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1 year ago

Cheese sticks that were previously priced at "10 for $1" are now "2 for $1". Find each percent change.

Mathematics

1 answer:

AysviL [449]1 year ago

7 0

Answer:

The percent decrease in the number of cheese sticks you can buy for $1 is 80%.

Step-by-step explanation:

Given : Cheese sticks that were previously priced at "10 for $1" are now "2 for $1".

To find : The percent decrease in the number of cheese sticks you can buy for $1 ?

Solution :

The formula used to find percent decrease is given by,

$\%\text{ change}=\frac{\text{Amount of change}}{\text{Original amount }}\times 100$

The price change from 10 to 2,

$\%\text{ change}=\frac{10-2}{10}\times 100$

$\%\text{ change}=\frac{8}{10}\times 100$

$\%\text{ change}=80\%$

The percent decrease in the number of cheese sticks you can buy for $1 is 80%.

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Use Euler's formula to derive the identity. (Note that if a, b, c, d are real numbers, a + bi = c + di means that a = c and b =

storchak [24]

Answer with Step-by-step explanation:

We have to prove that

$sin 2\theta=2sin\theta cos\theta$ by using Euler's formula

Euler's formula : $e^{i\theta}=cos\theta+isin\theta$

$e^{i(2\theta)}=(e^{i\theta})^2$

By using Euler's identity, we get

$cos2\theta+isin2\theta=(cos\theta+isin\theta)^2$

$cos2\theta+isin2\theta=(cos^2\theta-sin^2\theta+2isin\theta cos\theta)$

$(a+b)^2=a^2+b^2+2ab, i^2=-1$

$cos2\theta+isin2\theta=cos2\theta+i(2sin\theta cos\theta)$

$cos2\theta=cos^2\theta-sin^2\theta$

Comparing imaginary part on both sides

Then, we get

$sin2\theta=2sin\theta cos\theta$

Hence, proved.

8 0

2 years ago

Mr. Gonzales has only $42.50 to spend at a clothing store. He wants to buy a shirt that costs $29, including tax, and some brace

lisabon 2012 [21]

Answer: The equation is $29 + x*$4.50 = $42.50, and the solution is x = 3

Step-by-step explanation:

The data we have is:

Gonzales has $42.50

He wants to buy:

a shirt that costs $29

some bracelets that cost $4.50 each.

The equation that we need to solve is:

Total cost = money that Gonzales has.

The cost is $29 + x*$4.50

where x is the number of bracelets he can buy.

The equation that we need to solve is:

$29 + x*$4.50 = $42.50

to solve it we must isolate x:

x*$4.50 = $42.50 - $29 = $13.50

x = 13.50/4.50 = 3

So we have that Mr. Gonzales can buy a total of 3 bracelets.

8 0

1 year ago

Read 2 more answers

9. Ms. Ortiz sells tomatoes wholesale. The function p(x)=–80x2 + 320x – 10, graphed below,

labwork [276]

Answer:

<u>The correct answer is B. $310 at $2 per kilogram</u>

Step-by-step explanation:

1. Let's find out the maximum profit Ms. Ortiz can make with a loaf of tomatoes:

function p(x)=–80x² + 320x – 10

For x = 1, price per kilogram of tomatoes would be: 4 - x = 3

–80x² + 320x – 10, replacing with x = 1

-80 * 1² + 320 * 1 - 10 = -80 + 320 - 10 = 230

For x = 2 price per kilogram of tomatoes would be: 4 - x = 2

–80x² + 320x – 10, replacing with x = 2

-80 * 2² + 320 * 2 - 10 = -320 + 640 - 10 = 310

For x = 3, price per kilogram of tomatoes would be: 4 - x = 1

–80x² + 320x – 10, replacing with x = 3

-80 * 3² + 320 * 3 - 10 = -720 + 960 - 10 = 230

<u>The correct answer is B. $310 at $2 per kilogram</u>

7 0

1 year ago

On the first day a total of 40 items were sold for $356. Define the variables and write a system of equations to find the number

Alina [70]

<h3><u><em>Question:</em></u></h3>

On the first day, a total of 40 items were sold for $356. Pies cost $10 and cakes cost $8. Define the variables, write a system of equations to find the number of cakes and pies sold, and state how many pies were sold.

<h3><em><u>Answer:</u></em></h3>

The variables are defined as:

"c" represent the number of cakes sold and "p" represent the number of pies sold

The system of equations used are:

c + p = 40 and 8c + 10p = 356

18 pies and 22 cakes were sold

<h3><em><u>Solution:</u></em></h3>

Let "c" represent the number of cakes sold

Let "p" represent the number of pies sold

Cost of 1 pie = $ 10

Cost of 1 cake = $ 8

Given that total of 40 items were sold

number of cakes + number of pies = 40

c + p = 40 ------ eqn 1

<u><em>Given items were sold for $356</em></u>

number of cakes sold x Cost of 1 cake + number of pies sold x Cost of 1 cake = 356

$c \times 8 + p \times 10 = 356$

8c + 10p = 356 ----- eqn 2

<u><em>Let us solve eqn 1 and eqn 2</em></u>

From eqn 1,

p = 40 - c ---- eqn 3

Substitute eqn 3 in eqn 2

8c + 10(40 - c) = 356

8c + 400 - 10c = 356

-2c = - 44

c = 22

<em>Substitute c = 22 in eqn 3</em>

p = 40 - c

p = 40 - 22

p = 18

Thus 18 pies and 22 cakes were sold

3 0

2 years ago

If a is an arbitrary nonzero constant, what happens to a/b as b approaches 0

Stolb23 [73]

It depends on how b approaches 0

If b is positive and gets closer to zero, then we say b is approaching 0 from the right, or from the positive side. Let's say a = 1. The equation a/b turns into 1/b. Looking at a table of values, 1/b will steadily increase without bound as positive b values get closer to 0.

On the other side, if b is negative and gets closer to zero, then 1/b will be negative and those negative values will decrease without bound. So 1/b approaches negative infinity if we approach 0 on the left (or negative) side.

The graph of y = 1/x shows this. See the diagram below. Note the vertical asymptote at x = 0. The portion to the right of it has the curve go upward to positive infinity as x approaches 0. The curve to the left goes down to negative infinity as x approaches 0.

7 0

2 years ago