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3 years ago

Aaron wants to divide the trail mix equally into 6 bags to give to his friends.How much trail mix will be in each bag?

Mathematics

2 answers:

emmainna [20.7K]3 years ago

8 0

1 is the answer because if you divided 6 by 6 you get 1

Nataly_w [17]3 years ago

6 0

Wouldn't you divide each ingredient by 6?

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Can somebody help with geometry?

GaryK [48]

Step-by-step explanation:

You can look this answers up off the paper and find the worksheet

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2 years ago

A 7 pound bag of dog food costs $21 What is the unit price of the dog food?

DiKsa [7]

I feel like it the answer is A

8 0

3 years ago

Read 2 more answers

If the tigers of their 45 games and the bearcats won 12 of their 30 games which team had the better record (write with denominat

julia-pushkina [17]

How many games did the tiger won?? It's complete!!

Complete your question, then I would answer you in inbox:)

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3 years ago

solve this problem by using a 5d process or writing and solving an equation. No matter which method you use, be sure to define y

marysya [2.9K]

Answer:

width of the rectangle=x=4in and length of the rectangle=11in

Step-by-step explanation:

Let the width of the rectangle be x, then according to the question, length is one less than three times its width that is l=3x-1.

Now, width=b=x and length=l=3x-1, thus

Perimeter of the rectangle= 2(l+b)

⇒ $30=2(3x-1+x)$

⇒ $15=(3x-1+x)$

⇒ $15=4x-1$

⇒ $16=4x$

⇒ $x=4in$

Thus, width of the rectangle=x=4in and length of the rectangle=l=3x-1=3(4)-1=12-1=11in.

5 0

3 years ago

Using these complex zeros (1,1,-1/2,2+i,2-i) factor f(x)=-2x^5 +11x^4 -22x^3 +14x^2 +4x -5

Marianna [84]

$\bf \begin{cases}
x=1\implies &x-1=0\\
x=1\implies &x-1=0\\
x=-\frac{1}{2}\implies 2x=-1\implies &2x+1=0\\
x=2+i\implies &x-2-i=0\\
x=2-i\implies &x-2+i=0
\end{cases}
\\\\\\
(x-1)(x-1)(2x+1)(x-2-i)(x-2+i)=\stackrel{original~polynomial}{0}
\\\\\\
(x-1)^2(2x+1)~\stackrel{\textit{difference of squares}}{[(x-2)-(i)][(x-2)+(i)]}$

$\bf (x^2-2x+1)(2x+1)~[(x-2)^2-(i)^2]
\\\\\\
(x^2-2x+1)(2x+1)~[(x^2-4x+4)-(-1)]
\\\\\\
(x^2-2x+1)(2x+1)~[(x^2-4x+4)+1]
\\\\\\
(x^2-2x+1)(2x+1)~[x^2-4x+5]
\\\\\\
(x^2-2x+1)(2x+1)(x^2-4x+5)$

of course, you can always use (x-1)(x-1)(2x+1)(x²-4x+5) as well.

7 0

3 years ago