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3 years ago

If 2=6 3=12 4=20 5=30 6=42 then 9=?

1 answer:

8 0

$
2=6\\ \\ 2\cdot (2+1)=6\\ \\ 3=12\\ \\ 3\cdot (3+1)\cdot 12\\ \\ 4=20\\ \\ 4\cdot (4+1)=20\\ \\ 5=30\\ \\ 5\cdot (5+1)=30\\ \\ 6\cdot (6+1)=42\\ \\ x\cdot (x+1)={ x }^{ 2 }+x\\ \\ for\quad x=9\\ \\ 9\cdot (9+1)=\quad { 9 }^{ 2 }+9\quad =\quad 81+9\quad =\quad 90$
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April worked 1 1/2 times as long on her math project as did Carl. Debbie worked 1 1/4 times as long as Sonia. Richard worked 1 3

vlada-n [284]

Answer:

Student Hours worked

April. $7\frac{7}{8} \ hrs$

Debbie. $8\frac{1}{8}\ hrs$

Richard. $7\frac{19}{24}\ hrs$

Step-by-step explanation:

Some data's were missing so we have attached the complete information in the attachment.

Given:

Number of Hours Carl worked on Math project = $5\frac{1}{4}\ hrs$

$5\frac{1}{4}\ hrs$ can be Rewritten as $\frac{21}{4}\ hrs$

Number of Hours Carl worked on Math project = $\frac{21}{4}\ hrs$

Number of Hours Sonia worked on Math project = $6\frac{1}{2}\ hrs$

$6\frac{1}{2}\ hrs$ can be rewritten as $\frac{13}{2}\ hrs$

Number of Hours Sonia worked on Math project = $\frac{13}{2}\ hrs$

Number of Hours Tony worked on Math project = $5\frac{2}{3}\ hrs$

$5\frac{2}{3}\ hrs$ can be rewritten as $\frac{17}{3}\ hrs.$

Number of Hours Tony worked on Math project = $\frac{17}{3}\ hrs.$

Now Given:

April worked $1\frac{1}{2}$ times as long on her math project as did Carl.

$1\frac{1}{2}$ can be Rewritten as $\frac{3}{2}$

Number of Hours April worked on math project = $\frac{3}{2}$ $\times$ Number of Hours Carl worked on Math project

Number of Hours April worked on math project = $\frac{3}{2}\times \frac{21}{4} = \frac{63}{8}\ hrs \ \ Or \ \ 7\frac{7}{8} \ hrs$

Also Given:

Debbie worked $1\frac{1}{4}$ times as long as Sonia.

$1\frac{1}{4}$ can be Rewritten as $\frac{5}{4}$ .

Number of Hours Debbie worked on math project = $\frac{5}{4}$ $\times$ Number of Hours Sonia worked on Math project

Number of Hours Debbie worked on math project = $\frac{5}{4}\times \frac{13}{2}= \frac{65}{8}\ hrs \ \ Or \ \ 8\frac{1}{8}\ hrs$

Also Given:

Richard worked $1\frac{3}{8}$ times as long as tony.

$1\frac{3}{8}$ can be Rewritten as $\frac{11}{8}$

Number of Hours Richard worked on math project = $\frac{11}{8}$ $\times$ Number of Hours Tony worked on Math project

Number of Hours Debbie worked on math project = $\frac{11}{8}\times \frac{17}{3}= \frac{187}{24}\ hrs \ \ Or \ \ 7\frac{19}{24}\ hrs$

Hence We will match each student with number of hours she worked.

Student Hours worked

April. $7\frac{7}{8} \ hrs$

Debbie. $8\frac{1}{8}\ hrs$

Richard. $7\frac{19}{24}\ hrs$

5 0

3 years ago

Read 2 more answers

Describe the end behavior of the function. f(x)=x^3+11x^2+35x+30

Lelechka [254]

To determine end behavior, we only have to look at the leading term.

First, the leading term is positive, so we won't have to negate anything.

The leading term has a power that is odd.

Since the exponent is odd, this means that the function goes to positive infinity as x goes to positive infinity.

This also means that the function goes to negative infinity as x goes to negative infinity.

Those are the end behaviors.

Have an awesome day! :)

4 0

3 years ago

What is 15% of 12 meters

melomori [17]

Think of the "of" as a multiplication sign:
(.15)(12)= 1.8 meters
There you go! hope this helps with this problem and future ones.

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3 years ago

Read 2 more answers

Use the discriminant to determine the number of solutions and types of solutions for the quadratic equation, below. Show your wo

maria [59]

Answer:

Step-by-step explanation:

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3 years ago

A caterer is planning a large dinner party for 1001 guests. Each table must seat an equal number of guests, more than 1 guest a

MArishka [77]

Answer:

The prime factorization of 1001 is 7*11*13, so with the constraint that the tables seat between 2 and 19 guests, we can use tables that seat either 7, 11, or 13 guests.

For tables that seat 7, the caterer would need 11*13 = 143 tables.

For tables that seat 11, the caterer would need 7*13 = 91 tables.

Step-by-step explanation:

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3 years ago