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const2013 [10]
3 years ago
10

What is the value of the expression shown below?

Mathematics
2 answers:
Ymorist [56]3 years ago
8 0
(8 + [7+1] 2 ÷ 4 × 1) ÷ 2^4

Let's focus on what's in parentheses first.

8 + 8 × 2 ÷ 4 × 1
8 + 16 ÷ 4 × 1
8 + 4 × 1
8 + 4
12

Now, to what is outside the parentheses:

Because of the way it is written, you will do 12 ÷ 2 first.
12 ÷ 2^4
6^4

Answer: 1296
MAXImum [283]3 years ago
5 0

Answer: l think the answer is 24 but, l'm not positive!

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Function f is graphed. According to the graph, is f even, odd, or neither?
miss Akunina [59]

Answer:

C

Step-by-step explanation:

f is neither even nor odd

3 0
3 years ago
The Saxena family plans to install a light to illuminate part of their rectangular yard. Nikki and Dylan each proposed a differe
monitta

Options:

a. Nikki's proposed placement will light a greater area than Dylan's placement.

b.  Dylan's proposed placement will light a greater area than Nikki's placement.

c.  Both proposed placements will light the same sized area.

d. Nikki's proposed placement will light more than half the yard.

e. Dylan's proposed placement will light more than half the yard.

f.  Dylan's proposed placement will light exactly half of the yard.

g. Nikki's proposed placement will light less than half of the yard.

Answer:

C) Both proposed placements will light the same sized area.

F) Dylan's proposed placement will light exactly half of the yard.

Step-by-step explanation:

The area of a triangle is (base x height) / 2, and both lights illuminate the same base and height = (60 x 38) / 2 = 1,140 sq ft

Both Dylan's and Nikki's proposed placement will lit exactly half of the yard. The yard's total area = 60 x 38 = 2,280 sq ft, which is twice the area lit by the lights.

3 0
3 years ago
your classmate is starting a new fitness program. He is planning to ride his bicycle 45 minutes every day. He burns 10 calories
Verizon [17]
Burn you mean to cook like pele
3 0
3 years ago
A biscuit recipe calls for 2 pounds of all purpose flour and .75 pound of butter, what would be the baker’s percentage for butte
Goryan [66]

Answer: 37.5%


Explanation:


Percentage is defined as parts of an amount  in 100 parts of a basis. This must be stated mathamatically for better understanding and for calculations. It is the ratio:

             % = (parts / base) × 100.


In this case you need to express the amount of butter as a part in 100 parts of all purpose flour, which, using the above definition, is:


        % butter = (0.75 pounds of butter / 2 pound of flour) × 100

         % butter = 0.75 × 100 / 2 = 75 / 2 = 37.5


The answer is expressed as 37.5%.

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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