Fountain drinks, regularly $.99, are now 50% off. what is the sale price for one fountain drink?

Mathematics

1 answer:

sergij07 [2.7K]3 years ago

4 0

50 cents, since $.99 would have to be divided into $.49.5 and it's usually rounded up

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pat needs 14 sticks of pepperoni to make 28 pizzas.How many sticks of pepperoni are needed to make 36 pizzas

Marizza181 [45]

Answer:

18 sticks of pepperoni

Step-by-step explanation:

In order to solve this problem, we have to find out how many pepperoni sticks are needed for just 1 pizza. If it takes 14 sticks to make 28 pizzas, that means it takes only half a stick to make a single pizza.

Now that we have this information, we can multiply 1/2 by 36 to find out how many sticks of pepperoni are needed to make 36 pizzas.

$\frac{1}{2} * 36\\\\\frac{1}{2} * \frac{36}{1} \\ =\\\frac{36}{2} \\ =\\\frac{18}{1}$

Therefore, it takes 18 sticks of pepperoni to make 36 pizzas. I hope this helps! Have a lovely day!! :)

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1 year ago

The price of a car was increased from $3,600 to $4,500. What was the percent of the increase?

Nataly [62]

Answer:

25%

Step-by-step explanation:

.25(3,600)=900

3,600+900=4,500

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2 years ago

Read 2 more answers

Set up but do not solve for the appropriate particular solution yp for the differential equation y′′+4y=5xcos(2x) using the Meth

taurus [48]

Answer:

So, solution of the differential equation is

$y(t)=-\frac{5x^2}{4}\cot 2x\cdot \cos 2x+c_1e^{-2it}+c_2e^{2it}\\$

Step-by-step explanation:

We have the given differential equation: y′′+4y=5xcos(2x)

We use the Method of Undetermined Coefficients.

We first solve the homogeneous differential equation y′′+4y=0.

$y''+4y=0\\\\r^2+4=0\\\\r=\pm2i\\\\$

It is a homogeneous solution:

$y_h(t)=c_1e^{-2i t}+c_2e^{2i t}$

Now, we finding a particular solution.

$y_p(t)=A5x\cos 2x\\\\y'_p(t)=A5\cos 2x-A10x\sin 2x\\\\y''_p(t)=-A20\sin 2x-A20x\cos 2x\\\\\\\implies y''+4y=5x\cos 2x\\\\-A20\sin 2x-A20x\cos 2x+4\cdot A5x\cos 2x=5x\cos 2x\\\\-A20\sin 2x=5x\cos 2x\\\\A=-\frac{x}{4} \cot 2x\\$

we get

$y_p(t)=A5\cos 2x\\\\y_p(t)=-\frac{5x^2}{4}\cot 2x\cdot \cos 2x\\\\\\y(t)=y_p(t)+y_h(t)\\\\y(t)=-\frac{5x^2}{4}\cot 2x\cdot \cos 2x+c_1e^{-2it}+c_2e^{2it}\\$