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Svetllana [295]
3 years ago
15

Two trucks started toward each other at the same time from towns 500 km apart. One truck traveled at a rate of 65 km per hour wh

ile the other traveled at 60 km per hour. After how many hours did they meet? If x represents the number of hours, which of the following equations could be used to solve the problem?
Mathematics
1 answer:
ale4655 [162]3 years ago
5 0
They met after (4 hours).
(60+65)*x=500<span />
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A square floor tile has a diagonal 19 inches long. What is the area of the tile?
Anit [1.1K]

Answer:

C

Step-by-step explanation:

The answer is C because you plugin using the area formula.

3 0
3 years ago
3x-6y=9<br> x=2y+3<br><br> SOMEONE HELP ME SOLVE THIS STEP BY STEP TY
RideAnS [48]

Answer:

0=0

infinitely many solutions

Step-by-step explanation:

plug x=2y+3 as x into 3x-6y=9...so

3(2y+3)-6y=9, which has our x=2y+3 plugged in because that is what x equals

if you solve its

6y+9-6y=9

0y=0

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making it 0=0

you can't go further from this

6 0
3 years ago
7 less than a number k is 18​
Elena-2011 [213]

Answer:

k-7=18; k=25

Step-by-step explanation:

k(-7+7)=18+7\\18+7=25\\k=25

7 less than 25 (or K) is 18.

4 0
3 years ago
Read 2 more answers
Find the simple interest earned to the nearest cent for each principal, interest rate, and time.
Y_Kistochka [10]

Answer:

Interest earned at 3.9 percent rate is $31.2

Interest earned at 2 percent rate is $5.8

Step-by-step explanation:

A = P(1 + rt)

Where 'A' is the amount, 'r' is the rate and 't' is the time in years

When;

P = $1200

r = 3.9%

t = \frac{8}{12} years

Then,

A = $1200(1 + 0.039(\frac{8}{12}))

A = $1200 + $31.2 = $1231.2

Interest = Amount - Principal

Interest earned at 3.9 percent rate is $1231.2 - $1200 = $31.2

When;

P = $580

r = 2%

t = \frac{6}{12} years

Then,

A = $580(1 + 0.02(\frac{6}{12}))

A = $580 + $5.8 = $585.8

Interest earned = Amount - Principal

Interest earned at 2 percent rate = $585.8 - $580 = $5.8

4 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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