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motikmotik
3 years ago
5

by 8:00 a.m., the snack bar had made $3,256 in sales. By noon, the snack bar's total sales were $8,821. On average how much mone

y did the snack bar take in each hour between 8:00 a.m. and noon?
Mathematics
1 answer:
sammy [17]3 years ago
7 0

Answer:

$1,391.25

Step-by-step explanation:

Noon would be 12:00 PM which would be 4 hours ahead of 8:00 AM

We subtract $8,821 by $3,256 to get the amount earned between 8:00 AM and 12:00 PM

$8,821 - $3,256

= $5,565

The find the average taken in within those 4 hours we will have to divide the money earned by the hours

$5,565/4

= $1,391.25

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4m≥m+9&gt;-9+4m<br> Please answer urgent
Vsevolod [243]
Next time, please include the instructions.

This inequality could be broken into two parts:

4m\geq m + 9   AND   m + 9 > -9 + 4m

Focus first on the inequality m + 9 > -9 + 4m.  subtract m from both sides, obtaining:

9 > -9 + 3m.  Next, add 9 to both sides, obtaining:  18>3m, or m<6

Last step:  Determine whether the inequality <span>4m≥m+9 is true if m<6.</span>
5 0
3 years ago
The 17 students on the math team want to raise $483.62 to buy practice books they have already raised $218.25 of each students r
sertanlavr [38]

Answer:

Each stuydent must raise $15.61 to reach their goal.

Step-by-step explanation:

Their goal is 483.62, but they already have 218.25 so we subtract,

483.62 - 218.25 = 265.37

Since there are 17 students and they raise the same amount we need to divide by the amount of students to find out how much money each of them need to raise to add up to their goal.

265.37/17 = $15.61

4 0
3 years ago
Guys this makes no sense someone help!! The amount of money raised at a charity fundraiser is directly proportional to the numbe
svlad2 [7]

A)

The formula for direct variation is written as Y = kx, where k is the proportion you need to solve for.

Y would be the amount raised and X would be the number of attendees:

100 = k5

Divide both sides by 5:

k = 100/5

k = 20

B. the constant of variation is the value of k above which is 20

C) Using the formula from A: y = kx, replace k with 20 and x with 60 and solve for y:

y = 20 * 60

y = 1200

They will raise $1,200

2. If the relationship is proportional the ratio would be a constant number. If the relationship is non proportional the ratio would vary between the different values.

5 0
3 years ago
Algebraic expression: t divided by 3-6=14
Lisa [10]
T=7 I think that is the correct awnser
6 0
3 years ago
Determine the longest interval in which the given initial value problem is certain to have a unique twice-differentiable solutio
AnnZ [28]

Answer:

y'' + \frac{7}{t} y = 1

For this case we can use the theorem of Existence and uniqueness that says:

Let p(t) , q(t) and g(t) be continuous on [a,b] then the differential equation given by:

y''+ p(t) y' +q(t) y = g(t) , y(t_o) =y_o, y'(t_o) = y'_o

has unique solution defined for all t in [a,b]

If we apply this to our equation we have that p(t) =0 and q(t) = \frac{7}{t} and g(t) =1

We see that q(t) is not defined at t =0, so the largest interval containing 1 on which p,q and g are defined and continuous is given by (0, \infty)

And by the theorem explained before we ensure the existence and uniqueness on this interval of a solution (unique) who satisfy the conditions required.

Step-by-step explanation:

For this case we have the following differential equation given:

t y'' + 7y = t

With the conditions y(1)= 1 and y'(1) = 7

The frist step on this case is divide both sides of the differential equation by t and we got:

y'' + \frac{7}{t} y = 1

For this case we can use the theorem of Existence and uniqueness that says:

Let p(t) , q(t) and g(t) be continuous on [a,b] then the differential equation given by:

y''+ p(t) y' +q(t) y = g(t) , y(t_o) =y_o, y'(t_o) = y'_o

has unique solution defined for all t in [a,b]

If we apply this to our equation we have that p(t) =0 and q(t) = \frac{7}{t} and g(t) =1

We see that q(t) is not defined at t =0, so the largest interval containing 1 on which p,q and g are defined and continuous is given by (0, \infty)

And by the theorem explained before we ensure the existence and uniqueness on this interval of a solution (unique) who satisfy the conditions required.

7 0
4 years ago
Read 2 more answers
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