Cisnero discusses writing three stories in one weekend at Iowa, which later connects with friends workshopping stories in her apartment, to show readers how dedicated she has become to her craft. In this way, she creates the effects of a foreshadowing.
<h3>What is
foreshadowing?</h3>
In literature, foreshadowing refers to the subtle reference to an event that is yet to occur.
Foreshadowing could occur or be deployed in several ways. Some of them are:
- Through the use of dialogue
- Through the use of symbols
- Through the use of character reactions etc.
Learn more about Foreshadowing at:
brainly.com/question/96170
I don't understand or see for the matter the question what is it
Answer:
50.27in
Step-by-step explanation:
Given data
Diameter of small tire= 13in
Radius= 13/2= 6.5in
Circumference= 2πr
C= 2*3.142*6.5
C= 40.85 in
Diameter of Big tire= 29in
Radius= 29/2= 14.5in
Circumference= 2πr
C= 2*3.142*14.5
C= 91.118in
Hence the difference in distance between the two tires after on revolution is
=91.118-40.85
=50.27in
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