All categories

3 years ago

What is the equation of the line that passes through the point (3,-3) and has a slope of 0?

Mathematics

1 answer:

Blizzard [7]3 years ago

7 0

Answer:

given us,

slope= 0

(x1y1)= (3,-3)

here

(y-y1) =m(x-x1)

or, (y+3)=0(x-3)

or (y+3)= x-3

or, x+y= -3-3

x+y= -6

Step-by-step explanation:

x+y= -6

You might be interested in

if j is 21 inches, k is 12 inches, and angle M measures 62 degrees, then find m using the Law is Cosines. round your answer to t

AURORKA [14]

Answer:

18.7

Step-by-step explanation:

The Law of Cosines is $c^2=a^2+b^2-2ab cos(C)$ , where c is the unknown side length of a triangle, a and b are the remaining two side lengths, and C is the angle opposite to side c. To answer this question just plug in the known values:

$m^2=21^2+12^2-2(21)(12) cos(62)$

Simplify:

$m^2=441+144-504 cos(62)$

$m^2=585-504 cos(62)$

$m=\sqrt{585-504cos(62)}$

m≈18.665...

When rounded to the nearest tenth, m=18.7

6 0

3 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:

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