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lidiya [134]
2 years ago
7

Someone please help me on this question! no links plz!

Mathematics
2 answers:
Akimi4 [234]2 years ago
8 0
630, you need to find the area of the walls & then add the, together
Alexxx [7]2 years ago
6 0
26x9=234
26x9=234
18x9=162

If we add together, we get 630

So D would be your answer
You might be interested in
The graphs below are both quadratic functions. The equation of the red graph is
Natali5045456 [20]

Answer:

D. g(x)=\frac{1}{5}x^2

Step-by-step explanation:

  • The equation of the red quadratic graph is f(x)=x^2.
  • This is the parent quadratic function or the base of the quadratic functions.
  • This red quadratic graph has been dilated by a certain scale factor  to obtain the blue graph. Recall that, when the scale factor is a fraction, such that 0\: the graph is stretched horizontally.
  • The dilated graph then becomes wider than the parent function.
  • From the given function equations, the possible value of k can only be k=\frac{1}{5}.
  • Therefore the blue graph has equation: g(x)=\frac{1}{5}x^2
  • The correct answer is D.
6 0
2 years ago
Jon's coin collection consists of quarters, dimes and nickels. If the ratio of quarters to dimes is 5:2, and the ratio of dimes
Margarita [4]

Answer:

The ratio of quarters to nickels is:

  • <u>15:8</u>

Step-by-step explanation:

To solve the exercise, you must first consider how big is the ratio of each one to the other, when it is mentioned that the ratio of quarters to dimes is 5:2 it means that for every 5 quarters, Jon has 2 dimes, for the more quarters there are with respect to the dimes, we simply divide the larger value by the smaller:

  • 5/2 = 2.5

Thus, we know that there are 2.5 more quarters than dimes. Now the following proportion: the ratio of dimes to nickels is 3:4, which means that for every 3 dimes there are 4 nickels, as the nickels are greater in quantity, we are going to see how many times they are greater in proportion than the dimes making again a division of the major over the minor:

  • 4/3 = 1.33333333333 (<em>we are going to use the fractional number so as not to lose decimals</em>).

As we can see, the nickels are a little more than 1 with respect to the dimes, now we are going to create two formulas with those values:

  • <em>q = 2.5d </em>
  • <em>n = 4/3 d</em>

Where:

  • q = quarters
  • d = dimes
  • n = nickels

As you see. We only express the calculated proportions, at first glance you can identify that the number of quarters is greater, but to know how much greater with respect to nickels we are going to divide the two values ​​that accompany "d" in the formulas:

  • <u>2.5 / (4/3) = 15/8</u>

If we express it in ratio as it is in the exercise, we obtain:

  • <u>15:8</u>

So, <u>15:8 is the ratio of quarters to nickels</u>, <em>which means that for every 15 quarters, jon has 8 nickels.</em>

8 0
2 years ago
Plz..Help Me Solve This ...
puteri [66]
R = recycling and reuse industry jobs
w = waste management jobs

we know, those two jobs combined are 1,275,000
or
r + w = 1,275,000

we also know that, whatever the "w" jobs are, are really less than "r"
jobs by 1,025,000

"r" has 1,025,000 more than  "w"
or
r = w + 1,025,000
or
r-w = 1,025,000

so... there, just solve by elimination, simple, notice the "w",
is just waiting to be eliminated :)

\bf \begin{cases}&#10;r+\quad  w=1,275,000\\&#10;r-\quad w=1,025,000\\&#10;\textendash\textendash\textendash\textendash\textendash\textendash\textendash\textendash\textendash\textendash\textendash\textendash\textendash\textendash\textendash\textendash\\&#10;\boxed{?}+\boxed{?}=\boxed{?}&#10;\end{cases}
7 0
3 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
4 divided by 1/8 = <br> Please I’m taking math inventory lol
Angelina_Jolie [31]
Your answer is 32! 4 divided by 1/8 is 32.
6 0
2 years ago
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