answer the question below! and must use way!
1. a trader has a 2 1/2 quintal supply of super quality rice and 6 1/4 quintals of medium quality rice. if both types of rice are sold each 2 quintal, how much rice trader leftover?
2. Note the location of some of the following places!
The school is located to the north of the market and to the west of the city park. the terminal is located to the northeast of the market and to the southwest of the city park.
make a simple floor plan where the place mentioned above!
3. in a field will be built a rectangular fish pond. the size of the length and width of the pool on the floor in a row is 10 cm and 8 cm. if the scale is 1:45, what is the area of land needed to build the pool?
4. as much as 10 liters of water poured into a cube-shaped container. after the container is full of water. it still has 2 liters of water left. what is the cm length of the ribs of the container?
5. following is the data of fish harvest belonging to pak ali for 10 days (in kg)
120; 125; 110; 120; 120;
125; 110; 100; 120; 125;
Determine the biggest and lowest yield!
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1. a trader has a 2 1/2 quintal supply of super quality rice and 6 1/4 quintals of medium quality rice. if both types of rice are sold each 2 quintal, how much rice trader leftover?
Resolution:
the rest of trader's rice
= ( 2 1/2 - 2 ) + ( 6 1/4 - 2 )
= 1/2 + 4 1/4
= 2/4 + 4 1/4
= 6 3/4 kuintal
2.
note the location of some of the following places!
The school is located to the north of the market and to the west of the city park. the terminal is located to the northeast of the market and to the southwest of the city park.
make a simple floor plan where the place mentioned above!
Resolution: The drawing of the drawing is attached yes.
3. Given:
the length of the pool on the floor plan = 10 cm
width of the pool on the floor plan = 8 cm
scale = 1: 45
Asked: what is the area of land needed to build the pool ....
Resolution:
actual ground length
= 10: 1/45
= 10 x 45/1
= 450 cm = 4.5 m
actual ground width
= 8: 1/45
= 8 x 45/1
= 360 cm = 3.6 m
The amount of land needed to build the pond is
= 4.5 m x 3.6 m
= 16.2 m²
4. as much as 10 liters of water poured into a cube-shaped container. after the container is full of water. it still has 2 liters of water left. what is the cm length of the ribs of the container?
Resolution:
The volume of water poured into a cube-shaped container:
= 10 liters - 2 liters
= 8 liters = 8 dm³ = 8,000 cm³
The formula looks for cube volume = s x s x s
So s = ∛ volume
s = ∛8,000
s = ∛ (20 x 20 x 20)
s = 20 cm
5. following is the data of fish harvest belonging to pak ali for 10 days (in kg)
120; 125; 110; 120; 120;
125; 110; 100; 120; 125;
Determine the largest and lowest yield!
Resolution:
Sorted from the fewest:
100 kg ⇒ 1
110 kg ⇒ 2
125 kg ⇒ 3
120 kg ⇒ 4
So, the difference between the largest and lowest yields = 120 - 100 = 20
Hope it helps :)
Resolution:
the rest of trader's rice
= ( 2 1/2 - 2 ) + ( 6 1/4 - 2 )
= 1/2 + 4 1/4
= 2/4 + 4 1/4
= 6 3/4 kuintal
2.
note the location of some of the following places!
The school is located to the north of the market and to the west of the city park. the terminal is located to the northeast of the market and to the southwest of the city park.
make a simple floor plan where the place mentioned above!
Resolution: The drawing of the drawing is attached yes.
3. Given:
the length of the pool on the floor plan = 10 cm
width of the pool on the floor plan = 8 cm
scale = 1: 45
Asked: what is the area of land needed to build the pool ....
Resolution:
actual ground length
= 10: 1/45
= 10 x 45/1
= 450 cm = 4.5 m
actual ground width
= 8: 1/45
= 8 x 45/1
= 360 cm = 3.6 m
The amount of land needed to build the pond is
= 4.5 m x 3.6 m
= 16.2 m²
4. as much as 10 liters of water poured into a cube-shaped container. after the container is full of water. it still has 2 liters of water left. what is the cm length of the ribs of the container?
Resolution:
The volume of water poured into a cube-shaped container:
= 10 liters - 2 liters
= 8 liters = 8 dm³ = 8,000 cm³
The formula looks for cube volume = s x s x s
So s = ∛ volume
s = ∛8,000
s = ∛ (20 x 20 x 20)
s = 20 cm
5. following is the data of fish harvest belonging to pak ali for 10 days (in kg)
120; 125; 110; 120; 120;
125; 110; 100; 120; 125;
Determine the largest and lowest yield!
Resolution:
Sorted from the fewest:
100 kg ⇒ 1
110 kg ⇒ 2
125 kg ⇒ 3
120 kg ⇒ 4
So, the difference between the largest and lowest yields = 120 - 100 = 20
Hope it helps :)
Attachments:
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