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3 years ago

9

The Natural History Museum has a 1:75 scale model of an apatosaurus dinosaur. The length of the model is 30 centimeters. Find th

e actual length x (in meters) of an apatosaurus.
x= ? meters

1 answer:

lora16 [44]3 years ago

8 0

Answer:

x = 22.50m

Step-by-step explanation:

(75/1) = (x / 30)

x = 30 * (75/1)

x = 2250 cm

we know that,

1m = 100cm

We have then:

x = 2250 * (1/100) = 22.50m

yo welcome!

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Gold’s gym charges $30 membership fee plus $3 for each workout, while Titletown charges $40 membership fee plus $1 for each work

ollegr [7]

Answer:

5 workouts should be performed to equalize the costs between both gyms.

Step-by-step explanation:

Given that Gold’s gym charges $ 30 membership fee plus $ 3 for each workout, while Titletown charges $ 40 membership fee plus $ 1 for each workout, to determine how many workouts does a person have to do until the total costs are equal, the following calculation must be performed:

Gold's Gym: y = 30 + 3x

Titletown Fitness: y = 40 + x

Gold's Gym: y = 30 + 3x2 = 36

Titletown Fitness: y = 40 + 2 = 42

Gold's Gym: y = 30 + 3x3 = 39

Titletown Fitness: y = 40 + 3 = 43

Gold's Gym: y = 30 + 3x4 = 42

Titletown Fitness: y = 40 + 4 = 44

Gold's Gym: y = 30 + 3x5 = 45

Titletown Fitness: y = 40 + 5 = 45

Thus, 5 workouts should be performed to equalize the costs between both gyms.

3 0

2 years ago

The County Fair has to 2 ticket options.Option 1 has an entry fee of $5 and charges $0.65 per ride per ride. Option 2 has an ent

Ugo [173]

For 25 tickets total cost of option one and option two to be the same.

Step-by-step explanation:

Let us assume the total number of rides = m

for which BOTH options cost same.

Case: 1

The cost of entry fee = $5

The cost per ride = $0.65

So, the cost of m rides = m x ( Cost of 1 ride )

= m x ($0.65 ) = 0.65 m

Cost of purchasing m tickets in first ride = Entry Fee + Per ticket cost

= 5 +0.65 m ..... (1)

Case: 2

The cost of entry fee = $10

The cost per ride = $0.45

So, the cost of m rides = m x ( Cost of 1 ride )

= m x ($0.45 ) = 0.45 m

Cost of purchasing m tickets in second ride = Entry Fee + Per ticket cost

= 10 +0.45 m ..... (2)

Now, equating (1) and (2), we get:

5 +0.65 m = 10 +0.45 m

or, 0.20 m = 5

or, m = 5/0.20 = 25

or, m = 25

Hence, for 25 tickets total cost of option one and option two to be the same.

5 0

3 years ago

Solve |x+2| >5-(x-1)^2 by graphing

Anna35 [415]

Answer:

x

<

−

1

or

x

>

2

Step-by-step explanation:

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3 years ago

Solve for the variable x<br><br> m = -12g + x

mario62 [17]

Answer:

x=m+12g

Step-by-step explanation:

m=-12g+x

making x subject of formula, we have,

x=m+12g

3 0

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