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Masteriza [31]
3 years ago
13

Round 3,101 to the nearest hundred.

Mathematics
1 answer:
ololo11 [35]3 years ago
6 0

Answer:

it is 3,100

Step-by-step explanation:

bc you cant round up 0 so your answer is 3.100

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There are 16 juniors and 8 seniors in the Chess Club. If the club members decide to send 9 juniors to a tournament, how many dif
Licemer1 [7]

Answer:

(16 over 9) = 16!/(9!*7!) = 11440

3 0
4 years ago
Help this is timed!!!
nataly862011 [7]

Answer:

C. (4,2)

Step-by-step explanation:

The point where the lines intersect is (4, 2)

4 0
3 years ago
Read 2 more answers
2000 is blank 10 times as much what is the answer and 2000 of 1/10
Molodets [167]
2,000 is 10 times as much as 200. To find the answer we can divide 2,000 by 10:

2,000 / 10 = 200

It means that 200 is also 1/10 of 2,000.

If you want to find out what 2,000 is 1/10 of, then just multiply it by 10:

2,000 * 10 = 20,000

2,000 is 1/10 of 20,000.
7 0
3 years ago
Are the following systems equivalent? How do you know?
Nadya [2.5K]

Answer:

<h3>No, they are not equivalent</h3>

Step-by-step explanation:

Are the following systems equivalent? How do you know?

For the equation;

2x+y= -4 ... 1

7x+7y=0 ... 2

Let us find the solution of the equation;

From 1;

y = 4 - 2x

Susbtitutw into 2;

7x + 7(4-2x) = 0

7x + 28 - 14x = 0

-7x + 28 = 0

7x = 28

x = 28/7

x = 4

Recall that y = 4-2x

y = 4 - 2(4)

y = 4 - 8

y = -4

The solution is (4, -4)

For the other equations;

5x+6y= 4 ... 1 * 1

4x+2y= -4 ..... 2 * 3

___________________

5x+6y= 4

12x+6y= -12

Subtract

5x-12x = 4 + 12

-7x = 16

x = -16/7

Since the solutions to both simultaneous equations are different, hence the systems of equations are not equivalent.

5 0
3 years ago
Find the minimum and maximum of f(x,y,z)=x2+y2+z2 subject to two constraints, x+2y+z=7 and x−y=6.
sweet-ann [11.9K]
Use the method of Lagrange multipliers. We have Lagrangian

L(x,y,z,\lambda_1,\lambda_2)=x^2+y^2+z^2+\lambda_1(x+2y+z-7)+\lambda_2(x-y-6)

with partial derivatives (set equal to 0) of

L_x=2x+\lambda_1+\lambda_2=0
L_y=2y+2\lambda_1-\lambda_2=0
L_z=2z+\lambda_1=0
L_{\lambda_1}=x+2y+z-7=0
L_{\lambda_2}=x-y-6=0

As x+2y+z=7, and x-y=6, we can obtain

\dfrac12L_x+L_y+\dfrac12L_z=0\implies3\lambda_1-\dfrac12\lambda_2=-7
L_x-L_y=0\implies\lambda_1-2\lambda_2=12
\begin{cases}3\lambda_1-\frac12\lambda_2=-7\\\lambda_1-2\lambda_2=12\end{cases}\implies\lambda_1=-\dfrac{40}{11},\lambda_2=-\dfrac{86}{11}

From this, we find a single critical point:

2x-\dfrac{40}{11}-\dfrac{86}{11}=0\implies x=\dfrac{63}{11}
\dfrac{63}{11}-y=6\implies y=-\dfrac3{11}
\dfrac{63}{11}-\dfrac6{11}+z=7\implies z=\dfrac{20}{11}

At this point, we have a value of

f\left(\dfrac{63}{11},-\dfrac3{11},\dfrac{20}{11}\right)=\dfrac{398}{11}

To determine what kind of extremum occurs at this point, we check the Hessian of f(x,y,z):

\mathbf H(x,y,z)=\begin{bmatrix}f_{xx}&f_{xy}&f_{xz}\\f_{yx}&f_{yy}&f_{yz}\\f_{zx}&f_{zy}&f_{zz}\end{bmatrix}=\begin{bmatrix}2&0&0\\0&2&0\\0&0&2\end{bmatrix}

We observe that \det\mathbf H(x,y,z)=8>0 at any point (x,y,z), and that the eigenvalues of this matrix are all positive (2 with multiplicity 3), so \mathbf H is positive definite. By the second partial derivative test, this means f(x,y,z) attains a minimum at this critical point. Meanwhile, f has no maximum value.
5 0
3 years ago
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