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3 years ago

Which of the following is a polynomial with roots negative square root of 5, square root of 5, and 3?

1 answer:

6 0

The polynomial with those roots would be:

y = (x + √5)(x - √5)(x - 3) 
▬▬▬▬▬▬▬▬▬▬▬ multiply this out to get 
y = (x² - 5)(x - 3) 

y = x³ - 3x² - 5x +15

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An entity with two multivalued attributes will be mapped as how many relations

grigory [225]

There are so many relationships between the original entity and new relation.
The relation mapped from an entity involved in a 1:1 unary relationship contains a foreign key that correspond to its own primary key. 1 to one foreign key will be unique.

5 0

3 years ago

Please help me 50 points

Afina-wow [57]

angle 3 = angle 6 = 60°

hope it helps...!!!

3 0

2 years ago

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Write each of the following as a function of theta. 1.) sin(pi/4 - theta) 2.) tan(theta+30°)

Paladinen [302]

Step-by-step explanation:

Let x represent theta.

$\sin( \frac{\pi}{4} - x )$

Using the angle addition trig formula,

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

$\sin( \frac{\pi}{4} ) \cos(x) - \cos( \frac{\pi}{4} ) \sin(x)$

$( \frac{ \sqrt{2} }{2}) \cos(x) - (\frac{ \sqrt{2} }{2} )\sin(x)$

Multiply one side at a time

Replace theta with x , the answer is

$\frac{ \sqrt{2} \cos(x) }{2} - \frac{ \sin(x) \sqrt{2} }{2}$

2. Convert 30 degrees into radian

$\frac{30}{1} \times \frac{\pi}{180} = \frac{\pi}{6}$

Using tangent formula,

$\tan(x + y) = \frac{ \tan(x) + \tan(y) }{1 - \tan(x) \tan(y) }$

$\frac{ \tan(x) + \tan( \frac{\pi}{6} ) }{1 - \tan(x) \tan( \frac{\pi}{6} ) }$

Tan if pi/6 is sqr root of 3/3

$\frac{ \tan(x) + ( \frac{ \sqrt{3} }{3} ) }{1 - \tan(x) (\frac{ \sqrt{3} }{3} ) }$

Since my phone about to die if you later simplify that,

you'll get

$\frac{(3 \tan(x) + \sqrt{3} )(3 + \sqrt{3} \tan(x) }{3(3 - \tan {}^{2} (x) }$

Replace theta with X.

4 0

2 years ago

35-4-2-13 O C5 ( 42) 13 o D( 64) -2-13

Radda [10]

Hmmm it’s D I think?

7 0

2 years ago

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M is the midpoint of AN, A has coordinates

maksim [4K]

Step-by-step explanation:

According to Question , M is the midpoint of AN, A has coordinates (-7,4), and M has coordinates (-4, -1).

Figure :-

$\setlength{\unitlength}{1 cm}\begin{picture}(20,12) \put(4,0.2){\line(0,-1){0.4}}\put(1,0){\line(1,0){6}} \put(3.8,-0.6){$\bf M(-4,-1) $} \put(1,-0.6){$\bf A (-7,4)$} \put(6.8,-0.6){$\bf N(x,y)$} \end{picture}$

Let the coordinates of N be ( x , y )

Now , according to Midpoint Formula , the midpoint of points say A(x , y) and B (x' , y') is given by ,

$\boxed{\blue{ \bf Midpoint =\bigg( \dfrac{x+x'}{2} , \dfrac{y+y'}{2} \bigg) }}$

On using this formula ,

$=> Midpoint = \bigg( \dfrac{x+x'}{2} , \dfrac{y+y'}{2} \bigg) \\ \\ => (-4, 1) = \bigg( \dfrac{x -7 }{2} , \dfrac{y + 4 }{2}\bigg) \\ \\ => -4 = \dfrac{x-7}{2} \\\\ => x - 7 = -8 \\\\ => x = 7 -8 \\\\ \boxed{\red{\sf=> x = -1 }} \\\\ => \dfrac{y+4}{2}=-1 \\\\=> y +4 = -2 \\ \\ y = -4-2 \\\\\boxed{\red{\sf=> y = -6 }}$

8 0

3 years ago

Read 2 more answers