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2 years ago

10

Ari thinks the perfect milkshake has 3 ounces of caramel for every 5 scoops of ice cream. Freeze Zone makes batches of milkshake

s with 6 ounces of caramel and 8 scoops of ice cream.
What will Ari think about Freeze Zone's milkshakes?
Choose 1 answer:

A
They have too much caramel.

B
They have too little caramel.

C
They are just right.

2 answers:

Andrew [12]2 years ago

8 0

B because it has to much Carmel

aksik [14]2 years ago

3 0

A. They have too much caramel

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Choose the option that best completes the statement below. In finding the number of permutations for a given number of items, __

vodomira [7]

Let’s look at the permutations of the letters “ABC.” We can write the letters in any of the following ways:
ABC
ACB
BAC
BCA
CBA
CAB
Since there are 3 choices for the first spot, two for the next and 1 for the last we end up with (3)(2)(1) = 6 permutations. Using the symbolism of permutations we have: $3 P_{3}=(3)(2)(1)=6$ . Note that the first 3 should also be small and low like the second one but I couldn’t get that to look right.

Now let’s see how this changes if the letters are AAB. Since the two As are identical, we end up with fewer permutations.
AAB
ABA
BAA
To make the point a bit better let’s think of one A are regular and one as bold A.
ABA and ABA look different now because we used bold for one of the As but if we don’t do this we see that these are actual the same. If they represented a word they would be the same exact word.

So in this case the formula would be $\frac{3 P_{3} }{2!}= \frac{(3)(2)(1)}{(2)(1)}= \frac{6}{2}=3$ . We use 2! In the denominator because there are 2 repeating letters. If there were three we would use 3!

Hopefully, this is enough to let you see that the answer is A. The number of permutations is limited by the number of items that are identical.

7 0

2 years ago

A conical pile of road salt has a diameter of 112 feet and a slant height of 65 feet. After a storm, the linear dimensions of th

QveST [7]

we know that

the volume of a cone is equal to

$V= \frac{1}{3} \pi r^{2}h$

in this problem

the radius is equal to

$r= \frac{112}{2}= 56ft$

1) <u>Find the height of the cone before the storm</u>

Applying the Pythagorean Theorem find the height

$h^{2} = l^{2}-r^{2}$

$l=65 ft$

$h^{2} = 65^{2}-56^{2}$

$h^{2} = 1,089$

$h=33 ft$

2) <u>Find the volume before the storm</u>

$V= \frac{1}{3}*\pi* 56^{2}*33$

$V=34,496\pi\ ft^{3}$

3) <u>Find the volume after the storm</u>

After a storm, the linear dimensions of the pile are 1/3 of the original dimensions

so

r=(56/3) ft

h=(33/3)=11 ft

$V= \frac{1}{3}*\pi* (56/3)^{2}*11$

$V= 1,277.63\pi\ ft^{3}$

<u>4) Find how this change affect the volume of the pile</u>

Divide the volume after the storm by the volume before the storm

$\frac{1,277.63 \pi }{34,496 \pi } = \frac{1}{27}$

therefore

<u>the answer part a) is</u>

The volume of the pile after the storm is $\frac{1}{27}$ times the original volume

<u>Part b)</u> Estimate the number of lane miles that were covered with salt

5) <u>Find the amount of salt that was used during the storm</u>

$=34,496 \pi - 1,277.63 \pi \\= 33.218.37 \pi \\= 104,358.59\ ft^{3}$

6) <u>Find the pounds of road salt used</u>

$104,358.59*80=8,348,687.2\ pounds$

7) <u>Find the number of lane miles that were covered with salt</u>

$8,348,687.2/350=23,853.39 \ lane\ miles$

therefore

<u>the answer part b) is</u>

the number of lane miles that were covered with salt is $23,853.39 \ lane\ miles$

<u>Part c) </u>How many lane miles can be covered with the remaining salt? Round your answer to the nearest lane mile

the remaining salt is equal to $1,277.63\pi\ ft^{3}$

$1,277.63\pi\ ft^{3}=4,013.79\ ft^{3}$

8) <u>Find the pounds of road salt </u>

$4,013.79*80=321,103.20\ pounds$

9) <u>Find the number of lane miles </u>

$321,103.20/350=917.44 \ lane\ miles$

therefore

<u>the answer part c) is</u>

the number of lane miles is $917 \ lane\ miles$

7 0

2 years ago

Three years ago Tom was twice as old as Jean, and in two years the sum of their ages will be 28 years. Find their present ages.

lubasha [3.4K]

Answer:

Jean is 9 and Tom is 15.

Step-by-step explanation:

3 years ago, Tom was 12 and Jean was 6, hence Tom was twice as old as Jean.

Since that was their age 3 years ago, they are currently 15 (Tom) and 9 (Jean).

Add 2 years to each of these ages, you get 17 and 11.

17 + 11 = 28

3 0

2 years ago

Read 2 more answers

In parallelogram ABCD , diagonals AC⎯⎯⎯⎯⎯ and BD⎯⎯⎯⎯⎯ intersect at point E, AE=x2−16 , and CE=6x .

MAXImum [283]

Answer:

<h3>AC=96 units.</h3>

Step-by-step explanation:

We are given a parallelogram ABCD with diagonals AC and BD intersect at point E.

$AE=x^2-16.$ , and CE=6x .

<em>Note: The diagonals of a parallelogram intersects at mid-point.</em>

Therefore, AE = EC.

Plugging expressions for AE and EC, we get

$x^2-16=6x.$

Subtracting 6x from both sides, we get

$x^2-16-6x=6x-6x$

$x^2-6x-16=0$

Factoriong quadratic by product sum rule.

We need to find the factors of -16 that add upto -6.

-16 has factors -8 and +2 that add upto -6.

Therefore, factor of $x^2-6x-16=0$ quadratic is (x-8)(x+2)=0

Setting each factor equal to 0 and solve for x.

x-8=0 => x=8

x+2=0 => x=-2.

We can't take x=-2 as it's a negative number.

Therefore, plugging x=8 in EC =6x, we get

EC = 6(8) = 48.

<h3>AC = AE + EC = 48+48 =96 units.</h3>

5 0

2 years ago

HELP PLS>>> The vector (3, - 7) ) is rotated by an angle of (3pi)/4 radians and then reflected across the y-axis. If th

horsena [70]

Answer:

$a \approx -2.826$ , $b \approx 7.072$

Step-by-step explanation:

First, the vector must be transformed into its polar form:

$r = \sqrt{3^{2}+ (-7)^{2}}$

$r \approx 7.616$

$\theta \approx 2\pi - \tan^{-1}\left(\frac{7}{3} \right)$

$\theta \approx 1.629\pi$

Let assume that vector is rotated counterclockwise. The new angle is:

$\theta' = \theta + \frac{3\pi}{4}$

$\theta' = 2.379\pi$

Which is coterminal with $\theta'' = 0.379\pi$ . The reflection across y-axis is:

$\theta''' = \pi - \theta''$

$\theta''' = 0.621\pi$

The equivalent vector in rectangular coordinates is:

$a = 7.616\cdot \cos 0.621\pi$

$a \approx -2.826$

$b = 7.616\cdot \sin 0.621\pi$

$b \approx 7.072$

7 0

1 year ago