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stepan [7]
4 years ago
13

7 ones 2 tenths + 4 tenths = ____ tenths = _____

Mathematics
1 answer:
Natasha_Volkova [10]4 years ago
7 0
I would put 67 tenths I would
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The diameter of a circle is 4 cm. Which equation can be used to find its circumference?
Leya [2.2K]

Answer:

pi times 4

Step-by-step explanation:

Circumference: pi times diameter (pi)(d) or pi times 2 times the radius (pi)(2r)

5 0
3 years ago
Read 2 more answers
in a recent semester at a local university 520 students enrolled in both general chemistry and calculus 1. Of these students 88
IceJOKER [234]

Let Ch and C denote the events of a student receiving an A in <u>ch</u>emistry or <u>c</u>alculus, respectively. We're given that

P(Ch) = 88/520

P(C) = 76/520

P(Ch and C) = 31/520

and we want to find P(Ch or C).

Using the inclusion/exclusion principle, we have

P(Ch or C) = P(Ch) + P(C) - P(Ch and C)

P(Ch or C) = 88/520 + 76/520 - 31/520

P(Ch or C) = 133/520

6 0
3 years ago
Decompose 5/15 and 8/12
AnnZ [28]

Answer:

:)

Step-by-step explanation:

5/15 divided by 5 is equal to 1/3 and

8/12 divided by 2 is 4/6 divided by 2 is 2/3

You have to divide the numerator with the same number that you divide the denominator with.  

6 0
2 years ago
Find the number of ways of arranging the numbers
Doss [256]

First of all, note that all integers are either 0,1, or 2 modulo 3 (if you're not familiar with this terminology, it means that every integer is either a multiple of 3, or it is 1 or 2 away from a multiple of 3).

So, we can think of our numbers as

\begin{array}{c|c}x&x\mod 3\\0&0\\1&1\\2&2\\3&0\\4&1\\5&2\\6&0\\7&1\\8&2\\9&0\end{array}

In order to make sure that the sum of any three adjacent numbers is divisible by 3, we have to make sure that any group of 3 three adjacent numbers contains a 0, a 1 and a 2. This is possible only if we arrange our 9 numbers in 3 groups of 3 numbers containing 0,1 and 2 exactly once, repeating always the same pattern.

For example, we could arrange our numbers following the pattern

0,1,2,0,1,2,0,1,2

or

2,0,1,2,0,1,2,0,1

We have 3!=6 possible patterns. Suppose for example that we choose the pattern

0,1,2,0,1,2,0,1,2

One possible way of following this pattern would be the arrangement

3,1,2,6,4,5,9,7,8

In fact, we substituted every '0' with a multiple of 3 (3, 6 or 9), every '1' with a number 1 away from a multiple of 3 (1, 4 or 7) and every '2' with a number 2 away from a multiple of 3 (2, 5 or 8).

This means that, once we fix a patter, we have 3 choices for the first 3 slots, 2 choices for the next 3 slots, and the final slot will be fixed. So, we have

3\cdot 3\cdot 3\cdot 2 \cdot 2 \cdot 2 = 216

possible ways of following a fixed pattern. Since the number of patterns was 6, we have

216\cdot 6 = 1296

possible arrangements.

7 0
3 years ago
Read 2 more answers
Solve by factoring: x^4-12x^2= 64
USPshnik [31]
The final solution is all the values that make <span><span><span><span><span>(<span>x+4</span>)</span><span>(<span>x<span>−4</span></span>)</span></span><span>(<span><span>x2</span>+4</span>)</span></span>=0</span><span><span><span><span>x+4</span>⁢<span>x<span>-4</span></span></span>⁢<span><span>x2</span>+4</span></span>=0</span></span> true.<span>x=<span>−4</span>,4,<span>2i</span>,<span><span>−2</span><span>i</span></span></span>
7 0
3 years ago
Read 2 more answers
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