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Elis [28]
3 years ago
15

A rectangular swimming pool measures 50 meters in length and 25 meters in width. If the length in the scale drawing is 10 cm, wh

at is the width of the scale drawing?​
Mathematics
1 answer:
mash [69]3 years ago
5 0

Answer:

The 25 m width of the actual swimming pool in the scale drawing is 5 cm

Step-by-step explanation:

The dimensions of the actual swimming pool given are

Length = 50 meters

Width = 25 meters

The length of the swimming pool on the scale drawing = 10 cm

The scale of the scale drawing is therefore;

10 cm on the scale drawing is equivalent to 50 meters of the actual swimming pool

Which gives;

10 cm:50 m

However, 50 m = 5,000 cm, therefore;

10 cm:5,000 cm or 10 : 5,000

Which is 10 to 5000 or 10/10 to 5000/10 = 1 to 500

The scale of the scale drawing is 1 to 500

or 1 cm of the scale drawing = 5 meters of the actual swimming pool

Therefore, the width of 25 meters of the actual swimming pool can be given as follows;

5 m = 1 cm

1 m = 1 cm/5

25 m = 25× 1 m = 25 × 1 cm/5 = 5 cm

Therefore, the 25 m width of the actual swimming pool in the scale drawing = 5 cm.

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A cylinder has a height h, and a radius (3h-2)
tekilochka [14]

Answer:

V= 4\pi (9h^2 -12h + 4)

Step-by-step explanation:

Given

Shape: Cylinder

Radius = 3h - 2

Height = h

Base\ Area = \pi (3h - 2)^2

Required

Simplify the volume

The volume is calculated as:

Volume = Base\ Area * Height

Substitute values for Base Area and Height

V= \pi(3h-2)^2*h

Expand the bracket

V= \pi(3h-2)(3h-2)*h

Open brackets

V= \pi (9h^2 -6h - 6h + 4) *h

V= \pi (9h^2 -12h + 4) *h

V= 4\pi (9h^2 -12h + 4)

6 0
2 years ago
PLEASE PLEASE PLEASE HELP ME WITH ALGEBRA 1 PLEASE!?!?!?!?!!!?!?!?!?!?! I WILL GIVE LOTS OF POINTS AND BRAINLEST
Damm [24]

<span>1.1 What are the different ways you can solve a system of linear equations in two variables? 

</span><span>A linear equation can be written in many ways, so the way we can use to solve a system of linear equations depends on the form the system is written. There are several methods to do this, the more common are: Method of Equalization, Substitution Method, Elimination Method. 

<span>1.2. Method of Equalization

</span>a. Write the two equations in the style [variable = other terms] either variable x or y.<span>
</span>b. Equalize the two equations 
c. Solve for the other variable and then for the first variable.

</span>

<span> 1.3. Substitution Method

a. Write one of the equations (the one that looks the simplest equation) in the style [variable = other terms] either variable x or y.
b. Substitute that variable in the other equation and solve using the usual algebra methods.
c. Solve the other equation for the other variable.<span>

1.4. Elimination Method

Eliminate</span> means to remove, so this method works by removing variables until there is just one left. The process is:
<span>
</span>a. Multiply an equation by a constant "a" such that there is a term in one equation like [a*variable] and there is a term in the other equation like [-a*variable]
b. Add (or subtract) an equation on to another equation so the aim is to eliminate the term (a*variable)
c. Solving for one variable and the for the other.
</span>

2.1 Benefits of Method of Equalization<span>
</span>


It's useful for equations that are in the form [variable = other terms] (both equations). In this way, it is fast to solve the system of linear equation by using this method. The limitations come up as neither of the equations are written in that way, so it would be necessary to rewrite the two equations to achieve our goal


2.2. Benefits and limitations of Substitution Method


This method is useful for equations that at least one of them is in the form [variable = other terms]. So unlike the previous method, you only need one equation to be expressed in this way. Hence, it is fast to solve the system of linear equation by using this method. The limitations are the same that happens with the previous method, if neither of the equations is written in that way, we would need to rewrite one equation to achieve our goal.


2.3. Benefits and limitations of Elimination Method


It's useful for equations in which the term [a*variable] appears in one equation and the term [-a*variable] appears in the other one, so adding (or subtracting). 


3.1. What are the different types of solutions


When the number of equations is the same as the number of variables there is likely to be a solution. Not guaranteed, but likely.

One solution: It's also called a consistent system. It happens as each equation gives new information, also called Linear Independence.

An infinite number of solutions: It's also called a consistent system, but happens when the two equations are really the same, also called Linear Dependence.

No solutions: It happens when they are actually parallel lines. 


3.2. Graph of one solution system


See figure 1, so there must be two straight lines with different slopes.


3.3. Graph of infinite number of solutions system


See figure 2. So there must be two straight lines that are really the same.


3.4. Graph of no solution system


See figure 3. So there must be two straight lines that are parallel.


4. Explain how using systems of equations might help you find a better deal on renting a car?


Q. A rental car agency charges $30 per day plus 10 cents per mile to rent a certain car. Another agency charges $20 per day plus 15 cents per mile to rent the same car. If the car is rented for one day in how many miles will the charge from both agencies be equal?

A. Recall: 10 cents = 0.1$, 15 cents = 0.15$

If you rent a car from the first car agency your cost for the rental will be:

(1) 30D+0.1M

<span>If you rent a car from the second car agency the total amount of money to pay is:

(2) 20D+0.15M</span>

Given that the problem says you want to rent the car just for one day, then D = 1, therefore:

First agency: 30(1)+0.1M=&#10;\boxed{30+0.1M}

Second agency: 20(1)+0.15M=&#10;\boxed{20+0.15M}

<span>At some number of miles driven the two costs will be the same:

30+0.1M=20+0.15M

Solving for M:

10=0.05M
M=200mi

There is a representation of this problem in figure 4. Up until you drive 200 miles you would save money by going with the second company.</span>

<span>

5. Describe different situations in the real world that could be modeled and solved by a system of equations in two variables </span>

<span>
</span>

For example, if you want to choose between two phone plans. The plan with the first company costs a certain price per month with calls costing an additional charge in cents per minute. The second company offers another plan at a certain price per month with calls costing an additional charge in cents per minute. So depending on the minutes used you should one or another plan.


3 0
3 years ago
What’s the mean of 55,60,60,40,60,75,40,25
Mazyrski [523]

Answer:

51.625

Step-by-step explanation:

you add them all up to get 415 then you divide by eight because there are eight numbers that you add up to get the answer 51.625

8 0
3 years ago
Read 2 more answers
if y = 2x + 7 were changed to y = 1/2 x + 7 , how would the graph of the new function compare with the original ?
iris [78.8K]
Y = 2x + 7.....the slope is 2 and the y int is 7
y = 1/2x + 7...the slope is 1/2 and the y int is 7

In comparison, the only difference is the slope. And if the original equation has a slope of 2, and it is changed to a slope of 1/2, then it will have a " less steep " slope. In other words, the slope will not be as steep....but the y intercepts will be the same.
8 0
3 years ago
Read 2 more answers
25 students are going to a museum and then to lunch. Each student much pay an entrance fee to the museum and $7.25 for lunch. Th
mrs_skeptik [129]
$2.45. Thats it i think but idk the equasion part. Sorry
4 0
3 years ago
Read 2 more answers
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