Answer:
2x^2=3x+2
2x^2-3x-2=0
(x-2)(x+1/2)=0
x=2 or x=-1/2
The sum of 2 solution 2+-1/2=3/2
Step-by-step explanation:
In a boxplot, there are 5 figures that are denoted, these are known as the 5-figure summary:
(1) the black dot to the very left of the boxplot is the minimum, or in other words the lowest data value
(2) the second black dot from the left is Q1 (quartile 1) and the area between the minimum and Q1 represents 25% of the data
(3) the middle dot is the median (or sometimes called Q2), which is the middle value of the data set - in this case it is 24
(4) the black dot to the right of the median represents Q3
(5) the black dot to the very right is the maximum value, or the highest data value
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