1/4 - 1/6 = (6-4)/24 = 2/24 = 1/12
1/3 - 1/4 = (4-3)/12 = 1/12
Look on Google or ask your teacher
Answer:
20
Step-by-step explanation:
The distance between two points is found by
d = sqrt( ( x2-x1)^2+(y2-y1)^2
= sqrt( ( 10- -6)^2 + (-4 -8)^2 )
= sqrt( (16^2 + (-12 )^2)
= sqrt(256+144)
= sqrt(400)
= 20
I think I can.
Let's do it together:
Let's call the number of buses 'B', and the number of vans 'V'.
(Pretty clever so far, don't you think ?)
OK. What do we know ?
-- Each bus holds 51 passengers. The number of passengers in 'B' buses is 51B.
-- A van holds 10 passengers. The number of passengers in 'V' vans is 10V.
-- The total number of passengers ... (51B + 10V) ... is 142.
-- The total number of vehicles ... (B + V) ... is 6, because there are 6 drivers.
Can you make a system of equations out of that information yet ?
How about this:
51B + 10V = 142
B + V = 6
I really think you can handle it from here.
=======================================
Multiply the 2nd equation by 10 :
51B + 10V = 142
10B + 10V = 60
Subtract the 2nd equation from the 1st one:
41B = 82
Divide each side by 41: <em> B = 2 buses</em>
2 buses . . . . 2 drivers . . . . 2 x 51 = 102 passengers
4 vans . . . . . 4 drivers . . . . . 4 x 10 = 40 passengers
6 vehicles . . . 6 drivers . . . . . 102 + 40 = 142 passengers
Step-by-step explanation:
The ratios are;
\dfrac{BC}{AB} = \dfrac{3}{5}
AB
BC
=
5
3
\dfrac{AC}{AB} = \dfrac{4}{5}
AB
AC
=
5
4
\dfrac{BC}{AC} = \dfrac{3}{4}
AC
BC
=
4
3
\dfrac{DE}{AD} = \dfrac{3}{5}
AD
DE
=
5
3
\dfrac{AE}{AD} = \dfrac{4}{5}
AD
AE
=
5
4
\dfrac{DE}{AE} =\dfrac{3}{4}
AE
DE
=
4
3
koGiven that the lengths of the sides are;
\overline {AB}
AB
= 20
\overline {BC}
BC
= 12
\overline {AC}
AC
= 16
\overline {AD}
AD
= 10
\overline {DE}
DE
= 6
\overline {AE}
AE
= 8
The ratios are;
\dfrac{Length \ opposite \ \angle A}{Hypothenus} = \dfrac{BC}{AB} = \dfrac{12}{20} = \dfrac{3}{5}
Hypothenus
Length opposite ∠A
=
AB
BC
=
20
12
=
5
3
\dfrac{Length \ adjacent\ \angle A}{Hypothenus} = \dfrac{AC}{AB} = \dfrac{16}{20} = \dfrac{4}{5}
Hypothenus
Length adjacent ∠A
=
AB
AC
=
20
16
=
5
4
\dfrac{Length \ opposite \ \angle A}{Length \ adjacent \ \angle A} = \dfrac{BC}{AC} = \dfrac{12}{16} = \dfrac{3}{4}
Length adjacent ∠A
Length opposite ∠A
=
AC
BC
=
16
12
=
4
3
\dfrac{Length \ opposite \ \angle A}{Hypothenus} = \dfrac{DE}{AD} = \dfrac{6}{10} = \dfrac{3}{5}
Hypothenus
Length opposite ∠A
=
AD
DE
=
10
6
=
5
3
\dfrac{Length \ adjacent\ \angle A}{Hypothenus} = \dfrac{AE}{AD} = \dfrac{8}{10} = \dfrac{4}{5}
Hypothenus
Length adjacent ∠A
=
AD
AE
=
10
8
=
5
4
\dfrac{Length \ opposite \ \angle A}{Length \ adjacent \ \angle A} = \dfrac{DE}{AE} = \dfrac{6}{8} = \dfrac{3}{4}
Length adjacent ∠A
Length opposite ∠A
=
AE
DE
=
8
6
=
4
3