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3 years ago

14

Jillian ran 40 miles in one week. Her friend Lacey ran the distance that Jillian ran. How many miles did Lacey run? Express your

answer in simplest form.

1 answer:

vichka [17]3 years ago

3 0

I'm SURE you left something out of the space between "ran" ... and ...
"the distance". But here in the Math category, it's our job to answer the
questions, not to fix them.

The question tells us that Jillian ran 40 miles, and that Lacey ran
the distance that Jillian ran. We can logically conclude, therefore,
that Lacey ran 40 miles.

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Which rule describes the relationship between the x- and y- coordinates on line m?

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Answer:

y = 2x - 1

Step-by-step explanation:

None of those would be correct, the line on the graph is y = 2x - 1 because the slope of the line is 2 and the y -intercept -1.

You can also check, we know that on the graph the coordinates where the dots are located are: (2,3),(3,5),(4,7) where the first number is x and second is y coordinate.

Check answer:

Let x=2 if this is correct y should be equal to 3 because of the first point (2,3).

y = 2(2)-1 = 4 - 1 = 3

check the other points if you want to confirm.

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Given: Line segment N M is parallel to line segment P O. and Angle 1 is-congruent-to angle 3 Prove: Line segment N M is parallel

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Step-by-step explanation:

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3 years ago

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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

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3 years ago

To describe a specific arith-

Bezzdna [24]

Answer:

$f(x) = 7x + 30$

Step-by-step explanation:

We need at least two points to write the equation of a straight line.

The recursive formula that Elijah wrote is:

$f(0) = 30$

$f(n + 1) = f(n) + 7$

When we substitute n=0, we get:

$f(0 + 1) = f(0) + 7$

$f(1) = 30 + 7$

$f(1) = 37$

The points (0,30) and (1,37) lies on this line.

The equation of this line is of the form:

$f(x) = mx + b$

where b =30 is the y-intercept and m=7 is the slope.

We plug in these values to get:

$f(x) = 7x + 30$

Note that the slope of the line is equal to the common difference of the Arithmetic Sequence.

You could also use the two points to find the slope:

$m = \frac{37 - 30}{1 - 0} = 7$

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3 years ago

If a quadrilateral has vertices a(-2,2) b(2,3) c (3,-4) d(-3,-2) find the slope of AB,BC,CD,DA

FrozenT [24]

Slope AB=0.25
BC=-7
CD=-1/3
DA=4

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3 years ago