I need help with # 8,10,12,14,16,18

Mathematics

1 answer:

zheka24 [161]3 years ago

5 0

8) 81 has diploma

9)54

10)54

11)27

12) 843

13)1,456

14)433

15)353

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PLEASE HELP!!

Illusion [34]

Answer:

8 and 10

Step-by-step explanation:

Our first step is to set x as the other side length, and y as the hypotenuse. We get that 6 + x + y = 24, and x + y = 18. The area gives us that 6x = 24(2) = 48. We then divide both sides by 6 and we get x = 8. We have x = 8 so we plug that into the equation x + y = 18 and we get that y = 10. So the other side lengths are 8 and 10.

5 0

2 years ago

Factor the polynomial function over the complex numbers. f(x)=x^4-x^3-2x−4 Enter your answer in the box. f(x) = The answer is: (

Vladimir79 [104]

Answer:

Factors: $(x+i\sqrt2)(x-i\sqrt2)(x+1)(x-2)=0$

Step-by-step explanation:

We are given a polynomial:

$f(x)=x^4-x^3-2x-4$

We have to factor the given polynomial into its complex factors.

The factorization can be done as follows:

$f(x)=x^4-x^3-2x-4 = 0\\x^4-x^3-2x-4 = 0\\x^4-4-x^3-2x=0\\\text{Identity: }a^2-b^2 = (a+b)(a-b)\\(x^4-4)-(x^3+2x) = 0\\(x^2+2)(x^2-2)-x(x^2+2) = 0\\(x^2+2)(x^2-2-x) = 0\\(x^2+2)(x^2-x-2) = 0\\(x^2+2)(x^2-2x-+x-2) = 0\\(x^2+2)((x(x-2)+1(x-2))=0\\(x^2+2)(x+1)(x-2)=0\\\text{Identity: }a^2-b^2 = (a+b)(a-b)\\(x^2-(\sqrt{-2})^2)(x+1)(x-2)=0\\(x+i\sqrt2)(x-i\sqrt2)(x+1)(x-2)=0$