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2 years ago

Acellus

Mathematics

1 answer:

aksik [14]2 years ago

7 0

9514 1404 393

Answer:

{-8, -4, -2, -1, 1, 2, 4, 8}

Step-by-step explanation:

Possible rational roots are positive and negative divisors of the constant 8:

{-8, -4, -2, -1, 1, 2, 4, 8}

_____

<em>Additional comment</em>

This polynomial has no rational roots. Its two real roots are both negative and irrational.

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Simplify these expressions

Xelga [282]

$\frac{10a}{2} \\ \\ 5a \\ \\$

The answer is: 5a

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$\frac{36b}{6b} \\ \\ \frac{36}{6} \\ \\ 6 \\ \\$

The answer is: 6

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$\frac{16c^2}{8c} \\ \\ 2c^{2-1} \\ \\ 2c^1 \\ \\ 2c \\ \\$

The answer is: 2c

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$\frac{18ab^2}{2b} \\ \\ 9ab^{2-1} \\ \\ 9ab^1 \\ \\ 9ab \\ \\$

The answer is: 9ab

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$\frac{40b^2c^2}{5bc} \\ \\ 8b^{2-1} c^{2-1} \\ \\ 8b^1c^1 \\ \\ 8bc^1 \\ \\ 8bc \\ \\$

The answer is: 8bc

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$\frac{63ab^2c^3}{7b^2c^2} \\ \\ \frac{63ac^3}{7c^2} \\ \\ 9ac^{3-2} \\ \\ 9ac^1 \\ \\ 9ac \\ \\$

The answer is: 9ac

6 0

3 years ago

..................

sveticcg [70]

Answer:

y = 100°

Step-by-step explanation:

x = 40° (vertical angles are congruent)

y is an exterior angel of a triangle that has two opposite internal angles, x (40°) and 60°.

According to the exterior angle of a triangle, thus:

y = 40 + 60

y = 100°

3 0

2 years ago

if your bicycle tire has a diameter of 9 inches. how many rotations does each tire make if you travel 300 feet?

Vesnalui [34]

Step One
Calculate the number of feet traveled in 1 rotation.

Formula
C = π*d
P = 3.14
d = 9 inches = 9/12 feet = 3/4 of a foot.

C = 3.14 * 3/4 = 2.355 So that means that every time the tire turns around 1 complete turn, the distance traveled on the ground is 2.355 feet.

Step Two
Figure out the number of revolutions.
1 revolution = 2.355 feet
x revolutions = 300 feet.

1/x = 2.355/ 300 Cross multiply
2.355 feet * x = 1 rev * 300 feet
2.355 x = 300 rev Divide by 2.355
x = 300 / 2.355
x = 127.39 revolutions. <<<< Answer

8 0

3 years ago

Suppose that f: R --> R is a continuous function such that f(x +y) = f(x)+ f(y) for all x, yER Prove that there exists KeR su

Pachacha [2.7K]

<h2>Answer with explanation:</h2>

It is given that:

f: R → R is a continuous function such that:

$f(x+y)=f(x)+f(y)------(1)$ ∀ x,y ∈ R

Now, let us assume f(1)=k

Also,

f(0)=0

( Since,

f(0)=f(0+0)

i.e.

f(0)=f(0)+f(0)

By using property (1)

Also,

f(0)=2f(0)

i.e.

2f(0)-f(0)=0

i.e.

f(0)=0 )

Also,

f(2)=f(1+1)

i.e.

f(2)=f(1)+f(1) ( By using property (1) )

i.e.

f(2)=2f(1)

i.e.

f(2)=2k

Similarly for any m ∈ N

f(m)=f(1+1+1+...+1)

i.e.

f(m)=f(1)+f(1)+f(1)+.......+f(1) (m times)

i.e.

f(m)=mf(1)

i.e.

f(m)=mk

Now,

$f(1)=f(\dfrac{1}{n}+\dfrac{1}{n}+.......+\dfrac{1}{n})=f(\dfrac{1}{n})+f(\dfrac{1}{n})+....+f(\dfrac{1}{n})\\\\\\i.e.\\\\\\f(\dfrac{1}{n}+\dfrac{1}{n}+.......+\dfrac{1}{n})=nf(\dfrac{1}{n})=f(1)=k\\\\\\i.e.\\\\\\f(\dfrac{1}{n})=k\cdot \dfrac{1}{n}$

Also,

when x∈ Q

i.e. $x=\dfrac{p}{q}$

Then,

$f(\dfrac{p}{q})=f(\dfrac{1}{q})+f(\dfrac{1}{q})+.....+f(\dfrac{1}{q})=pf(\dfrac{1}{q})\\\\i.e.\\\\f(\dfrac{p}{q})=p\dfrac{k}{q}\\\\i.e.\\\\f(\dfrac{p}{q})=k\dfrac{p}{q}\\\\i.e.\\\\f(x)=kx\ for\ all\ x\ belongs\ to\ Q$