ellie brought 60 cookies to class today. 30% of them have mint filing. how many cookies have the mint filling

Mathematics

2 answers:

Pepsi [2]3 years ago

8 0

60 cookies X 0.30 = 18 mint filling

Elan Coil [88]3 years ago

5 0

So simple.

18.

If you need help to find why, here's a link

http://www.geteasysolution.com/what-is-30-percent-of-60

You might be interested in

Distribute then combine like terms in the following: 2(2x - 3y) - 3(x - 2y - 4) 4

nikdorinn [45]

I think its:

2(2x-3y)-3(x-2y-4)4

4x-6y-3x-6y(4)

4x-6y-3x-24y

x-30y

3 0

2 years ago

Read 2 more answers

I WILL GIVE BRAINLIEST

Travka [436]

Answer:

Step-by-step explanation:

Ape x.

5 0

1 year ago

Read 2 more answers

Find a polynomial P(x) with real coefficients having a degree 4, leading coefficient 4, and zeros 3-i and 4i.

Alisiya [41]

now, this polynomial has roots of 3-i and 4i, namely 3 - i and 0 + 4i.

let's bear in mind that a complex root never comes all by her lonesome, her sibling is always with her, the conjugate, so if 3 - i is there, 3 + i is also coming along, likewise if 0 + 4i is there, her sibling 0 - 4i is also there.

$\bf \begin{cases} x=3-i\implies &x-3+i=0\\ x=3+i\implies &x-3-i=0\\ x=4i\implies &x-4i=0\\ x=-4i\implies &x+4i=0 \end{cases}\\\\[-0.35em] ~\dotfill\\\\ (x-3+i)(x-3-i)(x-4i)(x+4i)=\stackrel{y}{0} \\[2em] \underset{\textit{difference of squares}}{[(x-3)+i][(x-3)-i]}\underset{\textit{difference of squares}}{[x-4i][x+4i]}=0$

$\bf [(x-3)^2-i^2][x^2-(4i)^2]=y\implies [(x-3)^2-(-1)][x^2-(4^2i^2)]=0 \\[2em] [(x-3)^2-(-1)][x^2-(16(-1))]=0\implies [(x-3)^2+1][x^2+16]=0 \\[2em] [(x^2-6x+9)+1][x^2+16]=y\implies (x^2-6x+10)(x^2+16)=0 \\\\\\ x^4-6x^3+10x^2+16x^2-96x+160=0 \\\\\\ x^4-6x^3+26x^2-96x+160=0 \\\\\\ \stackrel{\textit{multiplying both sides by 4}}{4(x^4-6x^3+26x^2-96x+160)=4(0)} \\\\[-0.35em] \rule{34em}{0.25pt}\\\\ ~\hfill 4x^4-24x^3+104x^2-384x+640=y~\hfill$