3.9x+0.8y is the answer to your question.
Answer:
Step-by-step explanation:
1
the initial level is below normal so it is represented by a negative number. the level continued to decrease in June and July so those changes are also represented by negative numbers. find the sum of these values to find what the new level was with respect to the normal level
2
write the mixed numbers as improper fractions
3
get common denominator
4
add the numerators since all the numbers are negative and have a common denominator. write the improper fraction as a mixed number
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
Answer: it's not necessary so i can't round to the nearest tenth
Area of a kite = (1/2) (d^1 * (d^2)
d^1 = 7
d^2 = 2
A = (1/2)2 * 7
1/2 = 1
1 * 7 = 7
A =7cm^2