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Eva8 [605]
3 years ago
13

Katherine and her children went into a bakery and where they sell cupcakes for $2.50 each and donuts for $1.25 each. Katherine h

as $25 to spend and must buy at least 14 cupcakes and donuts altogether. If xx represents the number of cupcakes purchased and yy represents the number of donuts purchased, write and solve a system of inequalities graphically and determine one possible solution.
Mathematics
1 answer:
yan [13]3 years ago
6 0

Answer:the ration to x and b that equals c is about 3 dollars

Step-by-step explanation:

You might be interested in
Ken has read 1/3 of his book what fraction of his book is left to read?​
Stells [14]

Answer:

2/3

Step-by-step explanation:

fraction of book already read = 1/3

fraction which represents whole book = 1 = 3/3

fraction of remaining book = fraction of whole book - fraction of book already read

= 1 - 1/3

= 3/3 - 1/3

= (3-1) /3

= 2/3

8 0
3 years ago
A researcher wants to determine if socioeconomic status (low, moderate, high) is related to smoking (yes or no). The Chi-Square
julsineya [31]

Answer:

The chi-square null hypothesis for the study that "socioeconomic status is related to smoking behavior" is False

Step-by-step explanation:

The chi-square null hypothesis is false because the chi-square null hypothesis states that no relationship exists on the categorical variables in a population, they are all independent of each other.

3 0
3 years ago
EXAMPLE 5 Find the maximum value of the function f(x, y, z) = x + 2y + 11z on the curve of intersection of the plane x − y + z =
Taya2010 [7]

Answer:

\displaystyle x= -\frac{10}{\sqrt{269}}\\\\\displaystyle y= \frac{13}{\sqrt{269}}\\\\\displaystyle z = \frac{23\sqrt{269}+269}{269}

<em>Maximum value of f=2.41</em>

Step-by-step explanation:

<u>Lagrange Multipliers</u>

It's a method to optimize (maximize or minimize) functions of more than one variable subject to equality restrictions.

Given a function of three variables f(x,y,z) and a restriction in the form of an equality g(x,y,z)=0, then we are interested in finding the values of x,y,z where both gradients are parallel, i.e.

\bigtriangledown  f=\lambda \bigtriangledown  g

for some scalar \lambda called the Lagrange multiplier.

For more than one restriction, say g(x,y,z)=0 and h(x,y,z)=0, the Lagrange condition is

\bigtriangledown  f=\lambda \bigtriangledown  g+\mu \bigtriangledown  h

The gradient of f is

\bigtriangledown  f=

Considering each variable as independent we have three equations right from the Lagrange condition, plus one for each restriction, to form a 5x5 system of equations in x,y,z,\lambda,\mu.

We have

f(x, y, z) = x + 2y + 11z\\g(x, y, z) = x - y + z -1=0\\h(x, y, z) = x^2 + y^2 -1= 0

Let's compute the partial derivatives

f_x=1\ ,f_y=2\ ,f_z=11\ \\g_x=1\ ,g_y=-1\ ,g_z=1\\h_x=2x\ ,h_y=2y\ ,h_z=0

The Lagrange condition leads to

1=\lambda (1)+\mu (2x)\\2=\lambda (-1)+\mu (2y)\\11=\lambda (1)+\mu (0)

Operating and simplifying

1=\lambda+2x\mu\\2=-\lambda +2y\mu \\\lambda=11

Replacing the value of \lambda in the two first equations, we get

1=11+2x\mu\\2=-11 +2y\mu

From the first equation

\displaystyle 2\mu=\frac{-10}{x}

Replacing into the second

\displaystyle 13=y\frac{-10}{x}

Or, equivalently

13x=-10y

Squaring

169x^2=100y^2

To solve, we use the restriction h

x^2 + y^2 = 1

Multiplying by 100

100x^2 + 100y^2 = 100

Replacing the above condition

100x^2 + 169x^2 = 100

Solving for x

\displaystyle x=\pm \frac{10}{\sqrt{269}}

We compute the values of y by solving

13x=-10y

\displaystyle y=-\frac{13x}{10}

For

\displaystyle x= \frac{10}{\sqrt{269}}

\displaystyle y= -\frac{13}{\sqrt{269}}

And for

\displaystyle x= -\frac{10}{\sqrt{269}}

\displaystyle y= \frac{13}{\sqrt{269}}

Finally, we get z using the other restriction

x - y + z = 1

Or:

z = 1-x+y

The first solution yields to

\displaystyle z = 1-\frac{10}{\sqrt{269}}-\frac{13}{\sqrt{269}}

\displaystyle z = \frac{-23\sqrt{269}+269}{269}

And the second solution gives us

\displaystyle z = 1+\frac{10}{\sqrt{269}}+\frac{13}{\sqrt{269}}

\displaystyle z = \frac{23\sqrt{269}+269}{269}

Complete first solution:

\displaystyle x= \frac{10}{\sqrt{269}}\\\\\displaystyle y= -\frac{13}{\sqrt{269}}\\\\\displaystyle z = \frac{-23\sqrt{269}+269}{269}

Replacing into f, we get

f(x,y,z)=-0.4

Complete second solution:

\displaystyle x= -\frac{10}{\sqrt{269}}\\\\\displaystyle y= \frac{13}{\sqrt{269}}\\\\\displaystyle z = \frac{23\sqrt{269}+269}{269}

Replacing into f, we get

f(x,y,z)=2.4

The second solution maximizes f to 2.4

5 0
3 years ago
The length of a rectangle is 11 units more than its width. The perimeter of the rectangle is 182 units. Solve for the dimensions
alekssr [168]

Answer:

2160 units square

Step-by-step explanation:

We are given the perimeter of 182 units and that the lengths are 11 more units than the width. So the length is W+11, eleven more than the width. There is also 2 width sides so the sum of the lengths are 2W+22. There is also 2 width sides so if you add that to the equation the perimeter equals 4W+22.

4W+22=182

4W=160

W=40

So the width is 40 units, and the length is 11 more so the length is 54 units. Multiply these two numbers to get the area. 44*54=2160 square units.

3 0
3 years ago
Please help!! <br> I will mark brainiest
katovenus [111]

Answer:

30 degrees

Step-by-step explanation:

6 0
3 years ago
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