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anastassius [24]

3 years ago

8

The table shows the cost to have various numbers of pizzas delivered from Papas Slice of Italy pizzeria. Is the relationship bet

ween the cost and the number of pizzas proportional? Explain.

2 answers:

Cloud [144]3 years ago

6 0

Answer:

12.5x-5(x-1)

Step-by-step explanation:

you can find the pattern easily

it simplifies as

12.5x-5x+5=7.5x+5

lorasvet [3.4K]3 years ago

5 0

Answer: No

Step-by-step explanation:

We know that the equation for direct proportion between two quantities x ( independent variable) and y (dependent variable) is given by :-

$y=kx$ , where k is the proportionality constant.

if the relationship between the cost (dependent variable ) and the number of pizzas (independent variable ) is proportional, then k in the above equation remains constant for each row.

Let $k=\dfrac{\text{Cost }}{\text{Number of pizzas}}$

1) $k=\dfrac{12.50}{1}=12.50$

2) $k=\dfrac{20}{2}=10$

But 10≠12.50, thus the value for k is not fixed.

Hence, the relationship between the cost and the number of pizzas is not proportional.

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Please help prove these identities!

lord [1]

<h3>Hi! It will be a pleasure to help you to prove these identities, so let's get started:</h3>

<h2>PART a)</h2>

We have the following expression:

$tan(\theta)cot(\theta)-sin^{2}(\theta)=cos^2(\theta)$

We know that:

$cot(\theta)=\frac{1}{cot(\theta)}$

Therefore, by substituting in the original expression:

$tan(\theta)\left(\frac{1}{tan(\theta)}\right)-sin^{2}(\theta)=cos^2(\theta) \\ \\ \\ Simplifying: \\ \\ 1-sin^2(\theta)=cos^2(\theta)$

We know that the basic relationship between the sine and the cosine determined by the Pythagorean identity, so:

$sin^2(\theta)+cos^2(\theta)=1$

By subtracting $sin^2(\theta)$ from both sides, we get:

$\boxed{cos^2(\theta)=1-sin^2(\theta)} \ Proved!$

<h2>PART b)</h2>

We have the following expression:

$\frac{cos(\alpha)}{cos(\alpha)-sin(\alpha)}=\frac{1}{1-tan(\alpha)}$

Here, let's multiply each side by $cos(\alpha)-sin(\alpha)$ :

$(cos(\alpha)-sin(\alpha))\left(\frac{cos(\alpha)}{cos(\alpha)-sin(\alpha)}\right)=(cos(\alpha)-sin(\alpha))\left(\frac{1}{1-tan(\alpha)}\right) \\ \\ Then: \\ \\ cos(\alpha)=\frac{cos(\alpha)-sin(\alpha)}{1-tan(\alpha)}$

We also know that:

$tan(\alpha)=\frac{sin(\alpha)}{cos(\alpha)}$

Then:

$cos(\alpha)=\frac{cos(\alpha)-sin(\alpha)}{1-\frac{sin(\alpha)}{cos(\alpha)}} \\ \\ \\ Simplifying: \\ \\ cos(\alpha)=\frac{cos(\alpha)-sin(\alpha)}{\frac{cos(\alpha)-sin(\alpha)}{cos(\alpha)}} \\ \\ Or: \\ \\ cos(\alpha)=\frac{\frac{cos(\alpha)-sin(\alpha)}{1}}{\frac{cos(\alpha)-sin(\alpha)}{cos(\alpha)}} \\ \\ Then: \\ \\ cos(\alpha)=cos(\alpha).\frac{cos(\alpha)-sin(\alpha)}{cos(\alpha)-sin(\alpha)} \\ \\ \boxed{cos(\alpha)=cos(\alpha)} \ Proved!$

<h2>PART c)</h2>

We have the following expression:

$\frac{cos(x+y)}{cosxsiny}=coty-tanx$

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$cos(x+y)=cos(x)cos(y)-sin(x)sin(y)$

Substituting this in our original expression, we have:

$\frac{cos(x)cos(y)-sin(x)sin(y)}{cosxsiny}=coty-tanx$

But we can also write this as follows:

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We have the following expression:

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