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2 years ago

Is (2, 1) a solution to this system of equations? 4x + 12y = 20 x + y = 3 yes no

Mathematics

2 answers:

dedylja [7]2 years ago

8 0

Answer:

yes

I hope to help you

my dear

Amanda [17]2 years ago

6 0

Answer:

Step-by-step explanation:

4(2)+12(1)=20

8 + 12 =20

20 = 20

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If f Subscript x(a,b)equals f Subscript y(a,b)equals0, does it follow that f has a local maximum or local minimum at (a,b)?

Anna [14]

Answer:

Step-by-step explanation:

The point must be a critical point but it could be a saddle point. If the point is a saddle point it would not be neither a maximum nor a minimum. So it must be critical but it does not follow directly that it has a local maximum or local minimum.

Therefore D. (a,b) would be a candidate, but is not necessarily a maximum or minimum. It could be a saddle.

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3 years ago

Bridget drives at a speed of 60 miles per hour. How long will it take to drive 270 miles?

Mademuasel [1]

Answer:

4 hours and 30 minutes.

Step-by-step explanation:

270÷60=4.5

3 0

3 years ago

Read 2 more answers

What is the slope of the line l described by the equation below? y=-2x+4

Masteriza [31]

Answer:

The slope of the line would be -2.

Step-by-step explanation:

Because the equation is a line, the slope is the same throughout the entire graph on the equation.

Lets let f(x)=y=-2x+4.

f(0) = -2(0) + 4 = 4

f(0)=4

f(1) = -2(1) + 4 = 2

f(1)=2

Use the equation to find slope.

m(slope) = $\frac{2-4}{1-0}$ = $\frac{-2}{1}$ = -2

The slope would be -2.

Or:

Use the equation y = mx + b where m=slope.

In the equation y=-2x+4, m=-2.

The slope is -2.

6 0

3 years ago

The north pole is in the middle of the ocean. A person at sea level at the north pole would be 3,949 miles from the center of Ea

damaskus [11]

Answer:

I LOVE YOU

Step-by-step explanation:

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3 years ago

Read 2 more answers

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