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3 years ago

Monique is factoring the expression 4x + 16xy. Her work is shown below.

Mathematics

2 answers:

agasfer [191]3 years ago

6 0

Answer:

D. Monique made and error when writing the factored expression.

Step-by-step explanation:

Edge 2020

PtichkaEL [24]3 years ago

4 0

Answer:

Step-by-step explanation:

She made an error writing the expression

it should be 4(x + 4y)

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Use the method of lagrange multipliers to find

Yanka [14]

Answer:

a) The function is: f(x, y) = x + y.

The constraint is: x*y = 196.

Remember that we must write the constraint as:

g(x, y) = x*y - 196 = 0

Then we have:

L(x, y, λ) = f(x, y) + λ*g(x, y)

L(x, y, λ) = x + y + λ*(x*y - 196)

Now, let's compute the partial derivations, those must be zero.

dL/dx = λ*y + 1

dL/dy = λ*x + 1

dL/dλ = (x*y - 196)

Those must be equal to zero, then we have a system of equations:

λ*y + 1 = 0

λ*x + 1 = 0

(x*y - 196) = 0

Let's solve this, in the first equation we can isolate λ to get:

λ = -1/y

Now we can replace this in the second equation and get;

-x/y + 1 = 0

Now let's isolate x.

x = y

Now we can replace this in the last equation, and we will get:

(x*x - 196) = 0

x^2 = 196

x = √196 = 14

then the minimum will be:

x + y = x + x = 14 + 14 = 28.

b) Now we have:

f(x) = x*y

g(x) = x + y - 196

Let's do the same as before:

L(x, y, λ) = f(x, y) + λ*g(x, y)

L(x, y, λ) = x*y + λ*(x + y - 196)

Now let's do the derivations:

dL/dx = y + λ

dL/dy = x + λ

dL/dλ = x + y - 196

Now we have the system of equations:

y + λ = 0

x + λ = 0

x + y - 196 = 0

To solve it, we can isolate lambda in the first equation to get:

λ = -y

Now we can replace this in the second equation:

x - y = 0

Now we can isolate x:

x = y

now we can replace that in the last equation

y + y - 196 = 0

2*y - 196 = 0

2*y = 196

y = 196/2 = 98

The maximum will be:

x*y = y*y = 98*98 = 9,604

6 0

2 years ago

How much can 0.75 go into 4.5

Nutka1998 [239]

Answer:

6 times

Step-by-step explanation:

well just divide 4.5 by 0.75 to get

8 0

2 years ago

Read 2 more answers

What is the equation of the line that is parallel to the

erastovalidia [21]

Answer:

Step-by-step explanation:

A parallel line will have the same slope as the reference line. In this case, I don't see the "given line" as promised in the question. If it does appear, and it looks like y = 5x + 3, for example, the slope is 5 and the new line will have the same slope.

<h3><u>If this slope is correct</u>, we can start the equation for the parallel line that goes through point (-3,2) by starting with:</h3><h3 /><h3>y = 5x + b</h3><h3 /><h3>We need a value of b that forces the line to go through point (-3,2). We can do that by using the given point in the equation and solving for b:</h3><h3>y = 5x + b</h3><h3>2 = 5(-3) + b</h3><h3>b = 17</h3><h3 /><h3>The parallel line to y=5x+3 is</h3><h3>y = 5x + 17</h3><h3 /><h3>See attachment.</h3><h3 /><h3 /><h3 />