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Ainat [17]
2 years ago
9

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:

no

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)equals​0, does it follow that f has a local maximum or local minimum at​ (a,b)?
Anna [14]

Answer:

D.

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.

5 0
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

<em>The slope would be -2.</em>

<em />

Or:

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

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

<em>The slope is -2.</em>

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:

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Step-by-step explanation:

3 0
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
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