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Stells [14]
3 years ago
15

If my car averages 27 miles to the gallon, how many miles can I drive on3/4 tank of gas

Mathematics
1 answer:
AURORKA [14]3 years ago
7 0

Answer:

20.25 miles

Step-by-step explanation:

If your car travels 27 miles on a full tank of gas, then your car will travel 27\cdot\frac{3}{4} miles on \frac{3}{4} tank of gas.

27 \cdot 0.75 = 20.25

I hope this helped!

You might be interested in
Two cars leave from a town at the same time traveling in opposite directions. One travels 5 mph faster than the other. In 3 hour
Doss [256]

Let the slower cars speed equal X.

The faster cars speed would be X+5 ( 5 mph faster).


They traveled for 3 hours

Multiply the time of travel by speed to equal the number of miles traveled.

So you have:

3X + 3(X+5) = 267 miles

Simplify the left side:

3X + 3X+15 = 267

Combine like terms:

6x + 15 = 267

Subtract 15 from each side"

6x = 252

Divide each side by 6:

x = 252 / 6

X = 42


The slower car was traveling at 42 mph and the faster car was traveling at 47 mph.




7 0
3 years ago
You can buy DVDs at a local store for $15.49 each. You can buy them at an online store for $13.99 each plus $6 for shippimg. How
NeTakaya
15.49x = 13.99x + 6
15.49x - 13.99x = 6
1.50x = 6
x = 6 / 1.50
x = 4 <=== u can buy 4 DVD's for the same amount

15.49 * 4 = 61.96
13.99(4) + 6 = 61.96
6 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
How do I do this my hwk id due tomoorw
nexus9112 [7]

The perimeter of the given circle is 25.1 square cm.

Step-by-step explanation:

  • From the given diagram, the radius of the given circle is 4 cm.
  • The perimeter of any given circle is calculated by multiplying 2π with the radius. The radius of this circle is 4 cm and it says π = 3.142
  • The Perimeter for the given circle = 2π × r = 2 × 3.142 × 4 cm = 6.284 × 4 cm = 25.136 cm.                                                       Rounding this off to one decimal 25.136 square cm, the perimeter equals 25.1 square cm.

6 0
3 years ago
Carl is paid $10 plus $8 an hour he was paid $66 how many hours did he work?
loris [4]
10+8h=66
minus 10 both sides
8h=56
divide both sides by 8
h=7
worked 7 hours
6 0
3 years ago
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