This is a quadratic equation, i.e. an equation involving a polynomial of degree 2. To solve them, you must rearrange them first, so that all terms are on the same side, so we get
i.e. now we're looking for the roots of the polynomial. To find them, we can use the following formula:
where 
is a compact way to indicate both solutions 
and 
, while 
are the coefficients of the quadratic equation, i.e. we consider the polynomial 
.
So, in your case, we have 
Plug those values into the formula to get
So, the two solutions are
Since triangle ABC is isosceles, then angles 1 and 9 are congruent.
4x + 3 = 9x - 47
-5x = -50
x = 10
m<1 = 4x + 3 = 40 + 3 = 43
m<1 = m<9 = 43
Triangle DBE is equilateral, so it is also equiangular.
m< 4 = m<5 = m<6 =60
Angles 3 and 8 are supplementary to congruent angles that measure 60 deg, so 180 - 60 = 120, so
m<3 = m<8 = 120
For angles 2 and 7, use the sum of the measures of the angles of a triangle is 180 deg.
m<1 + m<2 + m<3 = 180
43 + m<2 + 120 = 180
m<2 + 163 = 180
m<2 = 17
m<2 = m<7 = 17
Answer:
379.75
Step-by-step explanation:
15.5 x 2450
100
= 379.75
H(1) = 4
h(1)= 5 x 1 -1
=5-1
=4
hope that helps <3
The answer is ae=12 that what i got right on my test
