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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 is
x+8y=35
<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 is
x+8y=35
4, 5, and 7 are mutually coprime, so you can use the Chinese remainder theorem right away.
We construct a number such that taking it mod 4, 5, and 7 leaves the desired remainders:
- Taken mod 4, the last two terms vanish and we have
so we multiply the first term by 3.
- Taken mod 5, the first and last terms vanish and we have
so we multiply the second term by 2.
- Taken mod 7, the first two terms vanish and we have
so we multiply the last term by 7.
Now,
By the CRT, the system of congruences has a general solution
or all integers ,
, the least (and positive) of which is 27.