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Vedmedyk [2.9K]
3 years ago
12

How to use algebra tiles to show 4(2x+1)=8x+4

Mathematics
1 answer:
eimsori [14]3 years ago
5 0
Here is the answer 4
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Explain how you know 7/12 is greater than 1/3 but less than 2/3
Roman55 [17]
7/12
1/3=4/12
2/3=8/12

7/12>1/3
<span>7/12<2/3
</span>

convert all of these to a common denominator:
7/12 , 1/3 , 2/3
---
7/12 = 7/12
1/3 = 4/12
2/3 = 8/12
---
place into compound inequality:
---
(1/3=4/12) < (7/12=7/12) < (2/3=8/12)


3 0
3 years ago
Read 2 more answers
A catering service offers 12 appetizers, 8 main courses, and 4 desserts. A customer is to select 9 appetizers, 3 main courses, a
valentina_108 [34]

Answer:

73920

Step-by-step explanation:

Number of ways to choose 9 appetizers from 12: ₁₂C₉

Number of ways to choose 3 main courses from 8: ₈C₃

Number of ways to choose 2 desserts from 4: ₄C₂

The total number of ways is:

₁₂C₉ × ₈C₃ × ₄C₂

= 220 × 56 × 6

= 73920

5 0
3 years ago
What is the standard form of 365.05​
pantera1 [17]

9514 1404 393

Answer:

  it depends. 365.05  or  3.6505×10²

Step-by-step explanation:

In the US, 365.05 is already in standard form.

In the UK, "standard form" is the same as "scientific notation", so the number would be ...

  3.6505×10²

4 0
3 years ago
A number a is a root of P(x) if and only if the remainder, when diving the polynomial by x -a, equals zero True or false
Ilia_Sergeevich [38]
True.  This is a great way to determine whether a given number is a root of a polynomial.

8 0
3 years ago
How many 4-digit numbers divisible by 5, all of the digits of which are odd, are there
HACTEHA [7]

Answer:

So we must create 4-digit numbers with the 5 odd digits in the image.

First, a number is only divisible by 5 if the last digit is 0 or 5, and we only can use the last digit equal to 5 (because 0 is not in the image).

if each odd number can be used only once, we have:

Now, for the other 3 digits we have 4 options.

So for the first one we have 4 options,

for the second we have 3 options (because one is already taken)

for the last one we have only 2 options

then the number of combinations is:

C = 4*3*2 = 24.

If the numbers can be repeated (for example, 5555 is allowed, then)

we still have our last digit fixed in 5, and for the first digit we have 5 options, for the second we also have 5 options, and for the third we also have 5 options, then we have a total of:

C = 5*5*5 = 25*5 = 125 combinations.

3 0
3 years ago
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