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Mandarinka [93]
3 years ago
8

30 POINTS ANSWER ASAP...

Mathematics
2 answers:
rewona [7]3 years ago
5 0

Answer:

The middle one in the second section

Step-by-step explanation:

I have had this question before

wolverine [178]3 years ago
3 0

Answer:

The 2nd graph is the corret answer

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This is due tomorrow if someone could help me that would be great <3
kifflom [539]

<u>Hope this helped!</u>

<u>First Row</u>

\frac{5}{18}

\frac{10}{21}

\frac{9}{40}

6

5\frac{1}{3}

(this one is already done)

<u>Second Row</u>

4\frac{4}{9}

5\frac{1}{16}

\frac{13}{18}

6\frac{1}{4}

11\frac{2}{3}

5 0
2 years ago
What number has 62 hundreds and 5 tens
fomenos

Answer:

6250

Step-by-step explanation:

Definition of the question

6 0
2 years ago
Read 2 more answers
at a party at the middle school over on third avenue next to the bank, $60\%$ of the students are seventh graders. these student
Vaselesa [24]

There are 600 students including the seventh and eighth graders at the party.

This problem uses the concept of percentages to define the conditions that are laid in front of us.

Let the original  number of students be S , and the number of seventh graders be   = 0.60S

We know that percent is used to convey the mathematical term of a fraction multiplied by 100.

Total students  after 20 eighth graders arrive = S + 20

And we have that

Number of seventh graders / total number of students  =  58%

.60S  / [ S + 20 ]  =   .58    we multiply both sides by  S +  20

0.60S  =0 .58 [ S + 20]

.60S  = .58S + 11.6   we subtract  0.58S from  both the sides

0.02S = 11.6  we divide both the sides  by .02

S  = 11/6 / 0.02 =    580

So the total number of students = 580 + 20  = 600 .

Hence there are 600 students at the party at that time.

To learn more about students visit:

brainly.com/question/17332524

#SPJ1

5 0
1 year ago
The segments shown below could form a triangle ac=9. cb=8. ba=17
krek1111 [17]
The lengths of sides of a triangle have to satisfy the triangle inequality, which states that the sum of the two shorter sides must exceed the length of the third side.

Here 9+8=17 (not greater), so these segments do not form a triangle.
8 0
3 years ago
Read 2 more answers
Find a + b, if b^a = 64 and a^b = 81
jeyben [28]
\bf \begin{array}{llll}&#10;b^a=64\qquad \boxed{64=4^3}\qquad b^a=4^3&#10;\\\\\\&#10;a^b=81\qquad \boxed{81=3^4}\qquad a^b=3^4&#10;\end{array} \qquad &#10;\begin{cases}&#10;a=3\\&#10;b=4&#10;\end{cases}&#10;\\\\\\&#10;a+b\implies 3+4
3 0
3 years ago
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