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3 0

Ok. So she wants to advertise that there are 100 different dolls available. If she adds another category, then at most she already had 99 different types. If I'm right (don't take my word for it) but the minimum she would have to have is one.

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How can the average rate of change be identified for a function?

Digiron [165]

Answer:

$Rateofchange=\frac{f(x_2)-f(x_1)}{x_2-x_1}$

where x₁ and x₂ are values in the interval [x,y] respectively

Step-by-step explanation:

Well, first to determine the average rate of change of a function, you should have the interval of the values of x for the function.

So lets assume you have a function;

$f(x)=x^3-4x$

And the interval as [1,3]

Then the average rate of change for the function f(x) will be;

$=\frac{f(x_2)-f(x_1)}{x_2-x_1}$

where x₁ and x₂ are the interval coordinates x,y respectively. In this case x₁=1 and x₂=3

To find the average rate of change in this example will be;

$=\frac{f(x_2)-f(x_1)}{x_2-x_1} \\\\=f(x_2)=f(3)=3^3-4(3)=27-12=15\\\\=f(x_1)=f(1)=1^3-4(1)=1-4=-3\\\\\\=x_2-x_1=3-1=2\\\\\\=\frac{15--3}{2} \\\\=\frac{18}{2} =9$

8 0

3 years ago

The 8th grade is selling tickets to the 8th grade formal.The cost per ticket is $15.The decorations and food cost $600,and each

MArishka [77]

Answer:

Step-by-step explanation:

Each ticket is $15. The number of tickets is what we are trying to solve for. The class spends a certain amount of money to prepare for the formal. They hope that the money they make in ticket sales is MORE than what they spend. The expression that represents the number of tickets at $15 each is 15x, where x is the number of tickets. They hope that the sales are greater than what they spend, so what we have so far is

15x >

Greater than what, though? What do they spend? They spend 600 for the food, so

15x > 600...

but they also have to print a certain, unknown number of tickets at .50 each. The expression that represents the printing of each ticket is .5x (we can drop the 0; it doesn't change the answer or make it wrong if we drop it off). So the cost for this affair is the food + the printing.

15x > 600 + .5x

Solve this inequality for x. Begin by subtracting .5 from both sides to get

14.5x > 600 so

x > 41.3

Because we are not selling (or printing) .3 of a ticket, it's safe to say (and also correct!) that they need to sell (and print) 41 tickets. If they sell 41 tickets, the profit is found by

15(41) > 600 + .5(41)

615 > 600

This means that at 41 tickets, they make a profit. At 40 tickets, the inequality looks like this:

15(40) > 600 + .5(40) and

600 > 620. This is not true, so 40 tickets isn't enough.

8 0

3 years ago

Determine the values of the variables in the parallelograms below help me please

PolarNik [594]

Answer:

Part 1)

$x=-11$