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Ugo [173]
2 years ago
7

Ishaan and his friends have 40$ to spend at a restaurant.

Mathematics
1 answer:
Paul [167]2 years ago
3 0

The number of salads Ishaan and his friends can afford is 6 salads.

<h3>What are Word Problems?</h3>

Word Problems in mathematics involve crucial understanding and the use of variables and arithmetic operations to solve real-life cases.

From the given information;

  • The total amount with Ishaan and his friends = $40
  • If they decided to spend the money on salad(x) and one salad(x) cost = $6

Thus, the number of salads they can purchase is:

\mathbf{x = \dfrac{\$40}{\$6}}

x = 6 salads remaining $4

Therefore, we can conclude that the number of salads they can purchase with $40 is 6 salads with a remaining amount of $4.

Learn more about word problems here:

brainly.com/question/21405634

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Which shows a graph of a linear equation in standard form Ax + By = C, where A = 0, B is positive, and C is negative? A coordina
IrinaK [193]

Answer:

A coordinate plane with a horizontal line passing through (negative 4, negative 3), (0, negative 3) and (4, negative 3).

Step-by-step explanation:

Ax + By = C, where A = 0, B is positive, and C is negative

By= C or y= C/B which is negative

Since A is zero and C is negative, the line is parallel to x- axis and has negative value for y, so it is a horizontal line below zero

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A coordinate plane with a vertical line passing through (1, negative 4), (1, 0) and (1, 4).

  • no, the line is horizontal

A coordinate plane with a vertical line passing through (negative 2, negative 4), (negative 2, 0) and (negative 2, 4).

  • no, the line is horizontal

A coordinate plane with a horizontal line passing through (negative 4, 4), (0, 4) and (negative 4, 4).

  • no, the y-values should be negative

A coordinate plane with a horizontal line passing through (negative 4, negative 3), (0, negative 3) and (4, negative 3).

  • yes, this is horizontal line below zero
8 0
4 years ago
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The coefficient of -x is 1
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Answer:

it -1

Step-by-step explanation:

3 0
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Carolyn Shelly Jennifer and Sabrina live in different houses that are side by side.
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Your answer would turn out to be Shelby goes out her front door turns West and walks past the other three houses.
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What is the length of segment AB? (4 points)
Sveta_85 [38]

Answer:

10 units

Step-by-step explanation:

\sqrt{(0-6)^{2} +(10-2)^{2} }\\\\\sqrt{(-6)^{2}+(8)^{2} } \\\\\sqrt{36+64}\\\\\sqrt{100}\\\\10

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
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