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3 years ago

308 miles in 7 hours. Find the

Mathematics

1 answer:

Ronch [10]3 years ago

4 0

Answer:

They did 44 miles per hour

Step-by-step explanation:

308 / 7 = 44

You might be interested in

Circle the letter of the

laiz [17]

Step-by-step explanation:

the length of the third side could be 6

As in order to form a triangle The sum of the lengths of any two sides of a triangle is greater than the length of the third side.

so Here

8 + 3 > 6

11 > 6 ( True) ✔

8+ 6 > 3

14 > 3 ( True) ✔

6 + 3 > 8

9 > 8 ( True) ✔

Hope it will help you :)

5 0

3 years ago

How many complex roots does the polynomial equation have?

Olegator [25]

Answer:

Step-by-step explanation:

The aquatic equation suggest that 1=1+9-2+69=420 so if you divide that by 69 again you get 2

8 0

3 years ago

Read 2 more answers

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

"Lito had some marbles. He gave ½ of his marbles plus 1 to Felix, ½ of the remaining marbles plus 1 to Ces, and ½ of the last re

Nuetrik [128]

let

x-------> total amount of marbles at the beginning

we know that

1) He gave ½ of his marbles plus 1 to Felix

(1/2)*x+1=(x+2)/2 (Felix's marbles)

remaining=x-[(x+2)/2]-----> [2x-x-2]/2------> (x-2)/2

2)½ of the remaining marbles plus 1 to Ces

(1/2)*[(x-2)/2]+1=[(x-2)/4]+1-----> (x+2)/4 (Ce's marbles)

remaining=[(x-2)/2]-[(x+2)/4]-----> [2x-4-x-2]/4------> (x-6)/4

3) ½ of the last remaining marbles plus 1 to Pedro

(1/2)*[ (x-6)/4]+1=[(x-6)/8)+1------> (x+2)/8 (Pedro's marbles)

remaining=[(x-6)/4]-[(x+2)/8]------> [2x-12-x-2]/8-----> (x-14)/8

4)If Lito had 1 marble left for himself

the last remaining is equal to 1

(x-14)/8=1-----> x-14=8------> x=22 marbles

Verify

(x+2)/2 (Felix's marbles)------> (22+2)/2=12

(x+2)/4 (Ce's marbles)------> (22+2)/4=6

(x+2)/8 (Pedro's marbles)---> (22+2)/8=3

Lito's marbles------------------> 1

total=12+6+3+1=22--------> is ok

therefore

the answer is

the total amount of marbles at the beginning was 22

4 0

3 years ago

Can somebody please help me? I don't understand this!

Xelga [282]

Use the base of the triangle as the diameter, which in this case is 16, divide it by 2, which is 8, and then square it, which is 64, and multiply by Pi (64 times Pi is 200.96) so you can find the area of the whole circle which is 200.96ft^2. Since this is an equilateral triangle, the height might be 16, so using the area of a triangle, A=BH times 1/2, We multiply 16 by 16 which is 256. 256 times 1/2 is 128. So the area of the triangle will be 128ft^2. We then subtract our areas, 200.96-128 which is 72.96. Then therefore the area of the circle will be 72.96ft^2. I may be wrong, but this is what I think.

5 0

3 years ago