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3 years ago

Rodrigo made 6/8 of a pound of trail mix. If he puts 1/4 of a pound into each bag, how many bags can Rodrigo fill?

Mathematics

2 answers:

ValentinkaMS [17]3 years ago

7 0

Answer:

He can fill 3 bags

Step-by-step explanation:

You would first make them have a common denominator so you would have 2/8 instead of 1/4 then you would do 6/8 divided by 2/8 to get 3

1/4 = 2/8

6/8 divided by 2/8 = 3

ki77a [65]3 years ago

4 0

Answer:3 1/4=2/8 if it's 6/8 then you divide and get three.

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The landing for the steps leading up to a county courthouse is shaped like a trapezoid. The area of the landing is 1500 square f

77julia77 [94]

Answer:

The length of the longer base is 80 feet.

Step-by-step explanation:

We are given that county courthouse is shaped like a trapezoid.

Let the height of the court house be h

Since we are given that The shorter base of the trapezoid is 15 feet longer than the height.

So, the shorter base = h+15

We are also given that The longer base is 5 feet longer than 3 times the height

So, longer Base = 3h+5

Now we are also given that the area of the landing is 1500 square feet.

Formula of area of trapezium :

= $\frac{1}{2} *(\text{Length of longer base}+\text{Length of shorter base})*height$

Substituting the values .

$1500=\frac{1}{2} *(h+15+3h+5)*h$

$1500=\frac{1}{2} *(4h+20)*h$

$1500*2=(4h+20)*h$

$3000=4h^2+20h$

Taking 4 common from the equation.

$750=h^2+5h$

$h^2+5h-750=0$

$h^2-25h+30h-750=0$

$h(h-25)+30(h-25)=0$

$(h-25)(h+30)=0$

⇒ h = 25, -30

Neglect the negative value .

So, height = 25 feet.

Length of Longer base = 3h+5 =3*25+5=80 feet

Hence , the length of the longer base is 80 feet.

6 0

4 years ago

A six-sided die is rolled.

insens350 [35]

Answer:

a.) 1/3

Step-by-step explanation:

1/6 is a 2 is rolled

1/6 if a 4 is rolled

so 2/6 if either are rolled

which simplifies to 1/3

8 0

3 years ago

Read 2 more answers

Examine the pattern in the powers of i you wrote in the table, and create a rule for finding the value of large powers of i. Jus

denpristay [2]

Let $N$ be any integer. Then

$i^N=\begin{cases}1&\text{if }N\equiv0\mod4\\i&\text{if }N\equiv1\mod4\\-1&\text{if }N\equiv2\mod4\\-i&\text{if }N\equiv3\mod4\end{cases}$

In other words:

(1) If $N$ is a multiple of 4, then $i^N=1$ .

(2) If dividing $N$ by 4 leaves a remainder of 1, then $i^N=i$ , since $i^N=i^{4n+1}$ for some integer $n$ , and from (1) we know that $i^{4n+1}=i^{4n}i=i$ (because, obviously, $4n$ is a multiple of 4).

(3) If instead you get a remainder of 2, then $i^N=-1$ . This follows from (1) as well. $i^N=i^{4n+2}=i^{4n}i^2=(1)(-1)=-1$ .

(4) Finally, if you get a remainder of 3, then $i^N=i^{4n+3}=i^{4n}i^2i=(1)(-1)(i)=-i$ .