<img src="https://tex.z-dn.net/?f=%20%5Ctimes%20%20%2B%208%20%5Cgeqslant%20%209" id="TexFormula1" title=" \times + 8 \geqslant

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Arte-miy333 [17]

3 years ago

9" alt=" \times + 8 \geqslant 9" align="absmiddle" class="latex-formula">

Mathematics

1 answer:

Annette [7]3 years ago

8 0

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8. Which one of the following examples represents a repeating decimal?

A. 4.252525

B. 1.111114

C. 0.123123

D. 0.777777

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How do you construct the inscribed and circumscribed circle of a triangle?

suter [353]

Answer:

Draw the triangle.

Draw the perpendicular bisector to each side of the triangle. Draw the lines long enough so that you see a point of intersection of all three lines.

Draw the circle with radius at the intersection point of the bisectors that passes through one of the vertices

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shepuryov [24]

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3 years ago

Give that m/_ y=39, find the measures of angles a and b

Amanda [17]

Answer:

∠a = 39°

∠b = 129°

Step-by-step explanation:

Vertical angles are congruent. An exterior angle of a triangle is equal to the sum of the remote interior angles.

<h3>Application</h3>

Angle y and angle 'a' are vertical angles, so congruent.

∠a = ∠y = 39°

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∠b = 129°

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1 year ago

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Consider the implicit differential equation <img src="https://tex.z-dn.net/?f=%2849%20y%5E%7B3%7D%20%2B%2045%20xy%29%20dx%20%2B%

BaLLatris [955]

We're looking for an integrating factor $\mu(x,y)=x^py^q$ such that

$\mu\underbrace{(49y^3+45xy)}_M\,\mathrm dx+\mu\underbrace{(98xy^2+50x^2)}_N\,\mathrm dy=0$

is exact, which would require that

$(\mu M)_y=(\mu N)_x$
$(49x^py^{q+3}+45x^{p+1}y^{q+1})_y=(98x^{p+1}y^{q+2}+50x^{p+2}y^q)_x$
$49(q+3)x^py^{q+2}+45(q+1)x^{p+1}y^q=98(p+1)x^py^{q+2}+50(p+2)x^{p+1}y^q$
$\implies\begin{cases}49(q+3)=98(p+1)\\45(q+1)=50(p+2)\end{cases}\implies p=\dfrac52,q=4$

You can verify that $(\mu M)_y=(\mu N)_x$ if you'd like. With the ODE now exact, we have a solution $F(x,y)=C$ such that

$F_x=\mu M$
$F=\displaystyle\int(49y^3+45xy)x^{5/2}y^4\,\mathrm dx$
$F=10x^{9/2}y^5+14x^{7/2}y^7+f(y)$

$F_y=\mu N$
$50x^{9/2}y^4+98x^{7/2}y^6+f'(y)=98x^{7/2}y^2+50x^{9/2}y^4$
$f'(y)=0$
$\implies f(y)=C$

and so the general solution is

$F(x,y)=10x^{9/2}y^5+14x^{7/2}y^7=C$

8 0

3 years ago