In the plane, if a line is perpendicular to a circle's radius at its outer endpoint, then the line is _____(what) to that circle

Mathematics

1 answer:

denis-greek [22]2 years ago

8 0

Answer:

It is B i think i hope this helps

Step-by-step explanation:

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Assoli18 [71]

Answer:

75°

Step-by-step explanation:

Please see attached picture for full solution.

5 0

2 years ago

Which value must be added to the expression x2 – 3x to make it a perfect-square trinomial?

Vitek1552 [10]

Answer:

$\frac{9}{4}$

Step-by-step explanation:

Given

x² - 3x

To make the expression a perfect square

add ( half the coefficient of the x- term )²

x² + 2( - $\frac{3}{2}$ )x + $\frac{9}{4}$ , thus

x² - 3x + $\frac{9}{4}$

= (x - $\frac{3}{2}$ )² ← a perfect square

7 0

2 years ago

What is the polynomial function of lowest degree with lead coefficient 1 and roots 1 and 1 + i? f(x) = x2 – 2x + 2 f(x) = x3 – x

konstantin123 [22]

Answer:

$f(x)=x^{3}-3x^{2} +4x-2$

Step-by-step explanation:

we know that

The <u><em>conjugate root theorem</em></u> states that if the complex number a + bi is a root of a polynomial P(x) in one variable with real coefficients, then the complex conjugate a - bi is also a root of that polynomial

In this problem we have that

The polynomial has roots 1 and (1+i)

by the conjugate root theorem

(1-i) is also a root of the polynomial

therefore

The lowest degree of the polynomial is 3

$f(x)=a(x-1)(x-(1+i))(x-(1-i))$

Remember that

The leading coefficient is 1

a=1

$f(x)=(x-1)(x-(1+i))(x-(1-i))\\\\f(x)=(x-1)[x^{2} -(1-i)x-(1+i)x+(1-i^2)]\\\\f(x)=(x-1)[x^{2} -x+xi-x-xi+2]\\\\f(x)=(x-1)[x^{2} -2x+2]\\\\f(x)=x^{3}-2x^{2} +2x-x^{2} +2x-2\\\\f(x)=x^{3}-3x^{2} +4x-2$