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3 years ago

PLEASE HELP DUE SOON

Mathematics

1 answer:

Anika [276]3 years ago

4 0

Answer:

x = 36

Step-by-step explanation:

The total angles of a triangle is 180 degrees

(2x - 17) + (x + 53) + x = 180

2x - 17 + x + 53 + x = 180

4x + 36 = 180

Subtract 36 from both sides;

4x = 144

Divide both sides by 4;

x = 36

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Answer:

<h3>Q cuts the diagonal PA into 2 equal halves, since the diagonals of rhombus meet at right angles.</h3><h3>The value of x is 8.</h3>

Step-by-step explanation:

Given that Quadrilateral CAMP below is a rhombus. the length PQ is (x+2) units, and the length of QA is (3x-14) units

From the given Q is the middle point, which cut the diagonal PA into 2 equal halves.

By the definition of rhombus, diagonals meet at right angles.

Implies that PQ = QA

x+2 = 3x - 14

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dividing by 2 on both sides, we will get,

$x =\frac{16}{2}$

<h3>∴ x=8</h3><h3>Since Q cuts the diagonal PA into 2 equal halves, since the diagonals of rhombus meet at right angles we can equate x+2 = 3x-14 to find the value of x.</h3>

The line segment $\overrightarrow{PA}=\overrightarrow{PQ}+\overrightarrow{QA}$

$\overrightarrow{PA}=x+2+3x-14$

$=4x-12$

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Answer:

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

has critical points where the first derivatives vanish:

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

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

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(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

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

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

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

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

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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).