Answer:
C
Step-by-step explanation:
183-160=23 or 160+23=183
Hope this helps!
Answer:
Equation B
Step-by-step explanation:
Equation B because:
it is 7x + 3
so for the first one, it's 7 + 3
the second one is 14 + 3
the third one is 21 + 3
the fourth one is 28 + 3
Please Mark Brainliest
Answer:
(C) 220
Step-by-step explanation:
Let x represent the number of adult tickets sold and y represent the number of student tickets sold. With the information given, we can set up two equations:
(Since for every adult ticket sold, $5 is made and for every student ticket sold, $3 is made)
In the first equation, we can represent x in terms of y:

And then, we can substitute x in the second equation for 360 - y to get:
which simplifies to:
and therefore,
.
Hope this helps :)
Answer:
Step-by-step explanation:
From the given information:
r = 10 cos( θ)
r = 5
We are to find the the area of the region that lies inside the first curve and outside the second curve.
The first thing we need to do is to determine the intersection of the points in these two curves.
To do that :
let equate the two parameters together
So;
10 cos( θ) = 5
cos( θ) = 

Now, the area of the region that lies inside the first curve and outside the second curve can be determined by finding the integral . i.e









The diagrammatic expression showing the area of the region that lies inside the first curve and outside the second curve can be seen in the attached file below.
Answer:
x=2 y=1
Step-by-step explanation:
3 times 2 =6
plus 1 = 7