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Triss [41]
3 years ago
14

Please help me, I don't know

Mathematics
1 answer:
Mila [183]3 years ago
8 0

Answer:

3

4

5

Step-by-step explanation:

You might be interested in
Determine whether the two expressions are equivalent<br> 2(3c + 2) - 2c and 4c + 2
densk [106]

Answer:

The equations are not equivalent

Step-by-step explanation:

2(3c +2) -2c

6c + 4 - 2c

combine like terms

4c +4 does not equal 4c +2

6 0
3 years ago
Read 2 more answers
8 inch by 6 inch retangular photo inside a 10 inch by 8 inch rectangular fram how much greater is the perimeter of the frame
Flura [38]

The perimeter of frame is 1.29 greater

Step-by-step explanation:

Perimeter of a rectangular shape object is given by the formula;

2(l+w) when, l is length and w is width

Perimeter of the photo is;

2(8+6)= 2(14) =28 inches

Perimeter of the frame is;

2(10+8)= 2(18) =36 inches

Divide perimeter of frame by that of photo as;

36/28 =1.29

The perimeter of frame is 1.29 greater

Learn More

Perimeter of rectangular shaped object:brainly.com/question/12919591

Keywords : rectangular, photo, inside, frame, greater

#LearnwithBrainly

7 0
4 years ago
jim has three times as many comic books as charles charles has 2/3 as many books as bob bob has 27 books how many books do jim h
Alecsey [184]
Jim has 54 comic books.

Bob has 27
So Charles has 18 (27*2/3)
Which means Jim has 54 (18*3)
5 0
3 years ago
Read 2 more answers
Highest common factor of 72 and 96
Likurg_2 [28]

Answer:

24

Step-by-step explanation:

Prime Factorise 72:

72=2*36\\72=2*2*18\\72=2*2*2*9\\72=2*2*2*3*3\\\\72=2^3*3^2

Prime factorise 96:

96=2*48\\96=2*2*24\\96=2*2*2*12\\96=2*2*2*2*6\\96=2*2*2*2*2*3\\96=2^5*3

Multiply the common factors of 2^3*3^2 and 2^5*3:

2^3*3=24

4 0
4 years ago
Let A, B and C be sets. Prove that if A ⊆ B ∪ C and A ∩ B = ∅ then A ⊆ C
Studentka2010 [4]

Answer and Solution:

As per the question:

Given:

If A\subseteq B\cup C

A\cap B = \phi

To prove:

A\subseteq \C

Proof:

Suppose t\in A

As we know that:

A\subseteq B\cup C

Therefore,

t\in B or t\in C

Now, if we assume that t\in B

Then

t\in A\cap B

Since,

t\in A and t\in B

But

A and B are disjoint set and A\cap B = \phi

Therefore, this is contradictory.

Thus

t\notin B

So,

t\in C

Every element in the set A is also present in the set C

Therefore, A\subseteq \C

Hence, proved.

6 0
4 years ago
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