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never [62]
3 years ago
15

A roller coaster car goes above and below ground.Use the number line to show it changes in height.What is the height of the car

at the end of the ride

Mathematics
2 answers:
Ksju [112]3 years ago
6 0
The height of the car at the end of the ride is 4 meters
Gnesinka [82]3 years ago
6 0

Initial position of roller coaster= 1 meter above the ground=+1

Now, Position of the Roller coaster when it goes downwards by 4 meters

                                                                      =1-4

                                                                      = -3 meter

Now, it rises 13 meter above from it's final position.

So, Position of Roller coaster = -3 +13=10 meter above from ground level=+10

Now, The Roller coaster drops down by 6 meters from the last position it was,

 So, Position of Roller Coaster

   =10+(-6)

    =4 meters above the ground level  

Height of car from ground level=4 meter               

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25x2-49x4= -146

Step-by-step explanation:

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Step 1: 1x-21+3=7 step 2: 1x-21=7-3 step 3: 1x-21=4 what would the next step be to solve equation
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Add 21 to both sides
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3 years ago
Two competitive neighbours build rectangular pools that cover the same area but are different shapes. Pool A has a width of (x +
GenaCL600 [577]

<u>Answer: </u>

a)Dimensions of pool A are length = 6.667m and width = 3.667 m and dimension of pool B are length = 7.333m and width = 3.333m.

b) Area of pool A is equal to area of pool B equal to 24.44 meters.

<u> Solution: </u>

Let’s first calculate area of pool A .

Given that width of the pool A = (x+3)  

Length of the pool A is 3 meter longer than its width.

So length of pool A = (x+3) + 3 =(x + 6)

Area of rectangle = length x width

So area of pool A =(x+6) (x+3)        ------(1)

Let’s calculate area of pool B

Given that length of pool B is double of width of pool A.

So length of pool B = 2(x+3) =(2x + 6) m

Width of pool B is 4 meter shorter than its length,

So width of pool B = (2x +6 ) – 4 = 2x + 2

Area of rectangle = length x width

So area of pool B =(2x+6)(2x+2)        ------(2)

Since area of pool A is equal to area of pool B, so from equation (1) and (2)

(x+6) (x+3) =(2x+6) (2x+2)    

On solving above equation for x    

(x+6) (x+3) =2(x+3) (2x+2)  

x+6 = 4x + 4    

x-4x = 4 – 6

x = \frac{2}{3}

Dimension of pool A

Length = x+6 = (\frac{2}{3}) +6 = 6.667m

Width = x +3 = (\frac{2}{3}) +3 = 3.667m

Dimension of pool B

Length = 2x +6 = 2(\frac{2}{3}) + 6 = \frac{22}{3} = 7.333m

Width = 2x + 2 = 2(\frac{2}{3}) + 2 = \frac{10}{3} = 3.333m

Verifying the area:

Area of pool A = (\frac{20}{3}) x (\frac{11}{3}) = \frac{220}{9} = 24.44 meter

Area of pool B = (\frac{22}{3}) x (\frac{10}{3}) = \frac{220}{9} = 24.44 meter

Summarizing the results:

(a)Dimensions of pool A are length = 6.667m and width = 3.667 m and dimension of pool B are length = 7.333m and width = 3.333m.

(b)Area of pool A is equal to Area of pool B equal to 24.44 meters.

5 0
3 years ago
Find two acute angles that satisfy the equation sin(2x + 7) = cos(x + 23). Check that your answers make
Vera_Pavlovna [14]

Answer:

The smaller angle is 43 and the larger angle is 47

Step-by-step explanation:

3 0
2 years ago
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