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sertanlavr [38]
3 years ago
7

In a smoothie the ratio of ounces of blueberries to ounces of yogurt is 4 to 5.

Mathematics
1 answer:
KonstantinChe [14]3 years ago
8 0

Answer:

12.5

Step-by-step explanation:

for people who don't get another  quizzez assignment about ratios from their teachers, this question showed & the correct answer was 12.5 Hope this answer somehow helps.

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Find all the real cube roots of -343.
andre [41]

The Cube Root Of -343 is ( -7)

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3 years ago
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Put these fractions in order of size, smallest to largest.<br> 7/10<br> 2/3<br> 4/5<br> 11/15
Alla [95]

Answer:

11/15 7/10 4/5 2/3

Step-by-step explanation:

the bigger the fraction the smaller the number

4 0
3 years ago
In the past month, Amy rented 7 video games and 4 DVDs. The rental price for each video game was $3.30. The rental price for eac
Lera25 [3.4K]

Answer:

$39.90

Step-by-step explanation:

if each video game cost $3.30, and she rented 7, then you multiply $3.30 by 7, to get $21.10, and if each DVD was $4.20, and she rented 4, then its $4.20x4= $16.80

after that you just add the two total costs together, which is $23.10+$16.80=$39.90

8 0
3 years ago
How many different linear arrangements are there of the letters a, b,c, d, e for which: (a a is last in line? (b a is before d?
inna [77]
A) Since a is last in line, we can disregard a, and concentrate on the remaining letters.
Let's start by drawing out a representation:

_ _ _ _ a

Since the other letters don't matter, then the number of ways simply becomes 4! = 24 ways

b) Since a is before d, we need to account for all of the possible cases.

Case 1: a d _ _ _ 
Case 2: a _ d _ _
Case 3: a _ _ d _
Case 4: a _ _ _ d

Let's start with case 1.
Since there are four different arrangements they can make, we also need to account for the remaining 4 letters.
\text{Case 1: } 4 \cdot 4!

Now, for case 2:
Let's group the three terms together. They can appear in: 3 spaces.
\text{Case 2: } 3 \cdot 4!

Case 3:
Exactly, the same process. Account for how many times this can happen, and multiply by 4!, since there are no specifics for the remaining letters.
\text{Case 3: } 2 \cdot 4!

\text{Case 4: } 1 \cdot 4!

\text{Total arrangements}: 4 \cdot 4! + 3 \cdot 4! + 2 \cdot 4! + 1 \cdot 4! = 240

c) Let's start by dealing with the restrictions.
By visually representing it, then we can see some obvious patterns.

a b c _ _

We know that this isn't the only arrangement that they can make.
From the previous question, we know that they can also sit in these positions:

_ a b c _
_ _ a b c

So, we have three possible arrangements. Now, we can say:
a c b _ _ or c a b _ _
and they are together.

In fact, they can swap in 3! ways. Thus, we need to account for these extra 3! and 2! (since the d and e can swap as well).

\text{Total arrangements: } 3 \cdot 3! \cdot 2! = 36
7 0
3 years ago
Cho tam giác ABC đều tâm O, dựng ảnh của tam giác ABC qua phép quay tâm O góc quay 120 độ
Sidana [21]
The answer is going abc tam
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