Answer:
3 dogs and 5 cats were adopted out on Christmas eve.
Step-by-step explanation:
Let the number of dogs adopted out on Christmas eve be x and the number of cats be y.
From the question,
On Christmas Eve, the animal adopted out 8 total animals
That is, x + y = 8 ...... (1)
and brought in $400 in total adoption fees.
Also from the question,
Each dog adoption fee was $75 and each cat adoption fee was $35.
∴ On Christmas Eve,
75x + 35y = 400 ...... (2)
Now, to determine how many dogs and how many cats were adopted out on Christmas Eve, we will solve the two equations simultaneously.
From equation (1)
x + y = 8
Make x the subject of the relation.
∴ x = 8 - y ...... (3)
Substitute the value of x into equation (2)
75(8 - y) + 35y = 400
600 - 75y + 35y = 400
600 -40y = 400
600 - 400 = 40y
200 = 40y
∴ 40y = 200
y = 200/40
y = 5
Substitute the value of y into equation (3)
x = 8 - y
x = 8 - 5
x = 3
∴ x = 3 and y = 5
Hence, 3 dogs and 5 cats were adopted out on Christmas eve.
Answer:
Mixed Fraction
Step-by-step explanation:
A mixed fraction is a fraction with a whole number attached to it like 5 1/5 or 3 1/2
Answer:
Set up the equation. On a piece of paper, write the number being divided on the right, under the division symbol, and the number doing the division to the left on the outside. ...
Divide the first digit. ...
Divide the first two digits. ...
Enter the first digit of the quotient.
hope this helps :)
Answer:
Domain = {1,2,3,4,5,6} and Range = {42,49,56,63,70,77}
Step-by-step explanation:
The given function is :
y = 7x + 35
Where
x is no of members
y is the monthly cost
Your group has enough money for up to six members to joim fitness club. It means domain (the set of input values) are:
Domain = {1,2,3,4,5,6}
Put x = 1 to 6 one by one,
y = 7(1) + 35 = 42
y = 7(2) + 35 = 49
y = 7(3) + 35 = 56
y = 7(4) + 35 = 63
y= 7(5) + 35 = 70
y = 7(6) + 35 = 77
Range = {42,49,56,63,70,77}
