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Lady_Fox [76]
2 years ago
11

Multiply: (x+4)(x^2+2x+3) Simplify and write your answer in Standard Form.

Mathematics
1 answer:
Oxana [17]2 years ago
6 0

Answer:

x^3+6x^2+11x+12

Step-by-step explanation:

(x+4)(x^2+2x+3)

=x(x^2+2x+3)+4(x^2+2x+3) [by multiplying with bith sides]

=x^3+2x^2+3x+4x^2+8x+12

=x^3+2x^2+4x^2+3x+8x+12

=x^3+6x^2+11x+12

(please mark me brainliest)

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Using data from 2010 and projected to 2020, the population of the United Kingdom (y, in millions) can be approximated by the equ
Nimfa-mama [501]

Answer:

In year 2030 the population is predicted to be 71.75 million

Step-by-step explanation:

* <em>Lets explain how to solve the problem</em>

- Using data from 2010 and projected to 2020, the population of

 the United Kingdom (y, in millions) can be approximated by the

 equation  10.0 y − 4.55 x = 581

- x is the number of years after 2000

- We need to know in what year the population is predicted to be

  71.75 million

* <em>Lets substitute the value of y in the equation by 71,75</em>

∵ The equation of the population is 10.0 y - 4.55 x = 581

∵ y = 71.75

∴ 10.0(71.75) - 4.55 x = 581

∴ 717.5 - 4.55 x = 581

- Subtract 717.5 from both sides

∴ - 4.55 x = - 136.5

- Divide both sides by - 4.55

∴ x = 30

∵ x represents the number of years after 2000

∵ 2000 + 30 = 2030

∴ In year 2030 the population is predicted to be 71.75 million

6 0
3 years ago
Tamir wants to buy a snowboard. The original price is $760. How much will Tamir pay if he buys it during the sale?
grandymaker [24]
Depends on how much it’s on sale
4 0
3 years ago
Joshua has $500 to spend at a bicycle store for some new gear and biking outfits. Assume all prices listed include tax.
spayn [35]

Answer: 12

Step-by-step explanation:

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
If the point (x,square root 3/2) is on the unit circle, what. is x?
galina1969 [7]

Answer:

x=\pm\frac{1}{2}

Step-by-step explanation:

On a unit circle, (x,y)=(\cos\theta,\sin\theta), so \sin\theta=\frac{\sqrt{3}}{2} in this case. If you look at the attached circle, the only time that the y-coordinate is \frac{\sqrt{3}}{2} is when x=-\frac{1}{2} and x=\frac{1}{2}, which correspond to angles of \theta=\frac{2\pi}{3} and \theta=\frac{\pi}{3} respectively

3 0
3 years ago
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