We have that
<span>triangle ABC
where
A(-5, 5), B(1, 1), and C(3, 4) are the vertices
using a graph tool
see the attached figure
the hypotenuse is the segment AC
find the equation of the line AC
</span>A(-5, 5) C(3, 4)
<span>
step 1
find the slope m
m=(y2-y1)/(x2-x1)-----> m=</span>(4-5)/(3+5)-----> m=-1/8
step 2
with C(3,4) and m=-1/8
find the equation of a line
y-y1=m*(x-x1)-----> y-4=(-1/8)*(x-3)----> y=(-1/8)*x+(3/8)+4
y=(-1/8)*x+(3/8)+4----> multiply by 8----> 8y=-x+3+32
8y=-x+35
the standard form is Ax+By=C
so
x+8y=35
A=1
B=8
C=35
the answer isx+8y=35
Answer:
157
Step-by-step explanation:
8.00 +(0.06x) = 18.27
0.08x = 18.27 - 6.00
0.08x = 12.56
12.56/0.08=
157
Answer:x=3/9
Step-by-step explanation:
X directly proportional to Y
X=ky
Make k the subject of formula
k=x/y
K=27/81
Divided by 9
K=3/9
Answer:
3+16 divided by 4 is 4.75
4×7-2×6..... is5.33
168
The sine of any acute angle is equal to the cosine of its complement. The cosine of any acute angle is equal to the sine of its complement. of any acute angle equals its cofunction of the angle's complement. Yes, there is a "relationship" regarding the tangent of the two acute angles (A and B) in a right triangle.