A 2.5m long ladder leans against the wall of a building. The base of the ladder is 1.5m away from the wall. What is the height o
2 answers:
Answer:
2 m
Step-by-step explanation:
The model of the situation is a right triangle with legs 1.5 and h the height of the wall. The hypotenuse is the ladder.
Using Pythagoras' identity
h² + 1.5² = 2,5²
h² + 2.25 = 6.25 ( subtract 2.25 from both sides )
h² = 4 ( take the square root of both sides )
h = = 2
The height of the wall is 2 m
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Answer:
2598 square kilometers
Step-by-step explanation:
Hello
Step 1
year one
using a rule of three is possible to find how much is 8.75 od 4500 km2
Let
if
4500 km2 ⇒ 100$
x?km2 ⇒8.75
do the relation and isolate x
at the end of the year one, the area will be
4500-393.75=4106.25
this will be the initial area for the year 2.
Step 2
repite the step 1 with area initial =4106.25 km2
4106.25 km2 ⇒ 100$
x?km2 ⇒8.75
do the relation and isolate x
at the end of the year 2, the area will be
4106-359.29=3746.70
this will be the initial area for the year 3.
Step 3
repite the step 1 with area initial =4106.25 km2
3746.70 km2 ⇒ 100$
x?km2 ⇒8.75
do the relation and isolate x
at the end of the year 3, the area will be
3746.70-327.83=3419.09
this will be the initial area for the year 4.
Step 4
year four
repite the step 1 with area initial =3419.09 km2
3419.09 km2 ⇒ 100$
x?km2 ⇒8.75
do the relation and isolate x
at the end of the year 4, the area will be
3419.09-299.173=3119.82
this will be the initial area for the year 5.
Step 5
year five
repite the step 1 with area initial =3119.82 km2
3119.82 km2 ⇒ 100$
x?km2 ⇒8.75
do the relation and isolate x
at the end of the year 5, the area will be
3119.82-272.99=2846.92
this will be the initial area for the year 6.
Step 6
year six
repite the step 1 with area initial =2846.92km2
2846.92 km2 ⇒ 100$
x?km2 ⇒8.75
do the relation and isolate x
at the end of the year six, the area will be
2846.92-249.10=2597.82 square kilometers
Have a great day.