The graph of g(x) = (x + 2)2 is a translation of the

A town has a population of 32,000 in the year 2002; 35,200 in the year 2003; 38,720 in the year 2004; and 42,592 in the year 200

Romashka-Z-Leto [24]

The answer is 102,442

3 0

3 years ago

How to form a polynomial with given zeros and degree and multiplicity calculator

ohaa [14]

Answer:

Taking P(x) = x³-12x-16 as an example

Step-by-step explanation:

For a polynomial, if

x = a is a zero of the function, then (x − a) is a factor of the function.

We have two unique zeros:

−2 and 4. However, −2 has a multiplicity of 2, which means that the factor that correlates to a zero of −2 is represented in the polynomial twice.

Following how it's constructed

zero at -2, multiplicity 2

zero at 4, multiplicity 1

p(x)=x−(−2))²(x−4)¹

Thus,p(x)=(x+2)²(x−4)

Expand: p(x)=(x²+4x+4)(x−4)

p(x) =x³−12x−16

3 0

3 years ago

Hey again, could you help me with this and explain? thank you

Alex

Answer:

option 2

in the first equation y is greater than so it would be shaded above the line and in the second equation y is less than so it would be shaded below the line

Step-by-step explanation:

4 0

3 years ago

PLS HELP ME WITH 1, 2, and 3 (SHOW WORK!!!) MUST SHOW WORK! + LOTS OF POINTS!

Gekata [30.6K]

1)105 divided by 2 equals 52.5
2)294 divided by 6 equals 49 then, 49 times 8 equals 392
3)22 divided by 5 equals 4.4 then, 77 divided by 4.4 equals 17.5

4 0

3 years ago

Help! Prove the equality arccos √(2/3) - arccos (1+√6)/(2*√3) = π/6

stealth61 [152]

Answer:

Proof in the explanation

Step-by-step explanation:

Trigonometric Equalities

Those are expressions involving trigonometric functions which must be proven, generally working on only one side of the equality

For this particular equality, we'll use the following equation

$\displaystyle cos(x-y)=cos\ x\ cos\ y+sin\ x\ sin\ y$

The equality we want to prove is

$\displaystyle arccos\ \sqrt{\frac{2}{3}}-arccos\left(\frac{1+\sqrt{6}}{2\sqrt{3}}\right)=\frac{\pi}{6}$

Let's set the following variables:

$\displaystyle x=arccos\ \sqrt{\frac{2}{3}},\ y=arccos(\frac{1+\sqrt{6}}{2\sqrt{3}})$

And modify the first variable:

$\displaystyle x=arccos\ \frac{\sqrt{6}}{3}}=>\ cos\ x= \frac{\sqrt{6}}{3}}$

Now with the second variable

$\displaystyle y=arccos\ \frac{1+\sqrt{6}}{2\sqrt{3}}=>cos\ y=\frac{1+\sqrt{6}}{2\sqrt{3}}=\frac{\sqrt{3}+3\sqrt{2}}{6}$

Knowing that

$sin^2x+cos^2x=1$

We compute the other two trigonometric functions of X and Y

$\displaystyle sin \ x=\sqrt{1-cos^2\ x}=\sqrt{1-(\frac{\sqrt{6}}{3})^2}=\sqrt{1-\frac{6}{9}}=\frac{\sqrt{3}}{3}$

$\displaystyle sin\ y=\sqrt{1-cos^2y}=\sqrt{1-\frac{(\sqrt{3}+3\sqrt{2})^2}{36}}}$

$\displaystyle sin\ y=\sqrt{\frac{36-(3+6\sqrt{6}+18)}{36}}=\sqrt{\frac{15-6\sqrt{6}}{36}}$

Computing

$15-6\sqrt{6}=(3-\sqrt{6})^2$

Then

$\displaystyle sin\ y=\frac{3-\sqrt{6}}{6}$

Now we replace all in the first equality:

$\displaystyle cos(x-y)=\frac{\sqrt{6}}{3}.\frac{\sqrt{3}+3\sqrt{2}}{6}+\frac{\sqrt{3}}{3}.\frac{3-\sqrt{6}}{6}$

$\displaystyle cos(x-y)=\frac{3\sqrt{2}+6\sqrt{3}}{18}+\frac{3\sqrt{3}-3\sqrt{2}}{18}$

$\displaystyle cos(x-y)=\frac{9\sqrt{3}}{18}=\frac{\sqrt{3}}{2}=cos\ \pi/6$

Thus, proven

5 0

3 years ago