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Lisa [10]
3 years ago
8

Jill took a total of 6 quizzes over the course of 3 weeks. After attending 7 weeks of school this quarter, how many quizzes will

Jill have taken in total? Solve using unit rates.​
Mathematics
1 answer:
MissTica3 years ago
6 0

Answer:

14 quizzes

Step-by-step explanation:

Jill takes 6 quizzes every 3 weeks.

the unit rate is 3 quizzes every week, because 6 divided by 3 equals 2

now, take the number of quizzes per week and multiply it by 7

(2 x 7) = 14

So, Jill would take 14 quizzes in 7 weeks.

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4/9+r=-4/9<br> What is r
julsineya [31]

Answer:

r = - 8/9

Step-by-step explanation:

\frac{4}{9} +  r =  \frac{ - 4}{9}  \\ r =  \frac{ - 4}{9}  -  \frac{4}{9}  \\ r =  \frac{ - 8}{9}

6 0
3 years ago
Sally’s house is located at (5, −1) and her school is located at (−9, 7). Her best friend Molly lives at
prohojiy [21]

Answer:

B. (−2, 3)

Step-by-step explanation:

To find the midpoint of a segment, add the x-coordinates of the endpoints and divide by 2. Add the y-coordinates of the segments and divide by 2.

x: (5 + (-9))/2 = -2

y: (-1 + 7)/2 = 3

Midpoint: (-2, 3)

Answer: B. (−2, 3)

7 0
3 years ago
Read 2 more answers
Which expression has a positive value?<br> O (-5)(-9)<br> O 2-4<br> O 12+(-4)<br> O-3+(-7)
PolarNik [594]
(-5)(-9) because a negative times a negative equals a positive.
3 0
3 years ago
Read 2 more answers
50 points help Which number is a common factor of 24 and 136?
Tom [10]

Answer:

1, 2, 4, 8. 8 is the greatest common factor.

Step-by-step explanation:

Find factors of both.

24:  1, 2, 3, 4, 6, 8, 12, 24

136: 1, 2, 4, 8, 17, 34, 68, 136

Find common numbers.

24:  1, 2, 3, 4, 6, 8, 12, 24

136: 1, 2, 4, 8, 17, 34, 68, 136

Factors are 1, 2, 4, and 8.

5 0
3 years ago
Read 2 more answers
Using suitable algebraic identities, find the value of<br> 2004^2 - 2003^2 + 2002^2 - 2001^2
PSYCHO15rus [73]

(x+1)^2-x^2=(x^2+2x+1)-x^2=2x+1

Let x=2003. Then

2004^2-2003^2=(2003+1)^2-2003^2=2\cdot2003+1=4007

Similarly, if x=2001, then

2002^2-2001^2=2\cdot2001+1=4003

So

2004^2-2003^2+2002^2-2001^2=4007+4003=8010

Alternatively, you can consider the larger expression

(x+3)^2-(x+2)^2+(x+1)^2-x^2

Expanding each binomial product gives

(x^2+6x+9)-(x^2+4x+4)+(x^2+2x+1)-x^2=4x+6

Then if x=2001, we get

2004^2-2003^2+2002^2-2001^2=4\cdot2001+6=8010

3 0
3 years ago
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