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3 years ago

If a polynomial function f(x) has roots -4,2,1 and square root of 11 what must be also a root of f(x) ?

Mathematics

1 answer:

Zigmanuir [339]3 years ago

5 0

Answer:

- $\sqrt{11}$

Step-by-step explanation:

Radical roots occur in conjugate pairs

Thus if $\sqrt{11}$ is a root then

- $\sqrt{11}$ is also a root

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In 1955 an antique car that originally cost 3,943 is valued at 64,125 if in excellent condition, winch is 2 1/4 times as much as

nika2105 [10]

Answer:

The value of the car in very nice condition is $28,500.

Step-by-step explanation:

Consider the provided information.

In 1955 an antique car that originally cost 3,943 is valued at 64,125 if in excellent condition, winch is 2 1/4 times as much as a car in very nice condition.

The mixed fraction can be written as:

$2 \frac{1}{4}=\frac{9}{4}$

Let x is the value of car in very nice condition.

$64,125 =\frac{9}{4}x$

$x=\frac{64,125\times 4}{9}$

$x=28,500$

Hence, the value of the car in very nice condition is $28,500.

7 0

3 years ago

-. For the event of rolling three standard fair 6-sided dice, find the following probabilities. Give all

elixir [45]

Answer:

1/6 or 6/36

Step-by-step explanation:

Given the qualifications, there are six qualifying combinations out of the 36 combinations, which simplified, is 1 out of 6.

4 0

3 years ago

Help ASAP please! Will give brainliest if correct.

Neporo4naja [7]

Answer:

Step-by-step explanation:

5 0

3 years ago

Read 2 more answers

What is the period and midline?

OverLord2011 [107]

$\bf \qquad \qquad \qquad \qquad \textit{function transformations}
\\ \quad \\
% function transformations for trigonometric functions
\begin{array}{rllll}
% left side templates
f(x)=&{{ A}}cos({{ B}}x+{{ C}})+{{ D}}
\\ \quad \\

\end{array}\qquad$

$\bf \begin{array}{llll}
% right side info
\bullet \textit{ stretches or shrinks}\\
\quad \textit{horizontally by amplitude } |{{ A}}|\\
\bullet \textit{ horizontal shift by }\frac{{{ C}}}{{{ B}}}\\
\qquad if\ \frac{{{ C}}}{{{ B}}}\textit{ is negative, to the right}\\
\qquad if\ \frac{{{ C}}}{{{ B}}}\textit{ is positive, to the left}\\
\bullet \textit{vertical shift by }{{ D}}\\
\qquad if\ {{ D}}\textit{ is negative, downwards}\\
\qquad if\ {{ D}}\textit{ is positive, upwards}\\


\end{array}$

$\bf \begin{array}{llll}
\bullet \textit{function period}\\
\qquad \frac{2\pi }{{{ B}}}\ for\ cos(\theta),\ sin(\theta),\ sec(\theta),\ csc(\theta)\\
\qquad \frac{\pi }{{{ B}}}\ for\ tan(\theta),\ cot(\theta)
\end{array}
$

so if you notice yours $\bf \begin{array}{llll}
3.2cos&\left( \frac{5}{3}\theta \right)+&6.1\\
&\ \uparrow&\uparrow \\
&B&D 
\end{array}$

now.. normally the function $\bf 3.2cos&\left( \frac{5}{3}\theta \right)$
has a D value of 0, or no vertical shift, and the amplitude and the period simply make the wave taller and thinner, but the midline is still the x-axis

now, with D = 6.1, that moves the midline up vertically that much

now.. the period, well, B = 5/3, normal period of cosine is $2\pi$
so, the new period will be $\bf \cfrac{2\pi }{B}\implies \cfrac{2\pi }{\frac{5}{3}}\implies \cfrac{6\pi }{5}$

notice the picture below
the vertical shift by the D component, or 6.1, moved the midline to y = 6.1 :)

6 0

3 years ago

-15x=-4x... How to move the x to the left?

Delvig [45]

-15x=-4x

We know that adding zero wouldn’t change the value

-15x=-4x+0

Now we can add 4x (since that’s the opposite of the negative sign) to the other side

-11x=0

0 divided by 11 is 0

x=0

4 0

3 years ago