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Nady [450]
2 years ago
12

A coffee shop gives the customer a free cup of coffee after the purchase of 5 cups: a.) industrial marketing program b.) loyalty

marketing program c.) product sampling program d.) cooperative marketing program
Mathematics
1 answer:
AveGali [126]2 years ago
3 0
I think it’s a loyalty marking program. I think it’s that because I feel like it’s the only answer that really makes sense bc the customer is being loyal to the service in order to receive their “reward”
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Did i get almost an entire lesson on math done in a day? <br> Yes<br> Was it hard?<br> N o p e
lions [1.4K]
Good job!!!!!!!!!!!!!!!
3 0
3 years ago
Read 2 more answers
What is a simple way to solve for the sum and difference of 2 cubes? For example, 27+64<img src="https://tex.z-dn.net/?f=x%5E3"
salantis [7]

Answer:

(3 + 4x)(9 - 12x + 16x²) = 0

(4m - 1)(16m² + 4m + 1) = 0

Step-by-step explanation:

Here we have to solve the sum of two cubes which is 27 + 64x^{3}

Now, the equation is 27 + 64x^{3} = 0

⇒ 3³ + (4x)³ = 0

⇒ (3 + 4x)[3² - 3(4x) + (4x)²] = 0

⇒ (3 + 4x)(9 - 12x + 16x²) = 0

So, (3 + 4x) = 0 or (9 - 12x + 16x²) = 0

Therefore, from the above two relation we can solve for x.

One root will be - \frac{3}{4} and the others we will get by applying Sridhar Acharya Formula, which will give a pair of conjugate imaginary roots of the equation.  

Again, we have to solve the difference of two cubes which is 64m^{3} - 1

Now, the equation is 64m^{3} - 1 = 0

⇒ (4m)³ - 1³ = 0

⇒ (4m - 1)[(4m)² + 4m(1) + 1²] = 0

⇒ (4m - 1)(16m² + 4m + 1) = 0

So, (4m - 1) = 0 or (16m² + 4m + 1) = 0

Therefore, from the above two relation we can solve for m.

One root will be \frac{1}{4} and the others we will get by applying Sridhar Acharya Formula, which will give a pair of conjugate imaginary roots of the equation.  (Answer)

8 0
3 years ago
Helppppppppppp:)))))))))
Whitepunk [10]

Hi there!

We are given the set of ordered pairs below:

\large \boxed{(3, - 1),(2, - 2),(0,2),(2,1)}

1. What is the domain?

  • Domain is a set of all x-values in one set of ordered pairs. So what are the x-values that I am talking about? In ordered pairs, we define x and y which both have relation to each others which we can write as (x,y). That's right, the domain is set of all x-values from ordered pairs.

Therefore, we gather only x-values from (x,y). Hence, the domain is {3,2,0,2}. Whoops! Something is not right. As we learn in Set Theory that we don't write the same or repetitive in a set. Hence, <u>t</u><u>h</u><u>e</u><u> </u><u>a</u><u>c</u><u>t</u><u>u</u><u>a</u><u>l</u><u> </u><u>d</u><u>o</u><u>m</u><u>a</u><u>i</u><u>n</u><u> </u><u>i</u><u>s</u><u> </u><u>{</u><u>0</u><u>,</u><u>2</u><u>,</u><u>3</u><u>}</u>

2. What is the range?

  • Because domain is set of all x-values. Then what do you think the range is? That's right! The range is <u>s</u><u>e</u><u>t</u><u> </u><u>o</u><u>f</u><u> </u><u>a</u><u>l</u><u>l</u><u> </u><u>y</u><u>-</u><u>v</u><u>a</u><u>l</u><u>u</u><u>e</u><u>s</u><u>.</u> If you got this right before looking up the underlined words then a handclap for you! So how do we find range? Simple, we just do like finding the domain in the Q1, except we gather the y-values in (x,y) instead and make sure that we don't write same number!

Therefore, gather y-values from the ordered pairs. Hence, <u>t</u><u>h</u><u>e</u><u> </u><u>r</u><u>a</u><u>n</u><u>g</u><u>e</u><u> </u><u>i</u><u>s</u><u> </u><u>{</u><u>-</u><u>2</u><u>,</u><u>-</u><u>1</u><u>,</u><u>1</u><u>,</u><u>2</u><u>}</u>

3. Is the relation a function?

  • All functions are relations but not all relations are functions. Function is a set of ordered pairs where <u>d</u><u>o</u><u>m</u><u>a</u><u>i</u><u>n</u><u> </u><u>i</u><u>s</u><u> </u><u>n</u><u>o</u><u>t</u><u> </u><u>r</u><u>e</u><u>p</u><u>e</u><u>t</u><u>i</u><u>t</u><u>i</u><u>v</u><u>e</u><u> </u><u>o</u><u>r</u><u> </u><u>i</u><u>n</u><u> </u><u>a</u><u> </u><u>s</u><u>e</u><u>t</u><u>,</u><u> </u><u>t</u><u>h</u><u>e</u><u>r</u><u>e</u><u> </u><u>c</u><u>a</u><u>n</u><u>n</u><u>o</u><u>t</u><u> </u><u>b</u><u>e</u><u> </u><u>m</u><u>o</u><u>r</u><u>e</u><u> </u><u>t</u><u>h</u><u>a</u><u>n</u><u> </u><u>o</u><u>n</u><u>e</u><u> </u><u>s</u><u>a</u><u>m</u><u>e</u><u> </u><u>v</u><u>a</u><u>l</u><u>u</u><u>e</u><u>.</u> Consider the following relation: (1,1),(1,2) - Oh, looks like in a set of ordered pairs, there are two same domains which make it only a relation, and not a function. On the other hand, (1,1),(2,2) - Looking good! No same or repetitive domain, making it indeed a function.

Consider the domain from Q1 and see if there are two same values of x in a set. Looks like the relation is not a function since there are same x-values which are 2 in a set, making it only a relation. Hence, the relation is not a function.

These are all 3 answers along with an explanation. Let me know if you have any doubts regarding Relations and Functions.

<em>F</em><em>r</em><em>o</em><em>m</em><em> </em><em>t</em><em>h</em><em>e</em><em> </em><em>Q</em><em>1</em><em>'</em><em>s</em><em> </em><em>a</em><em>n</em><em>s</em><em>w</em><em>e</em><em>r</em><em>,</em><em> </em><em>t</em><em>h</em><em>e</em><em>r</em><em>e</em><em> </em><em>a</em><em>r</em><em>e</em><em> </em><em>t</em><em>w</em><em>o</em><em> </em><em>b</em><em>o</em><em>l</em><em>d</em><em> </em><em>t</em><em>e</em><em>x</em><em>t</em><em>s</em><em>,</em><em> </em><em>p</em><em>l</em><em>e</em><em>a</em><em>s</em><em>e</em><em> </em><em>c</em><em>h</em><em>o</em><em>o</em><em>s</em><em>e</em><em> </em><em>t</em><em>h</em><em>e</em><em> </em><em>s</em><em>e</em><em>c</em><em>o</em><em>n</em><em>d</em><em> </em><em>b</em><em>o</em><em>l</em><em>d</em><em> </em><em>t</em><em>e</em><em>x</em><em>t</em><em> </em><em>t</em><em>o</em><em> </em><em>a</em><em>n</em><em>s</em><em>w</em><em>e</em><em>r</em><em> </em><em>(</em><em>t</em><em>h</em><em>e</em><em> </em><em>o</em><em>n</em><em>e</em><em> </em><em>w</em><em>i</em><em>t</em><em>h</em><em> </em><em>u</em><em>n</em><em>d</em><em>e</em><em>r</em><em>l</em><em>i</em><em>n</em><em>e</em><em>)</em><em> </em><em>a</em><em>n</em><em>d</em><em> </em><em>n</em><em>o</em><em>t</em><em> </em><em>t</em><em>h</em><em>e</em><em> </em><em>f</em><em>i</em><em>r</em><em>s</em><em>t</em><em> </em><em>o</em><em>n</em><em>e</em><em> </em><em>(</em><em>t</em><em>h</em><em>e</em><em> </em><em>o</em><em>n</em><em>e</em><em> </em><em>w</em><em>i</em><em>t</em><em>h</em><em> </em><em>s</em><em>a</em><em>m</em><em>e</em><em> </em><em>2</em><em>'</em><em>s</em><em>)</em><em>.</em><em> </em>

Good luck on your assignment, have a nice day!

4 0
3 years ago
Please help please !!!
Kobotan [32]
<h3>Answer:  12</h3>

========================================================

Explanation:

You can use the AAS (angle angle side) theorem to prove that triangle ABD is congruent to triangle CBD.

From there, we can then say that AD and DC are the same length

AD = DC

3y+6 = 5y-18

3y-5y = -18-6

-2y = -24

y = (-24)/(-2)

y = 12

6 0
3 years ago
Noha, Sarah, and Karima are sisters. Noha is 3 years older than Sarah.
Virty [35]

Answer:

25 years old

Step-by-step explanation:

karima is 20 years old

Sarah is 2 years older than karima (20+2= 22)

Noha is 3 years older than Sarah

if sara is 22 then noha is 22+3=25

5 0
2 years ago
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