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Nastasia [14]
2 years ago
13

Work out the size of the angle marked t

Mathematics
2 answers:
vitfil [10]2 years ago
6 0
The triangle shown is an isosceles triangle, which means both the both angles are of the same size.

the sum of all the angles is 180°.

t ° + t ° + 110 ° = 180 °

2t ° = 180 ° - 110 °

2t ° = 70 °

t = 70°/2

t = 35 °
tiny-mole [99]2 years ago
3 0

Answer:

<h2>35°</h2>

Step-by-step explanation:

This is an isoceles triangle

And as we all know an isoceles triangle has two equal base angles.

And we also know that the sum of all interior angles of the triangle is 180°

So based on this we can form an algebraic equation of solve for angke t (Let's do it :))

110 + 2t = 180 \\ 2t = 180 - 110 \\ 2t = 70 \\ t  =  \frac{70}{2}  \\ t = 35

So the base angle is 35°

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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

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