Shawna bought a bag of candy. She ate of the bag on Friday and of the bag on Saturday. What portion of the bag is left?

Mathematics

1 answer:

vodka [1.7K]3 years ago

5 0

Answer:

I believe it would be A hope this helps!

Step-by-step explanation:

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ONCE I HIT 30 ILL GIVE YOU 100 Points if i get a good grade Maybe even more

N76 [4]

Answer:

The correct anwser is 2x+5<4

Step-by-step explanation:

8 0

3 years ago

Use Lagrange multipliers to find the maximum and minimum values of

marusya05 [52]

The Lagrangian is

$L(x,y,\lambda)=x+8y+\lambda(x^2+y^2-4)$

It has critical points where the first order derivatives vanish:

$L_x=1+2\lambda x=0\implies\lambda=-\dfrac1{2x}$

$L_y=8+2\lambda y=0\implies\lambda=-\dfrac4y$

$L_\lambda=x^2+y^2-4=0$

From the first two equations we get

$-\dfrac1{2x}=-\dfrac4y\implies y=8x$

Then

$x^2+y^2=65x^2=4\implies x=\pm\dfrac2{\sqrt{65}}\implies y=\pm\dfrac{16}{\sqrt{65}}$

At these critical points, we have

$f\left(\dfrac2{\sqrt{65}},\dfrac{16}{\sqrt{65}}\right)=2\sqrt{65}\approx16.125$ (maximum)

$f\left(-\dfrac2{\sqrt{65}},-\dfrac{16}{\sqrt{65}}\right)=-2\sqrt{65}\approx-16.125$ (minimum)

5 0

3 years ago

How to evaluate the limit

anzhelika [568]

$\displaystyle\lim_{x\to2}\frac{x^2-x+6}{x+2}$

Both the numerator and denominator are continuous at $x=2$ , which means the quotient rule for limits applies:

$\dfrac{\displaystyle\lim_{x\to2}(x^2-x+6)}{\displaystyle\lim_{x\to2}(x+2)}=\dfrac{2^2-2+6}{2+2}=\dfrac84=2$

Perhaps you meant to write that $x\to-2$ instead? In that case, you would have

$\displaystyle\lim_{x\to-2}\frac{x^2-x+6}{x+2}=\lim_{x\to-2}\frac{(x+2)(x-3)}{x+2}=\lim_{x\to-2}(x-3)=-2-3=-5$