IN the diagram Below AB is parallel to CD what is the value of y ? picture is below

Mathematics

1 answer:

satela [25.4K]2 years ago

7 0

Similar to the other question, because AB is parallel to CD, you know the 72º angle corresponds to the angle on top of y.

Because that angle above y and y together make a straight line (and a straight line measures 180º), you have
72 + y = 180

Solving that for y, you get y = 108º

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-27+9-3+1 using sigma notation

Vesna [10]

Do you have question choices if not then I'll try I guess...... -27+9-3+1 36-3+1 3

4 0

3 years ago

# 34 OK I'm stuck on it

Yuki888 [10]

There are two ways you can figure this out.

1: 7.95-3.2-1.05= 3.7

2: 3.2+1.05+(All the choices)=7.95
So you do that till the one of the choices equal 7.95.

So overall, it is D. 3.70

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2 years ago

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barxatty [35]

Answer:

umm

Step-by-step explanation:

banana

5 0

2 years ago

Use Lagrange multipliers to find the maximum and minimum values of f(x,y)=x+4y subject to the constraint x2+y2=9, if such values

Vesnalui [34]

The Lagrangian is

$L(x,y,\lambda)=x+4y+\lambda(x^2+y^2-9)$

with critical points where the partial derivatives vanish.

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

$L_y=4+2\lambda y=0\implies y=-\dfrac2\lambda$

$L_\lambda=x^2+y^2-9=0$

Substitute $x,y$ into the last equation and solve for $\lambda$ :

$\left(-\dfrac1{2\lambda}\right)^2+\left(-\dfrac2\lambda\right)^2=9\implies\lambda=\pm\dfrac{\sqrt{17}}6$

Then we get two critical points,

$(x,y)=\left(-\dfrac3{\sqrt{17}},-\dfrac{12}{\sqrt{17}}\right)\text{ and }(x,y)=\left(\dfrac3{\sqrt{17}},\dfrac{12}{\sqrt{17}}\right)$

We get an absolute maximum of $3\sqrt{17}\approx12.369$ at the second point, and an absolute minimum of $-3\sqrt{17}\approx-12.369$ at the first point.