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2 years ago

Which system of equations can be used to find the roots of the equation 4x^5-12x^4+6x=5x^3-2x?

Mathematics

1 answer:

Sveta_85 [38]2 years ago

7 0

$^{(1)}\ \ 4x^5-12x^4+6x=5x^3-2x\ \ ^{(2)}\\\\ \left\{\begin{array}{ccc}y=4x^5-12x^4+6x\ \ \ ^{(1)}\\y=5x^3-2x\ \ \ ^{(2)}\end{array}\right$

Answer: system nr 4.

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Nezavi [6.7K]

A: Their would be 24 red peppers because I did 18 divided by 3/4 and got 24 and 24 times 3/4 is 18
B:She bought 42 peppers all together because 18 plus 24 is 42

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2 years ago

In △ABC, m∠A=15 °, a=10 , and b=11 . Find c to the nearest tenth.

pshichka [43]

Answer:

The answer is:

$\bold{c\approx 20.2\ units}$

Step-by-step explanation:

Given:

In △ABC:

m∠A=15°

a=10 and

b=11

To find:

c = ?

Solution:

We can use cosine rule here to find the value of third side c.

Formula for cosine rule:

$cos A = \dfrac{b^{2}+c^{2}-a^{2}}{2bc}$

Where

a is the side opposite to $\angle A$

b is the side opposite to $\angle B$

c is the side opposite to $\angle C$

Putting all the values.

$cos 15^\circ = \dfrac{11^{2}+c^{2}-10^{2}}{2\times 11 \times c}\\\Rightarrow 0.96 = \dfrac{121+c^{2}-100}{22c}\\\Rightarrow 0.96 \times 22c= 121+c^{2}-100\\\Rightarrow 21.25 c= 21+c^{2}\\\Rightarrow c^{2}-21.25c+21=0\\\\\text{solving the quadratic equation:}\\\\c = \dfrac{21.25+\sqrt{21.25^2-4 \times 1 \times 21}}{2}\\c = \dfrac{21.25+\sqrt{367.56}}{2}\\c = \dfrac{21.25+19.17}{2}\\c \approx 20.2\ units$

The answer is:

$\bold{c\approx 20.2\ units}$

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2 years ago

Which of the following notations correctly describe the end behavior of the polynomial graphed below

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The answer is B
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2 years ago

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A bowl contains 55 mints. Customers eat 4 mints each hour. Which equation can be used to determine how many hours remain until 2

Nina [5.8K]

Answer:

55-4x=22

Step-by-step explanation:

5 0

1 year ago

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Trains A and B, 200 km apart on the same straight track, travel at speeds of 50 km/hr and 65 km/hr respectively. At the end of 1

Gwar [14]

Answer:

Ok, at the beginning the distance between the trains is 200km:

IPa(0s) - P(0s)I = 200km.

where Pa is the position of train A, and Pb is the position of train B.

Now, the speed of train A is:

Sa = 50km/h

Sb = 65km/h

Now, remember the relation:

Distance = Speed*time.

So, in one hour, the displacement of train A is:

Distance = 50km/h*1h = 50km

The displacement of train B is:

distance = 65km/h*1h = 65km.

Now, if at the beginning train A is 200km ahead of train B, then we have:

Pa - Pb = (Pa(0s) + 50km) - (Pb(0s) + 65km)

= (Pa(0s) - Pb(0s)) + (50km - 65km) = 200km - 15km = 185km.

Now, if train B was 200km ahead of train A, we have:

Pb - Pa = (Pb(0s) + 65km) - (Pa(0s) - 50km) =

(Pb(0s) - Pa(0s)) + (65km - 50km) = 200km + 15km = 215km

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2 years ago