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mojhsa [17]
3 years ago
9

There are several bowls containing various amounts of grapes on a table. When 12 of the bowls each had 8 more grapes added to th

em, the mean (average) number of grapes per bowl increased by 6.How many bowls of grapes are on the table?
Mathematics
1 answer:
frozen [14]3 years ago
4 0

Answer:

There are 16 bowls

Step-by-step explanation:

Number of bowls = x

Total grapes added = 12 x 8 = 96

The average increased by 6

So 96/6 = 16

There were 16 bowls of grapes on the table

OR

12 x 8 = 96

96/6=16

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Write your answer as a whole number and remainder.530/34
arsen [322]
Hey, Zalah. For my answer, I simplified 530/34. Which equals <span>15.58824. Hope this helps! :) </span> 
3 0
4 years ago
A bag contains yellow marbles and blue marbles, 37 in total. The number of yellow marbles is 7 more than 5 times the number of b
Mazyrski [523]

Answer:

32

Step-by-step explanation:

let b represent the number of blue marbles then

yellow = 5b + 7 and

b + 5b + 7 = 37, that is

6b + 7 = 37 ( subtract 7 from both sides )

6b = 30 ( divide both sides by 6 )

b = 5

Thus there are 5 blue and 5(5) + 7 = 25 + 7 = 32 yellow marbles

5 0
3 years ago
What is the answer for m^2-5m-14=0?
tensa zangetsu [6.8K]

Answer: m = 7 or -2

Step-by-step explanation:

Quadratic Formula!  \frac{{-b (+/-)\sqrt{(b^{2}-4ac }}}{2a}

Polynomial Form: ax^{2} +bx+c

So in this example a: 1  b: -5  c: -14

You get:\frac{{-(-5) (+/-)\sqrt{((-5)^{2}-4(1)(-14) }}}{2(1)}

m = 7 or -2

Check by plugging a value back in for m

(7)^{2} -5(7)-14=0?\\(-2)^{2} -5(-2)-14=0?\\

True for both.

3 0
4 years ago
I’ll give brainleyyyyyyy
Ira Lisetskai [31]

Answer:

0, -2, 2, -1

Step-by-step explanation:

You are trying to make it so that the one of the (  )= 0.

An example is (x+15) or (2x+3)

the first l one would be x= -15 since -15+15 would equal 0.

The second one is -3/2 since it would be -3+3 which would equal 0.

also since the equation starts with x( or -x it doesn't really matter) one of the zeros would also be 0.

Hope this helps!

7 0
3 years ago
What is the 23rd term of the arithmetic sequence where a1 = 8 and a9 = 48?
yKpoI14uk [10]
a_1=8
a_2=a_1+d=8+d
a_3=a_2+d=(8+d)+d=8+2d
a_4=a_3+d=(8+d)+2d=8+3d
...
a_9=8+8d=48\implies d=5
a_{10}=8+9d
...
a_{23}=8+22d=118
8 0
3 years ago
Read 2 more answers
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