Search results for "A shipment of 18 cars, some weighing 3,000 pounds, and the others weighing 5,000 pounds each. Together the shipment has a total weight of 30 tons. (60,000 lbs). Find the number of each kind of car."

Answers

Answer 1

Answer: The best way to solve a problem like this is to set up two equations. First assign a variable to each thing you are trying to find. In this case, it's two different kinds of cars. Let's call the cars that weigh 3,000 pounds x, and the ones that weigh 5,000 y. The two equations you should write are:

x+y=18 (because the problem tells you there were 18 cars in total)
3000x+5000y=60000 (because that is the total weight in the problem)

Next, you need to solve for one of the variables. I will solve for x first by subtracting y from both sides of the first equation.

x=18-y

Then you have to plug that into the other equation to get:

3000(18-y)+5000y=60000

Simplify and solve for y:

54000-3000y+5000y=60000
54000+2000y=60000
2000y=6000
y=3

Now that you know what y equals, you can put it into the equation we solved for x:

x=18-3
x=15

So there are 15 cars that weigh 3000 pounds and 3 that weigh 5000.

Answer 2

Answer:

Answer:

The best way to solve a problem like this is to set up two equations. First assign a variable to each thing you are trying to find. In this case, it's two different kinds of cars. Let's call the cars that weigh 3,000 pounds x, and the ones that weigh 5,000 y. The two equations you should write are:

x+y=18 (because the problem tells you there were 18 cars in total)

3000x+5000y=60000 (because that is the total weight in the problem)

Next, you need to solve for one of the variables. I will solve for x first by subtracting y from both sides of the first equation.

x=18-y

Then you have to plug that into the other equation to get:

3000(18-y)+5000y=60000

Simplify and solve for y:

54000-3000y+5000y=60000

54000+2000y=60000

2000y=6000

y=3

Now that you know what y equals, you can put it into the equation we solved for x:

x=18-3

x=15

So there are 15 cars that weigh 3000 pounds and 3 that weigh 5000.

Step-by-step explanation:

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Which statement is true about every rectangular pyramid?Each lateral face has an area less than the area of the base.
At least two of the lateral faces are congruent.
Each lateral face has the same height.
At least one of the faces has a whole number base or height.

Answers

Answer:

At least two of the lateral faces are congruent.

Step-by-step explanation:

Because the words "at least" are used this means that at a minimum 2 of the faces are congruent. Because the base is in the shape of a triangle, this will definitely be true.

hope this helps:)

Answer:

At least two of the lateral faces are congruent.

Step-by-step explanation:

Because the words "at least" are used this means that at a minimum 2 of the faces are congruent. Because the base is in the shape of a triangle, this will definitely be true.

hope this helps:)

in a sale all the normal price is reduced by 15% the sale price of a suit is £110.50 work out the normal price

Answers

Answer:

£130

Step-by-step explanation:

100%-15%=85%

110.5/85=1.3

1.3 is 1%

you need to find 100%, so:

1.3 x 100= 130

a square patio has an area of 225m2. the owner wants to put lights around the patio each string of lights is 25m long. how many strings of lights are needed?

Answers

ok....so look at the attached picture

Final answer:

To find the number of strings of lights needed for the square patio, divide the perimeter of the square by the length of each string of lights. Round up to the nearest whole number.

Explanation:

To find the number of strings of lights needed, we need to divide the perimeter of the square patio by the length of each string of lights. Since the patio is a square, all sides are equal in length. To find the side length, we can take the square root of the area.

In this case, the area is 225m², so the side length of the square patio is √225m = 15m. The perimeter of the square is 4 times the side length, which is 4 × 15m = 60m. Dividing the perimeter by the length of each string of lights, we get 60m ÷ 25m = 2.4. Since we can't have a fraction of a string of lights, we round up to the nearest whole number. Therefore, 3 strings of lights are needed.