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Mila [183]
2 years ago
9

I need help can some plz help

Mathematics
1 answer:
enot [183]2 years ago
5 0
Replace the x and y and the answer would be A
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courtney has 155 coins in her purse. all of the coins are quarters and dimes, worth a total of $30.20. how many quarters does sh
Assoli18 [71]

Answer: she would have 98 quarters

3 0
2 years ago
Solve for r -4(r+6)= -49
tatiyna

Answer: r= 25/3

Step-by-step explanation:

r-4(r+6)=-49 distribute the 4

r-4r-24=-49 combine like terms

-3r-24=-49 add 24 to both sides

-3r=-25 divide by -3

r= 25/3 answer

7 0
3 years ago
The following menu items were ordered at a pizza parlor on Friday. 3 salads, 8 pizzas, 4 sides of breadsticks, and 5 pepperoni c
den301095 [7]

Answer:

Using proportions, the pizza parlor should expect 20 breadsticks and 25 calzone orders if there were 100 total orders.

Step-by-step explanation:

Orders:

Salads                           3

Pizzas                           8

Sides of breadsticks   4

Pepperoni calzones    5

Total orders:              20


Proportion of sides breadsticks=4/20=0.2

Proportion of pepperoni calzones=5/20=0.25


For 100 total orders:

Sides of breadsticks=0.2(100)=20

Pepperoni calzones=0.25(100)=25


6 0
2 years ago
A school administrator will assign each student in a group of n students to one of m classrooms. If 3 < m < 13 < n, is
tatiyna

Answer:

Hence to get same number of students in each classroom,the sufficient condition is that assign 13n students to each classroom.

Step-by-step explanation:

Given:

There are m classrooms and n be the students

3<m<13<n.

To Find:

Whether it is possible to assign each of n students to one of m classrooms with same no.of students.

Solution:

This problem is related to p/q form  has to be integer in order to get same no of students assigned to the classroom.

As similar as ,n/m ratio

So 1st condition is that,

If it is possible to assign the n/m must be integer and n should be multiple of m,

when we assign 3n students to m classrooms ,we cannot say that 3n/m= integer so that  n is greater than 13 i.e n=14 and m=6

hence they are not multiple of each other so they will not make same students in each classrooms.

Otherwise,n=14 and m=7 they will give same number but this condition is not sufficient condition to assign the student.

So 2nd condition is that ,

When we assign 13n students to m classrooms, as 13 is prime number and

3<m<13 which implies the 13n/m to be integer so n and m must be multiple of each other.

Suppose n=20 and m=5 classrooms

then 13*20=260 ,

260/5=52 students in each classroom,

4 0
3 years ago
Find two positive numbers satisfying the given requirements.
Umnica [9.8K]

Answer:

8 and 19

Step-by-step explanation:

To some this, let's first list all the factors of 152. They are;

1, 2, 4, 8, 19, 38, 76, 152.

Now, let's arrange them to reflect being multiplied to get 152.

Thus;

1 × 152 = 152

2 × 76 = 152

4 × 38 = 152

8 × 19 = 152

Also, let's do the same for their sum;

1 + 152 = 153

2 + 76 = 78

4 + 38 = 42

8 + 19 = 27

Looking at the figures above, the ones that their product is 152 but have the least sum are 8 and 19

8 0
2 years ago
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