Since we need to find the point at which the 2 lines intersect at the x-axis, we need to find where the 2nd line intersects the x axis, by plugging in y=0.
x-2(0)=4
x=4
Therefore, at (4,0).
Now we plug in (4,0) at the first equation.
4b+3(0)=10
4b=10
b=5/2
Therefore, at b=5/2
we know that
Area of the circle is equal to
where
r is the radius
in this problem
therefore
the answer is the option
d. 113.04 sq. in.
1427 = (F) (+ F + 60) (2F - 50) (+ 3F)
<em>Each pair of brackets represents one of the classes - F meaning Freshmen Class. It has been split into brackets for demonstrational purposes - nothing is being multiplied.</em>
F is an unknown number and each other class size is based off of that so we put it in algebraic terms in order to work it out.
1 - Freshmen Class
2 - Sophomore Class (Freshmen Class + 60 more students)
3 - Junior Class ( Twice the size of the Freshmen Class - 50 students)
4 - Senior Class (Three times the size of the Freshmen Class)
All this can be simplified to 7F + 10 = 1427
1427 - 10 = 1417
1417/7 = 202.428....
Is there a mistake in the question?
Following the question with 202 as the answer - the number 1424 is reached
If increased to 203 - 1431 is reached.
The answer shouldn't include half a person.
Add
5
5
to both sides of the equation.
√
2
x
+
13
=
x
+
5
2
x
+
13
=
x
+
5
To remove the radical on the left side of the equation, square both sides of the equation.
(
√
2
x
+
13
)
2
=
(
x
+
5
)
2
(
2
x
+
13
)
2
=
(
x
+
5
)
2
Simplify each side of the equation.
2
x
+
13
=
x
2
+
10
x
+
25
2
x
+
13
=
x
2
+
10
x
+
25
Solve for
x
x
.
x
=
−
2
,
−
6
x
=
-
2
,
-
6
Exclude the solutions that do not make
√
2
x
+
13
−
5
=
x
2
x
+
13
-
5
=
x
true.
x
=
−
2
So, what’s the question, I probably can answer it if you include the question
