Use quadratic formula
if you had ax^2+bx+c=0, then
x=
a=1
b=?
c=34
subsitute
=5+/-3i
=5+/-3i
make 5+/-3 into fraction over 2,(10+/-6i)/2
=(10+/-6i)/2
multiply both sides by 2
=10+/-6i
we conclude that -b=10
b=-10
ok so equaton is
x^2-10x+34
Sin0= y/r
X= -4 and r= 9 (radius is also the hypotenuse)
In quadrant II y is positive
X^2 +y^2 = r^2
-4^2 + y^2 = 9^2
16 + y^2 = 81
Y^2 = 81-16
Y^2 = 65
Y= sqrt65
Sin0= sqrt65/9
"total" means "sum" more or less, so the total of m and 10 is m+10
The square of this is (m+10)^2
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If you want to expand this out, then you would get
(m+10)^2 = (m+10)(m+10)
(m+10)^2 = A(m+10) ... let A = m+10
(m+10)^2 = A*m + A*10
(m+10)^2 = m(A) + 10(A)
(m+10)^2 = m(m+10) + 10(m+10) .... plug in A = m+10
(m+10)^2 = m^2+10m+10m+100
(m+10)^2 = m^2+20m+100
Answer:
I know this is late, but for<em> future people</em>:
I believe the answer is <u>2b+3c</u>.
Step-by-step explanation:
I'm pretty sure all you have to do is combine like terms.
I'm also taking the test at the moment.
