The roosevelt middle school band is having a fundraiser. They sold a total of 300 hotdogs and hamburgers. Hot dogs were sold for $2 and hamburgers soldfor $3. They made a total of $780 how many individual hamburgers And hotdogs did they sell.

180 hamburgers

120 hotdogs

Step-by-step explanation:

In this question, we are asked to calculate the number of hamburgers and hotdogs sold by a company given the amount made by them and the total number of these snacks sold

We proceed as follows;

Let the amount of hotdogs sold be x and the amount of hamburgers sold be y.

We have a total of 300 snacks sold, mathematically;

x + y = 300 (I)

Now let’s look at the prices.

x number of hotdogs sold at $2, this give a total of $2x hotdogs

y number of hamburgers sold at $3, this give a total of $3y.

Adding both to give total, we have ;

2x + 3y = 780(ii)

This means we have two equations to solve simultaneously. From equation 1, we can say x = 300 -y

Now let’s insert this in the second equation;

2(300-y) + 3y = 780

600-2y + 3y = 780

y = 780-600 = 180

Recall; x + y = 300

x = 300 -y

x = 300-180 = 120

The School band sold a total of; 180 hamburgers and 120 hotdogs.

Step-by-step explanation: For a start we shall represent hotdogs by the letter d and hamburgers by the letter b. If they sold a total of 300 hotdogs and hamburgers, then we can express this as, d + b = 300.

Also if one hotdog was sold for $2 and one hamburger was sold for $3, and altogether realized $780,this can also be expressed as, 2d + 3b = 780.

We now have a pair of simultaneous equations as follows;

d + b = 300 ———(1)

2d + 3b = 780 ———(2)

From equation (1), make d the subject of the equation. Therefore d = 300 - b

Substitute for d into equation (2)

2(300 - b) + 3b = 780

600 - 2b + 3b = 780

Collect like terms

3b - 2b = 780 - 600

b = 180

We can now substitute for the value of b into equation (1)

d + b = 300

d + 180 = 300

Subtract 180 from both sides of the equation

d = 120

Therefore, the school band sold 180 hamburgers and 120 hotdogs

answer: ok

step-by-step explanation:

when functions are first introduced, you will probably have some simplistic "functions" and relations to deal with, usually being just sets of points. these won't be terribly useful or interesting functions and relations, but your text wants you to get the idea of what the domain and range of a function are. small sets of points are generally the simplest sorts of relations, so your book starts with those.

step-by-step explanation:

state the domain and range of the following relation. is the relation a function?

{(2, –3), (4, 6), (3, –1), (6, 6), (2, 3)}

the above list of points, being a relationship between certain x's and certain y's, is a relation. the domain is all the x-values, and the range is all the y-values. to give the domain and the range, i just list the values without duplication:

domain: {2, 3, 4, 6}

range: {–3, –1, 3, 6}

(it is customary to list these values in numerical order, but it is not required. sets are called "unordered lists", so you can list the numbers in any order you feel like. just don't duplicate: technically, repetitions are okay in sets, but most instructors would count off for this.)

while the given set does indeed represent a relation (because x's and y's are being related to each other), the set they gave me contains two points with the same x-value: (2, –3) and (2, 3). since x = 2 gives me two possible destinations (that is, two possible y-values), then this relation is not a function.

note that all i had to do to check whether the relation was a function was to look for duplicate x-values. if you find any duplicate x-values, then the different y-values mean that you do not have a function. remember: for a relation to be a function, each x-value has to go to one, and only one, y-value.

state the domain and range of the following relation. is the relation a function?

{(–3, 5), (–2, 5), (–1, 5), (0, 5), (1, 5), (2, 5)}

i'll just list the x-values for the domain and the y-values for the range: