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Luden [163]
3 years ago
12

Please help me I need help right away.

Mathematics
2 answers:
ra1l [238]3 years ago
8 0

Answer:

3 7/16

Step-by-step explanation:

o-na [289]3 years ago
4 0

Answer:

Step-by-step explanation:

improper fraction you take the whole number, multiply it by the denominator and add the numerator. That number goes on top and you keep the denominator the sam.

2 1/2 in an improper fraction is 5/2

1 3/8 in an improper fraction is 11/8

keep change flip

keep the first fraction

change multiplication to division and flip the second fraction

5/2 x 11/8

5/2 divided by 8/11

answer is 3 7/16

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Deshawn draws a regular pentagon and rotates it about its center.
salantis [7]

Answer:

f

Step-by-step explanation:

7 0
3 years ago
Read 2 more answers
Show that in a group of five people (where any two people are either friends or enemies), there are not necessarily three mutual
sveta [45]

Answer:

Lets call the people A,B,C,D and E. We need to find an example where there are not 3 mutual friends and not three mutual enemies.

Lets start by assuming that A has 2 friends, B and C, and 2 enemies, D and E.

Since we want A,B and C not to be mutually friends, then B and C neccesarily have to be enemies.

Now, since B and C are enemies, they cant be enemies at the same time with D, and they also cant be enemies at the same time with E. Therefore one of them has to be friends with D and one has to be friends with E.

Also, since D and E are enemies with A, they neccesarily need to be friends.

From the fact that D and E are friends, we coclude that they cant have a friend in common, as a consequence, B has to be friends with one of D and E and C has to be friend with the other. We can assume that B is friend with D and C is friend with E. If we use that, we have the following friendship configuration

---------------------

Friends of A: B, C

Enemies of A: D,E

-------------------------

Friends of B: A,D

Enemies of B: C, E

------------------------

Friends of C: A,E

Enemies of C: B,D

------------------------

Friends of D: B,E

Enemies of D: A,C

----------------------------

Friends of E: C,D

Enemies of E:  A,B

--------------------------

As you can check, there is not three mutual friends nor three mutual enemies.

8 0
3 years ago
Lisa typed a 165-word paragraph in three minutes. Use the ratio table to
JulsSmile [24]

Answer:

1 minute

Step-by-step explanation:

We can see that Lisa types 165 words in 3 mins.

So she types 55 words in 1 minute.

How did I work that out?

Well, all you really do is \frac {165}{3} which is equal to \frac{55}{1} if you divide top and bottom by 3. Anything divided by 1 is itself.

8 0
2 years ago
A whale is 24 meters long. a rhinoceros is 400cm long which is longer how much longer?
klemol [59]
The whale is longer
because 1 meter is = 100 cm
8 0
3 years ago
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A square carpet of side length 9 feet is designed with one large shaded square and eight smaller, congruent shaded squares, as s
Pepsi [2]

Answer:

3\sqrt{3} + 72

Step-by-step explanation:

I attached an image to aid the understanding of the question.

Looking at the image, we see that the 8 shaded parts are congruent, as affirmed in the question as well. And we are told that T = 3, this implies that the area of the square with T as it's side is 9ft². Since all the 8 squares are congruenrt, it means each square has its area to be 9. Therefore, the total area of the 8 shaded squares will be:

9 \times 8 = 72ft^{2}

It remains the area of the shaded square with side S.

From the question, we have the following ratio:

\frac{9}{S} = \frac{S}{T}[\tex]But[tex]\frac{9}{S} = \frac{S}{3}[\tex]Multiplying through by 3S we have27 = S² and this gives:[tex]S = \sqrt{27} = 3\sqrt{3}

I did not add ± because length is always positive, so the case of negative is eliminated.

Now the areas of S is 3\sqrt{3}

Therefore, the total area of the shaded squares is

3\sqrt{3} + 72

3 0
3 years ago
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