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2 years ago

Find the prime factors 204

Mathematics

2 answers:

Annette [7]2 years ago

4 0

2^2 • 3 • 17 because 204 is not a prime number. Hope this helps.

Lady bird [3.3K]2 years ago

3 0

2x2x3x17, factors are 1,2,3,4,6,12,17,34,51,68,102,204

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Help I will be marking brainliest!!!

baherus [9]

Answer:

Angle k is 34.5 degrees

Step-by-step explanation:

a circle makes 360 degrees total. okay so u have to consider the angles given. m arc go= 127

m<0 = 54.5

54.5 times 2 = 109

109 is arc gk

109 + 127+ 55= 291

360-291=69= arc OU

69 divided by 2 is 34.5

Angle k is 34.5 degrees

3 0

3 years ago

Find the least common multiple (LCM) of 8y^6+ 144y^5+ 640y^4 and 2y^4 + 40y^3 + 200y^2.

Mice21 [21]

Answer:

LCM = $8y^{4}(y+ 10)^{2}(y + 8)$

Step-by-step explanation:

Making factors of $8y^{6}+ 144y^{5}+ 640y^{4}$

Taking $8y^{4}$ common:

$\Rightarrow 8y^{4} (y^{2}+ 18y+ 80)$

Using factorization method:

$\Rightarrow 8y^{4} (y^{2}+ 10y + 8y + 80)\\\Rightarrow 8y^{4} (y (y+ 10) + 8(y + 10))\\\Rightarrow 8y^{4} (y+ 10)(y + 8))\\\Rightarrow \underline{2y^{2}} \times 4y^{2} \underline{(y+ 10)}(y + 8)) ..... (1)$

Now, Making factors of $2y^{4} + 40y^{3} + 200y^{2}$

Taking $2y^{2}$ common:

$\Rightarrow 2y^{2} (y^{2}+ 20y+ 100)$

Using factorization method:

$\Rightarrow 2y^{2} (y^{2}+ 10y+ 10y+ 100)\\\Rightarrow 2y^{2} (y (y+ 10) + 10(y + 10))\\\Rightarrow \underline {2y^{2} (y+ 10)}(y + 10) ............ (2)$

The underlined parts show the Highest Common Factor(HCF).

i.e. HCF is $2y^{2} (y+ 10)$ .

We know the relation between LCM, HCF of the two numbers 'p' , 'q' and the numbers themselves as:

$HCF \times LCM = p \times q$

Using equations (1) and (2): $\Rightarrow 2y^{2} (y+ 10) \times LCM = 2y^{2} \times 4y^{2}(y+ 10)(y + 8) \times 2y^{2} (y+ 10)(y + 10)\\\Rightarrow LCM = 2y^{2} \times 4y^{2}(y+ 10)(y + 8) \times (y + 10)\\\Rightarrow LCM = 8y^{4}(y+ 10)^{2}(y + 8)$

Hence, LCM = $8y^{4}(y+ 10)^{2}(y + 8)$

5 0

2 years ago

I need help with math

tatuchka [14]

Tell me the problems and i may be able to help

5 0

3 years ago

Quadratics need help making parabola

timama [110]

Answer:

The equation of the parabola is $y = \frac{1}{3}\cdot x^{2}-\frac{4}{3}\cdot x -4$ , whose real vertex is $(x,y) = (2, -5.333)$ , not $(x,y) = (2, -5)$ .

Step-by-step explanation:

A parabola is a second order polynomial. By Fundamental Theorem of Algebra we know that a second order polynomial can be formed when three distinct points are known. From statement we have the following information:

$(x_{1}, y_{1}) = (-2, 0)$ , $(x_{2}, y_{2}) = (6, 0)$ , $(x_{3}, y_{3}) = (0, -4)$

From definition of second order polynomial and the three points described above, we have the following system of linear equations:

$4\cdot a -2\cdot b + c = 0$ (1)

$36\cdot a + 6\cdot b + c = 0$ (2)

$c = -4$ (3)

The solution of this system is: $a = \frac{1}{3}$ , $b = - \frac{4}{3}$ , $c = -4$ . Hence, the equation of the parabola is $y = \frac{1}{3}\cdot x^{2}-\frac{4}{3}\cdot x -4$ . Lastly, we must check if $(x,y) = (2, -5)$ belongs to the function. If we know that $x = 2$ , then the value of $y$ is: