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katen-ka-za [31]
3 years ago
6

Explain Steps. Do not answer. #10 and 11Questions are in the Picture

Mathematics
1 answer:
svet-max [94.6K]3 years ago
8 0

Answer:

Step-by-step explanation:

For problem 10:

1. AE/ED=AC/CB (Since triangle ABC is similar to triangle ADE, we can determine that the ratio of AE to ED is equal to the ratio of AC to CB)

2. AE/ED=(AE+EC)/CB (Rewrite AC as the sum of the lengths forming it; This is sometimes referred to as the Partition Postulate)

3. 9/x=(9+6)/10 (Substitute the given values into this equation)

4. x=6 (Use algebra to solve for x)

For problem 11:

1. AG/AB=AE/AD (Use the same strategy as step one in problem 10, since the rectangles are similar we can create this equation)

2. AG/(AG+GB)=AE/(AE+ED) (Rewrite sides as the some of their parts)

3. 14/(14+7)=18/(18+x) (Substitute given values)

4. x=9 (Solve for x)

lmk if there are mistakes in my explanation, hope this helps :)

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2 years ago
Find a factorization of x² + 2x³ + 7x² - 6x + 44, given that<br> −2+i√√7 and 1 - i√/3 are roots.
Levart [38]

A factorization of x^4+2x^3+7x^2-6x+44 is (x^2+4x+11)(x^2-2x+4).

<h3>What are the properties of roots of a polynomial?</h3>
  • The maximum number of roots of a polynomial of degree n is n.
  • For a polynomial with real coefficients, the roots can be real or complex.
  • The complex roots of a polynomial with real coefficients always exist in a pair of conjugate numbers i.e., if a+ib is a root, then a-ib is also a root.

If the roots of the polynomial p(x)=ax^4+bx^3+cx^2+dx+e are r_1,r_2,r_3,r_4, then it can be factorized as p(x)=(x-r_1)(x-r_2)(x-r_3)(x-r_4).

Here, we are to find a factorization of p(x)=x^4+2x^3+7x^2-6x+44. Also, given that -2+i\sqrt{7} and 1-i\sqrt{3} are roots of the polynomial.

Since p(x)=x^4+2x^3+7x^2-6x+44 is a polynomial with real coefficients, so each complex root exists in a pair of conjugates.

Hence, -2-i\sqrt{7} and 1+i\sqrt{3} are also roots of the given polynomial.

Thus, all the four roots of the polynomial p(x)=x^4+2x^3+7x^2-6x+44, are: r_1=-2+i\sqrt{7}, r_2=-2-i\sqrt{7}, r_3=1-i\sqrt{3}, r_4=1+i\sqrt{3}.

So, the polynomial p(x)=x^4+2x^3+7x^2-6x+44 can be factorized as follows:

\{x-(-2+i\sqrt{7})\}\{x-(-2-i\sqrt{7})\}\{x-(1-i\sqrt{3})\}\{x-(1+i\sqrt{3})\}\\=(x+2-i\sqrt{7})(x+2+i\sqrt{7})(x-1+i\sqrt{3})(x-1-i\sqrt{3})\\=\{(x+2)^2+7\}\{(x-1)^2+3\}\hspace{1cm} [\because (a+b)(a-b)=a^2-b^2]\\=(x^2+4x+4+7)(x^2-2x+1+3)\\=(x^2+4x+11)(x^2-2x+4)

Therefore, a factorization of x^4+2x^3+7x^2-6x+44 is (x^2+4x+11)(x^2-2x+4).

To know more about factorization, refer: brainly.com/question/25829061

#SPJ9

3 0
1 year ago
Read 2 more answers
What is the solution to this system of equations?
BaLLatris [955]
The answer is B
X = 3 and 1/2= Y
3 + 2(1/2) = 4
3 + 1 = 4
Half of 2 = 1, making B the answer
6 0
3 years ago
Read 2 more answers
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