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krok68 [10]
3 years ago
11

I need help because this I did it but I haven’t done it in division

Mathematics
1 answer:
tigry1 [53]3 years ago
5 0
Can you give a clearer picture please?
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I need help please I don’t know what to it is due today
Kazeer [188]

Answer:

12 blue necklaces + 12 red necklaces = 24 necklaces. He will have 1 blue bead left over and 1 red bead left over.

Step-by-step explanation:

37/3 = 12.333 or 12 r1

25/2= 12.5 or 12 r1

1+1=2

6 0
3 years ago
Read 2 more answers
-9q = 63 what would be q
Snezhnost [94]

Answer:

-7

Step-by-step explanation:

-9q = 63

q= 63/-9 = -7

7 0
3 years ago
The average of a list of 4 numbers is 90.0. A new list of 4 numbers has the same first 3 numbers as the original list, but the f
Thepotemich [5.8K]

Answer:

the average of this new list of numbers is 94

Step-by-step explanation:

Hello!

To answer this question we will assign a letter to each number for the first list and the second list of numbers, remembering that the last number of the first list is 80 and the last number of the second list is 96

for the first list

\frac{a+b+c+80}{4} =90

for the new list

\frac{a+b+c+96}{4} =X

To solve this problem consider the following

1.X is the average value of the second list

2. We will assign a Y value to the sum of the numbers a, b, c.

a + b + c = Y to create two new equations

for the first list

\frac{y+80}{4} =90

solving  for Y

Y=(90)(4)-80=280

Y=280=a+b+c

for the second list

\frac{y+96}{4} =X\\

\frac{280+96}{4} =X\\x=94

the average of this new list of numbers is 94

4 0
3 years ago
Read 2 more answers
Solve for x: 2/3=8/x+6
Tresset [83]

Answer:

x = 6

Step-by-step explanation:

2/3=8/x+6

Using cross products

2 * (x+6) = 3*8

2(x+6) = 24

Divide by 2

x+6 =12

Subtract 6

x = 6

8 0
3 years ago
Read 2 more answers
Given the function f(x) = 5(x+4) − 6, solve for the inverse function when x = 19.
meriva

Answer:

5.74

Step-by-step explanation:

fx=5(x+4)-6

f(19)=5(19+4)-6

19f=5(23)-6

19f=115-6

19f=109

f=109÷19

f=5.74 or 19 14/19

7 0
3 years ago
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