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3 years ago

100 points plz help J qnswer this question

Mathematics

2 answers:

xeze [42]3 years ago

6 0

Answer:

$6^{8}$ = 1679616 nickels

Step-by-step explanation:

Since it wants the value of the 8th bracelet, you have to keep multiplying the value 8 times. A simpler way to put this is 6 to the 8th power. Simplifying this will give you 1679616 nickels.

Andrei [34K]3 years ago

6 0

Answer: 16796,16

Step-by-step explanation:

your are welcome i guess. you will have to keep multiple to get the number 6*6 =36 *2 =72 *36=2592*3=7776*216=1679616 is your answer. and the 8th bracelet will the number i have.

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Find x so that b ll e??????

tekilochka [14]

x=6

(6x+4) = (8x-8)

∠3 and ∠6 have to be equal since they are opposite interior angles (these angles are always equal to each other). Opposite interior angles prove that two lines are parallel if they are equal, so you just set them equal to each other and solve for x.

(6x+4) = (8x-8)

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12 = 2x

/2 /2

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(6x+4) = (8x-8)

(6[6]+4) = (8[6]-8)

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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.

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