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3 years ago

Please help me. 5(m-7)=-8(3m+8)

Mathematics

1 answer:

inessss [21]3 years ago

7 0

Answer:

m = -1

Step-by-step explanation:

Solve for m:

5 (m - 7) = -8 (3 m + 8)

Hint: | Write the linear polynomial on the left hand side in standard form.

Expand out terms of the left hand side:

5 m - 35 = -8 (3 m + 8)

Hint: | Write the linear polynomial on the left hand side in standard form.

Expand out terms of the right hand side:

5 m - 35 = -24 m - 64

Hint: | Move terms with m to the left hand side.

Add 24 m to both sides:

24 m + 5 m - 35 = (24 m - 24 m) - 64

Hint: | Look for the difference of two identical terms.

24 m - 24 m = 0:

24 m + 5 m - 35 = -64

Hint: | Group like terms in 24 m + 5 m - 35.

Grouping like terms, 24 m + 5 m - 35 = (5 m + 24 m) - 35:

(5 m + 24 m) - 35 = -64

Hint: | Add like terms in 5 m + 24 m.

5 m + 24 m = 29 m:

29 m - 35 = -64

Hint: | Isolate terms with m to the left hand side.

Add 35 to both sides:

29 m + (35 - 35) = 35 - 64

Hint: | Look for the difference of two identical terms.

35 - 35 = 0:

29 m = 35 - 64

Hint: | Evaluate 35 - 64.

35 - 64 = -29:

29 m = -29

Hint: | Divide both sides by a constant to simplify the equation.

Divide both sides of 29 m = -29 by 29:

(29 m)/29 = (-29)/29

Hint: | Any nonzero number divided by itself is one.

29/29 = 1:

m = (-29)/29

Hint: | Reduce (-29)/29 to lowest terms. Start by finding the GCD of -29 and 29.

The gcd of -29 and 29 is 29, so (-29)/29 = (29 (-1))/(29×1) = 29/29×-1 = -1:

Answer: m = -1

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Use Lagrange multipliers to find the maximum and minimum values of the function subject to the given constraint. (If an answer d

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The Lagrangian

$L(x,y,z,\lambda)=x^2+y^2+z^2+\lambda(x^4+y^4+z^4-13)$

has critical points where the first derivatives vanish:

$L_x=2x+4\lambda x^3=2x(1+2\lambda x^2)=0\implies x=0\text{ or }x^2=-\dfrac1{2\lambda}$

$L_y=2y+4\lambda y^3=2y(1+2\lambda y^2)=0\implies y=0\text{ or }y^2=-\dfrac1{2\lambda}$

$L_z=2z+4\lambda z^3=2z(1+2\lambda z^2)=0\implies z=0\text{ or }z^2=-\dfrac1{2\lambda}$

$L_\lambda=x^4+y^4+z^4-13=0$

We can't have $x=y=z=0$ , since that contradicts the last condition.

(0 critical points)

If two of them are zero, then the remaining variable has two possible values of $\pm\sqrt[4]{13}$ . For example, if $y=z=0$ , then $x^4=13\implies x=\pm\sqrt[4]{13}$ .

(6 critical points; 2 for each non-zero variable)

If only one of them is zero, then the squares of the remaining variables are equal and we would find $\lambda=-\frac1{\sqrt{26}}$ (taking the negative root because $x^2,y^2,z^2$ must be non-negative), and we can immediately find the critical points from there. For example, if $z=0$ , then $x^4+y^4=13$ . If both $x,y$ are non-zero, then $x^2=y^2=-\frac1{2\lambda}$ , and

$xL_x+yL_y=2(x^2+y^2)+52\lambda=-\dfrac2\lambda+52\lambda=0\implies\lambda=\pm\dfrac1{\sqrt{26}}$

$\implies x^2=\sqrt{\dfrac{13}2}\implies x=\pm\sqrt[4]{\dfrac{13}2}$

and for either choice of $x$ , we can independently choose from $y=\pm\sqrt[4]{\frac{13}2}$ .

(12 critical points; 3 ways of picking one variable to be zero, and 4 choices of sign for the remaining two variables)

If none of the variables are zero, then $x^2=y^2=z^2=-\frac1{2\lambda}$ . We have

$xL_x+yL_y+zL_z=2(x^2+y^2+z^2)+52\lambda=-\dfrac3\lambda+52\lambda=0\implies\lambda=\pm\dfrac{\sqrt{39}}{26}$

$\implies x^2=\sqrt{\dfrac{13}3}\implies x=\pm\sqrt[4]{\dfrac{13}3}$

and similary $y,z$ have the same solutions whose signs can be picked independently of one another.

(8 critical points)

Now evaluate $f$ at each critical point; you should end up with a maximum value of $\sqrt{39}$ and a minimum value of $\sqrt{13}$ (both occurring at various critical points).