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grigory [225]
3 years ago
8

Mrs. Onusko made 60 cookies for a bake sale. She sold 2/3 of them and gave 3/4 of the remaining cookies to the students working

at the sale. How many cookies did she have left?

Mathematics
1 answer:
yaroslaw [1]3 years ago
8 0

Answer:

Step-by-step explanation:

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I really don't understand this I need help
Salsk061 [2.6K]
In looking at it I would guess it is asking which of the points is on the line? To answer that you would have to take each point an plug it into the equation and see if it is true.

3x - 7 = y

A(9,20)  
3(9) - 7 = 20
27 - 7 = 20
20 = 20 True - this point is on the line

B(5,0)
3(5) - 7 = 0
15 - 7 = 0
8 = 0 False - this point is not on the line

C(3,2)
3(3) - 7 = 2
9 - 7 = 2
2 = 2  True - this point is on the line

I hope this helps
7 0
3 years ago
CAN SOMEONE SOLVE THIS PLEASE<br> −4x−(x−1)=3(x−5)
tatuchka [14]

Answer: x = 2

Hope this helps!

8 0
3 years ago
Read 2 more answers
Help me pls... ....
musickatia [10]

Step-by-step explanation:

A1

1 yard = 3 feet

1 foot = 12 inches

therefore,

100 yards = 100×3×12 inches = 3600 inches

A2

again, 1 yard = 3 feet

8 ft = 8/3 = 2.67 yards = 2 2/3 yards

A3

1 gallon = 8 pints

A4

1 liter = 1000 milliliters (remember, "mili" means 1/1000).

therefore,

2.2 liters = 2.2×1000 = 2200 milliliters

A5

again, 1 liter = 1000 milliliters

therefore

1 milliliter = 1/1000 liter

670 milliliters = 670 × 1/1000 liter = 670/1000 liter =

= 0.67 liter

B1

1 kg (kilogram) = 1000 grams ("kilo" means 1000).

we have 3 layers.

each layer weighs (cake plus icing)

1/2 kg + 100 g = 500 g + 100 g = 600 g

3 layers then weigh

3×600 g = 1800 g

the additional candle weighs 250 g.

so, the total is

1800 g + 250 g = 2050 g = 2.05 kg

B2

area is 110 m².

1 m = 100 cm ("centi" means 1/100).

and 1 m² is then 100×100 = 10000 cm².

so, the area in cm² is

110 × 10000 = 1,100,000 cm²

the top area of one brick = 10 × 20 = 200 cm²

so, we need

1,100,000 / 200 = 11,000 / 2 = 5500 bricks

6 0
2 years ago
Shureka Washburn has scores of 67​, 68​, 76​, and 63 on her algebra tests. a. Use an inequality to find the scores she must make
mash [69]
Average=(total number)/(number of items)
given that the final exam counts as two test, let the final exam be x. The weight of the final exams on the average is 2, thus the final exam can be written as 2x because any score Shureka gets will be doubled before the averaging.
Hence our inequality will be as follows:
(67+68+76+63+2x)/6≥71
(274+2x)/6≥71
solving the above we get:
274+2x≥71×6
274+2x≥426
2x≥426-274
2x≥152
x≥76

b] The above answer is x≥76, the mean of this is that if Shureka is aiming at getting an average of 71 or above, then she should be able to get a minimum score of 76 or above. Anything less than 76 will drop her average lower than 71.


8 0
3 years ago
What is the solution, if any, to the inequality |3x|2O?
Lera25 [3.4K]

Answer:

No solutions

Step-by-step explanation:

8 0
3 years ago
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