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2 years ago

Replace ∗ with a monomial so that the derived equality will be an identity: (15y + ∗)2 = 225y2+12x3y+0.16x6

Mathematics

1 answer:

kobusy [5.1K]2 years ago

6 0

Answer:

∗ = 0.4x³

Step-by-step explanation:

(15y + ∗)² = 225y²+12x³y+0.16x⁶

Note:

225y² = 15y * 15y = (15y)²

12x³y = 2 * 15y * 0.4x³

0.16x⁶ = 0.4x³ * 0.4x³ = (0.4x³)²

So, by factoring the right hand side:

225y²+12x³y+0.16x⁶ = (15y + 0.4x³)²

By comparing the left hand side with (15y + 0.4x³)²

So, ∗ should be replaced with the monomial 0.4x³

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2 years ago

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Answer:

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3 years ago

Centerville is the headquarters of Greedy Cablevision Inc. The cable company is about to expand service to two nearby towns, Spr

NeX [460]

Answer:

20.2057 Units.

Step-by-step explanation:

First, we determine the length of the cable.

Distance between Centerville (8,0) and point (x,0) is given as:

$\sqrt{(8-x)}^2=8-x$

Distance between point (x,0) and Springfield(0,7) is:

$\sqrt{(7-0)^2+(0-x)^{2}}=\sqrt{(7)^2+(x)^{2}}$

Distance between point (x,0) and Shelvyfield(0,-7) is:

$\sqrt{(-7-0)^2+(0-x)^{2}}=\sqrt{(7)^2+(x)^{2}}$

Therefore the Length of the Cable L(x)

$L(x)=(8-x)+2\sqrt{(7)^2+(x)^{2}}$

To find the critical point, we set the derivative of L(x)=0

$L^{'}(x)=-1+\frac{2x}{\sqrt{\left( 49 - x^{2}\right) }}$

$\frac{-\sqrt{\left( 49 - x^{2}\right)}+2x}{\sqrt{\left( 49 - x^{2}\right)}}=0\\-\sqrt{\left( 49 - x^{2}\right)}+2x=0\\\sqrt{\left( 49 - x^{2}\right)}=2x\\(\sqrt{\left( 49 - x^{2}\right)})^2=(2x)^2\\49 - x^{2}=4x^2\\49=5x^2\\x^2=\frac{49}{5}\\x= 3.1305$

To verify that L(x) has a minimum at this critical number we compute the second derivative L″(x) and find its value at the critical number.

$L^{''}(x)=\frac{\left( 98\right) \,\sqrt{\left( 49 - x^{2}\right) }}{{\left( 7 - x\right) }^{2}\,{\left( 7+x\right) }^{2}}\\At \:x=3.1305, L^{''}=0.3993$