She has a difference of $2.75, she may have subtracted something wrong while balancing out the check book.
f(x)=x^2+12x+26
Vertex form of f(x) = a(x-h)^2 +k
We change f(x) into vertex form using completing the square method
We take half of square of middle term
middle term is 12
Half of square of 12 is 
then 6^2 = 36
Add and subtract 36 to complete the square
The vertex form 
Answer:
Options (3) and (6)
Step-by-step explanation:
ΔABC is a dilated using a scale factor of 
to produce image triangle ΔA'B'C'.
Since, dilation is a rigid transformation,
Angles of both the triangles will be unchanged or congruent.
m∠A = m∠A' and m∠B = m∠B'
Since, sides of ΔA'B'C' = of the sides of ΔABC
Area of ΔA'B'C' = 
Area of ΔABC > Area of ΔA'B'C'
Since, angles of ΔABC and ΔA'B'C' are congruent, both the triangles will be similar.
ΔABC ~ ΔA'B'C'
Therefore, Option (3) and Option (6) are the correct options.
We will use the law of cosines
<span>side a² = b² + c² -2bc • cos(A)
</span><span>side a² = 729 + 196 -2*27*14 * cos (46)
</span><span>side a² = 925 -(756 * 0.69466)
</span>side a² = <span><span>399.83704
</span>
side a = </span><span><span><span>19.995925585
</span>
</span>
</span>
We could round that to 20
a = 20 b = 27 c =14
We can calculate a triangle's area when we know all 3 sides by using Heron's Formula
<span>area = square root (s • (s - a) • (s - b) • (s - c))
where s is the semi-perimeter </span>
semi-perimeter<span> = (side a + side b + side c) ÷ 2</span>
s = (20 + 27 + 14) / 2
s = 30.5
Now we use Heron's Formula
area = square root (s • (s - a) • (s - b) • (s - c))
area = square root (30.5 • (<span>30.5 - 20) • (</span><span>30.5 - 27) • (</span><span>30.5 - 14))</span>
area = square root (30.5 • (10.5) • (3.5) • (<span>16.5))</span>
<span>area = square root (18494.4375)
</span>
<span><span><span>area = 135.9942553934
</span>
</span>
</span>which rounds to
136 square feet
Source:
http://www.1728.org/triang.htm
