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2 years ago

33. A labourer is engaged for 20 days on the condition that he will receive 120 for each day he

Mathematics

1 answer:

DaniilM [7]2 years ago

6 0

Answer:

false

Step-by-step explanation:

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Solve the linear system using substitution

levacccp [35]

Hello there!

3x - y = 3
-x + y = 3

We need to solve 3x - y = 3 for y

Let start by adding -3x to both sides

3x - y = 3

3x - 3x - y = 3 - 3x

-y = 3 - 3x
We cannot leave the variable with a negative sign. We must multiply both sides by -1

-y * -1= (3 - 3x)-1

y = 3x - 3

Substitute 3x -3 for y in -x+y=3

-x + y = 3

-x + 3x - 3 = 3

2x - 3 = 3

2x = 3 + 3

2x = 6

x = 6/2

x = 3

Now substitute 3 for x in y=3x -3

y = 3x - 3

y = 3(3) - 3

y= 9 - 3

y= 6

Thus,

The answer is x=3 and y=6

It is always my pleasure to help students like you :)

As always, I am here to help!

7 0

4 years ago

Find the maximum and minimum values attained by f(x, y, z) = 5xyz on the unit ball x2 + y2 + z2 ≤ 1.

Allushta [10]

Check for critical points within the unit ball by solving for when the first-order partial derivatives vanish:
$f_x=5yz=0\implies y=0\text{ or }z=0$
$f_y=5xz=0\implies x=0\text{ or }z=0$
$f_z=5xy=0\implies x=0\text{ or }y=0$

Taken together, we find that (0, 0, 0) appears to be the only critical point on $f$ within the ball. At this point, we have $f(0,0,0)=0$ .

Now let's use the method of Lagrange multipliers to look for critical points on the boundary. We have the Lagrangian

$L(x,y,z,\lambda)=5xyz+\lambda(x^2+y^2+z^2-1)$

with partial derivatives (set to 0)

$L_x=5yz+2\lambda x=0$
$L_y=5xz+2\lambda y=0$
$L_z=5xy+2\lambda z=0$
$L_\lambda=x^2+y^2+z^2-1=0$

We then observe that

$xL_x+yL_y+zL_z=0\implies15xyz+2\lambda=0\implies\lambda=-\dfrac{15xyz}2$

So, ignoring the critical point we've already found at (0, 0, 0),

$5yz+2\left(-\dfrac{15xyz}2\right)x=0\implies5yz(1-3x^2)=0\implies x=\pm\dfrac1{\sqrt3}$
$5xz+2\left(-\dfrac{15xyz}2\right)y=0\implies5xz(1-3y^2)=0\implies y=\pm\dfrac1{\sqrt3}$
$5xy+2\left(-\dfrac{15xyz}2\right)z=0\implies5xy(1-3z^2)=0\implies z=\pm\dfrac1{\sqrt3}$

So ultimately, we have 9 critical points - 1 at the origin (0, 0, 0), and 8 at the various combinations of $\left(\pm\dfrac1{\sqrt3},\pm\dfrac1{\sqrt3},\pm\dfrac1{\sqrt3}\right)$ , at which points we get a value of either of $\pm\dfrac5{\sqrt3}$ , with the maximum being the positive value and the minimum being the negative one.

5 0

3 years ago

Pleasee! Which expression is equivalent to the given expression? Assume the denominator does not equal zero.

Ilia_Sergeevich [38]

Answer:

A. 2y^4 over x^2

Step-by-step explanation:

4x^4y^6 ÷ 7x^8y^2

First, you will find the GCF of the equation which is: 7x^4y^2 .

Then, you will divide both of the equation by the GCF which will become:

14x^4y^6 ÷ 7x^4y^2 = 2y^4

7x^8y^2 ÷ 7x^4y^2 = x^2

Hence, the final answer is 2y^4 over x^2

6 0

3 years ago

Read 2 more answers

Given: A, B, and C What is the value of X in the matrix equation AX + B = C?

Ksju [112]

Answer:

Option (2)

Step-by-step explanation:

Given expression is, AX + B = C

$A=\begin{bmatrix}-3 & -4\\ 1 & 0\end{bmatrix}$

$B=\begin{bmatrix}-7 & -9\\ 4 & -1\end{bmatrix}$

$C=\begin{bmatrix}-42 & -20\\ 5 & 4\end{bmatrix}$

AX + B = C

AX = C - B

C - B = $\begin{bmatrix}-42 & -20\\ 5 & 4\end{bmatrix}-\begin{bmatrix}-7 & -9\\ 4 & -1\end{bmatrix}$ = $\begin{bmatrix}-42+7 & -20+9\\ 5-4 & 4+1\end{bmatrix}$