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slamgirl [31]
3 years ago
6

Max has a Starbucks' gift card for $50. HIs favorite drink is the S'mores Frappuccino. Each drink costs $4.25. How many frappucc

inos can Max purchase with his gift card? Write an inequality to represent the number of frappuccinos Max can purchase. How many frappuccinos can Max purchase with his gift card?
Mathematics
1 answer:
hram777 [196]3 years ago
5 0

Answer:

4.25x ≤ 50

Number of drink = 11

Step-by-step explanation:

Given:

Amount of gift card = $50

Each drink cost = $4.25

Find:

Number of drink

Computation:

Assume;

Number of drink = x

So,

4.25x ≤ 50

x ≤ 11.76

so,

Number of drink = 11

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Consider the following. (A computer algebra system is recommended.) y'' + 3y' = 2t4 + t2e−3t + sin 3t (a) Determine a suitable f
drek231 [11]

First look for the fundamental solutions by solving the homogeneous version of the ODE:

y''+3y'=0

The characteristic equation is

r^2+3r=r(r+3)=0

with roots r=0 and r=-3, giving the two solutions C_1 and C_2e^{-3t}.

For the non-homogeneous version, you can exploit the superposition principle and consider one term from the right side at a time.

y''+3y'=2t^4

Assume the ansatz solution,

{y_p}=at^5+bt^4+ct^3+dt^2+et

\implies {y_p}'=5at^4+4bt^3+3ct^2+2dt+e

\implies {y_p}''=20at^3+12bt^2+6ct+2d

(You could include a constant term <em>f</em> here, but it would get absorbed by the first solution C_1 anyway.)

Substitute these into the ODE:

(20at^3+12bt^2+6ct+2d)+3(5at^4+4bt^3+3ct^2+2dt+e)=2t^4

15at^4+(20a+12b)t^3+(12b+9c)t^2+(6c+6d)t+(2d+e)=2t^4

\implies\begin{cases}15a=2\\20a+12b=0\\12b+9c=0\\6c+6d=0\\2d+e=0\end{cases}\implies a=\dfrac2{15},b=-\dfrac29,c=\dfrac8{27},d=-\dfrac8{27},e=\dfrac{16}{81}

y''+3y'=t^2e^{-3t}

e^{-3t} is already accounted for, so assume an ansatz of the form

y_p=(at^3+bt^2+ct)e^{-3t}

\implies {y_p}'=(-3at^3+(3a-3b)t^2+(2b-3c)t+c)e^{-3t}

\implies {y_p}''=(9at^3+(9b-18a)t^2+(9c-12b+6a)t+2b-6c)e^{-3t}

Substitute into the ODE:

(9at^3+(9b-18a)t^2+(9c-12b+6a)t+2b-6c)e^{-3t}+3(-3at^3+(3a-3b)t^2+(2b-3c)t+c)e^{-3t}=t^2e^{-3t}

9at^3+(9b-18a)t^2+(9c-12b+6a)t+2b-6c-9at^3+(9a-9b)t^2+(6b-9c)t+3c=t^2

-9at^2+(6a-6b)t+2b-3c=t^2

\implies\begin{cases}-9a=1\\6a-6b=0\\2b-3c=0\end{cases}\implies a=-\dfrac19,b=-\dfrac19,c=-\dfrac2{27}

y''+3y'=\sin(3t)

Assume an ansatz solution

y_p=a\sin(3t)+b\cos(3t)

\implies {y_p}'=3a\cos(3t)-3b\sin(3t)

\implies {y_p}''=-9a\sin(3t)-9b\cos(3t)

Substitute into the ODE:

(-9a\sin(3t)-9b\cos(3t))+3(3a\cos(3t)-3b\sin(3t))=\sin(3t)

(-9a-9b)\sin(3t)+(9a-9b)\cos(3t)=\sin(3t)

\implies\begin{cases}-9a-9b=1\\9a-9b=0\end{cases}\implies a=-\dfrac1{18},b=-\dfrac1{18}

So, the general solution of the original ODE is

y(t)=\dfrac{54t^5 - 90t^4 + 120t^3 - 120t^2 + 80t}{405}\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,-\dfrac{3t^3+3t^2+2t}{27}e^{-3t}-\dfrac{\sin(3t)+\cos(3t)}{18}

3 0
4 years ago
Re-write the quadratic function below in Standard Form<br> y = -(x – 3)2 +8
lisov135 [29]

Answer:

y = -x² + 6x - 1

General Formulas and Concepts:

<u>Algebra I</u>

  • Standard Form: ax² + bx + c = 0
  • Expanding by FOIL (First Inside Outside Last)
  • Combining like terms

Step-by-step explanation:

<u>Step 1: Define equation</u>

y = -(x - 3)² + 8

<u>Step 2: Rewrite</u>

  1. Expand [FOIL]:                    y = -(x² - 6x + 9) + 8
  2. Distribute -1:                        y = -x² + 6x - 9 + 8
  3. Combine like terms:           y = -x² + 6x - 1
4 0
3 years ago
The graph of a function f(x) passes through the following points:
Alex

The function is f(x) = -2x² + 2 if the function f(x) passes through the following points:(0,2), (1, 0), (-1,0)

<h3>What is a function?</h3>

It is defined as a special type of relationship, and they have a predefined domain and range according to the function every value in the domain is related to exactly one value in the range.

The question is incomplete.

The complete question is in the picture, please refer to the attached picture.

We have:

The graph of a function f(x) passes through the following points: (0,2), (1, 0), (-1,0)

Let

f(x) = ax² + bx + c

Plug x = 0, and y = 2

c = 2 ...(1)

Plug x = 1 and y = 0

a + b + c = 0 ..(2)

Plug x = -1 and y = 0

a - b + c = 0  ...(3)

After solving (1), (2), and (3)

a = -2

b = 0

c = 2

f(x) = -2x² + 2

Thus, the function is f(x) = -2x² + 2 if the function f(x) passes through the following points:(0,2), (1, 0), (-1,0)

Learn more about the function here:

brainly.com/question/5245372

#SPJ1

5 0
2 years ago
What is <br> (2n-9) -5 (-2.4n+4)
gtnhenbr [62]
2n-9-5-8n+4n
= 2n-14-4n
=2n-14
8 0
3 years ago
Read 2 more answers
The table shows the numbers y of students absent from school x days after a flu outbreak.
allsm [11]
  1. A function that models the data is given by this quadratic equation, y = -0.4908x² + 5.8845x + 1.3572.
  2. The number of students that are absent 10 days after the outbreak is equal to 11 students.

<h3>What is a scatter plot?</h3>

A scatter plot can be defined as a type of graph which is used for the graphical representation of the values of two (2) variables, with the resulting points showing any association (correlation) between the data set.

<h3>What is a quadratic function?</h3>

A quadratic function can be defined as a mathematical expression (equation) that can be used to define and represent the relationship that exists between two or more variable on a graph.

In Mathematics, the standard form of a quadratic equation is given by;

ax² + bx + c = 0

By critically observing the graph (see attachment) which models the data in the given table, we can infer and logically deduce that the quadratic function is given by:

y = -0.4908x² + 5.8845x + 1.3572

For the number of students that are absent 10 days after the outbreak, we have:

y = -0.4908(10)² + 5.8845(10) + 1.3572

y = -0.4908(100) + 58.845 + 1.3572

y = -49.08 + 58.845 + 1.3572

Number of students, y = 11.12 ≈ 11 students.

Read more on scatterplot here: brainly.com/question/6592115

#SPJ1

5 0
1 year ago
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