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3 years ago

10

Luna has 10 bags of potato chips. In each bag there is 18 potato chips if she shared them with 11 friends how much chips will ea

ch friend get will there be a remainder?

2 answers:

Leokris [45]3 years ago

8 0

Each friend will get 16 chips. There is a remainder of 4 chips.

borishaifa [10]3 years ago

3 0

16.36363636363 which include the remainder. <span />

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Please explain how to solve this step by step.

Phantasy [73]

Answer:

(a^8)/(b^9)

Step-by-step explanation:

Two rules of exponents come into play.

(a^b)/(a^c) = a^(b-c)

a^-b = 1/a^b

__

Applying the first rule, we have ...

(a^3)/(a^-5) × (b^-2)/(b^7) = a^(3 -(-5)) × b^(-2 -7) = a^8 × b^-9

Applying the second rule gives the simplified form ...

= (a^8)/(b^9)

5 0

1 year ago

Please help with this question<br> 2(2w-6)=6(w-6)+14<br> w=?

IRISSAK [1]

4w-12=6w-36+14
4w-12=6w-22
10=2w

5=w
(I think)

8 0

2 years ago

Get a number 100 by using four sevens (7’s) and a one (1)?

Oksanka [162]

<em><u>Answer 1: 177 – 77 = 100 </u></em>

<em><u>Answer 2: (7+7) * (7 + (1/7)) = 100</u></em>

8 0

2 years ago

Read 2 more answers

Which value is NOT a solution of 8x3 – 1 = 0?

Tpy6a [65]

<u><em>Note: As you may have unintentionally missed to add the value choices. But, I would make sure to explain the concept so that you may improve your understanding in terms of solving these type of questions.</em></u>

Answer:

Any value other than the values $x=\frac{1}{2},\:x=-\frac{1}{4}+i\frac{\sqrt{3}}{4},\:x=-\frac{1}{4}-i\frac{\sqrt{3}}{4}$ will not be a solution of $8x^3\:-\:1\:=\:0$ .

Step-by-step explanation:

Considering the equation

$8x^3\:-\:1\:=\:0$

Steps to solve the equation

$8x^3-1=0$

$\mathrm{Add\:}1\mathrm{\:to\:both\:sides}$

$8x^3-1+1=0+1$

$\mathrm{Simplify}$

$x^3=\frac{1}{8}$

$\mathrm{Divide\:both\:sides\:by\:}8$

$\frac{8x^3}{8}=\frac{1}{8}$

$\mathrm{Simplify}$

$x^3=\frac{1}{8}$

As

$\mathrm{For\:}x^3=f\left(a\right)\mathrm{\:the\:solutions\:are\:}x=\sqrt[3]{f\left(a\right)},\:\sqrt[3]{f\left(a\right)}\frac{-1-\sqrt{3}i}{2},\:\sqrt[3]{f\left(a\right)}\frac{-1+\sqrt{3}i}{2}$

$x=\sqrt[3]{\frac{1}{8}},\:x=\sqrt[3]{\frac{1}{8}}\frac{-1+\sqrt{3}i}{2},\:x=\sqrt[3]{\frac{1}{8}}\frac{-1-\sqrt{3}i}{2}$

So,

$x=\frac{1}{2},\:x=-\frac{1}{4}+i\frac{\sqrt{3}}{4},\:x=-\frac{1}{4}-i\frac{\sqrt{3}}{4}$

Therefore,

Any value other than the values $x=\frac{1}{2},\:x=-\frac{1}{4}+i\frac{\sqrt{3}}{4},\:x=-\frac{1}{4}-i\frac{\sqrt{3}}{4}$ will not be a solution of $8x^3\:-\:1\:=\:0$ .

Keywords: solution, value

Learn more about equation solution from brainly.com/question/1679491

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7 0

2 years ago

Form a quadratic equation with roots (1+√3) and (1-√3)

Alborosie

Answer:

hope this helps u

I guess

7 0

1 year ago