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lions [1.4K]
4 years ago
10

Chloe sorts her beads. The number of red beads she has is 5 5/6 times the number of green beads. If she has 60 green beads, how

many rec beads dose she have?
Mathematics
2 answers:
love history [14]4 years ago
5 0

Answer:

60x5 would equal 350, right? So then to break it down, just add this. 300+50=350 beads. Hope this helps.

Molodets [167]4 years ago
4 0
60*5=300+50=350 beads (I'm pretty sure this is right.)
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Please help! will give brainliest
Juliette [100K]

Answer:

yes, the point (-1, 5) is in the solution set

Step-by-step explanation:

y ≥ 2x + 3

Is 5 ≥ 2(-1) + 3?  Yes

y + x > 0

Is 5 + (-1) > 0?  Yes

4 0
3 years ago
Read 2 more answers
1. In a class there are 40 students. 90% of them take exams; 3/4 of those who took the exam passed. How many students graduated?
Tpy6a [65]

Answer:

A. 27

Step-by-step explanation:

There are 40 students and 90% of them take exams

      (40 students) * (0.9) = 36 students took exams

Of those 36 students, 75% (or 3/4) of them passed. Assuming passing exams means graduating:

      (36 students) * (0.75 ) = 27 students graduated

6 0
3 years ago
Can someone explain to me how to do this
-BARSIC- [3]
For every liter there is 1000 milliliters so 5*200=1000
You only need one because 2liters is 2000 milliliters
6 0
3 years ago
Solve image below:
ludmilkaskok [199]

Answer: x=\frac{6}{-3y+8}

Step-by-Step Explanation:

Let's solve for x.

\frac{x-2}{3y-5} =\frac{x}{3}

Step 1: Multiply both sides by 3y-5.

x-2=\frac{3xy-5x}{3}

Step 2: Multiply both sides by 3.

3x-6=3xy-5x

Step 3: Add -3xy to both sides.

3x-6+-3xy=3xy-5x+-3xy

-3xy+3x-6=-5x

Step 4: Add 5x to both sides.

-3xy+3x-6+5x=-5x+5x

-3xy+8x-6=0

Step 5: Add 6 to both sides.

-3xy+8x-6+6=0+6

-3xy+8x=6

Step 6: Factor out variable x.

x(-3y+8)=6

Step 7: Divide both sides by -3y+8.

\frac{x(-3y+8)}{-3y+8}=\frac{6}{y-3y+8}

x=\frac{6}{-3y+8}

Answer:

x=\frac{6}{-3y+8}

4 0
3 years ago
B/3 + 1/7=13 can someone tell me what b is , thanks :))
nikklg [1K]
B is 4 thats what i got
4 0
3 years ago
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