+y^3=10$

By solving:

$a=4/9\\b=-2/27\\x+y=4$

Second mathod:

Solve for x^2+y^2 = 7, x^3+y^3=10 first:

$x=\frac{1}{2} -\frac{\sqrt{13}}{2} \ or \ y=\frac{1}{2} +\frac{\sqrt{13}}{2} \\x=\frac{1}{2} +\frac{\sqrt{13}}{2} \ or \ y=\frac{1}{2} -\frac{\sqrt{13}}{2} \\x+y=-5\ or\ 1 \or\ 4$