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3 years ago

Find a polynomial of degree 2 with real number coefficients and zero 3-2i

1 answer:

4 0

Complex roots occur as conjugates so the other root is 3 + 2i

so we have the factored form

(x - (3 - 2i)(x - (3 + 2i)

= (x - 3 + 2i)(x - 3 - 2i)

= x^2 -3x - 2ix - 3x + 9 + 6i + 2ix - 6i- 4i^2

= x^2 -6x + 9 + 4

= x^2 - 6x + 13 Answer

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Please asap this is the only question and I'm also running out of time thank youuuuu

Wittaler [7]

One question, and no one answered it for you?

Simple,

the answer is....(6,5) Point of Intersection.

I triple checked my work. :)

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2 years ago

3. It will take half an hour for Matthew to travel to

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3 years ago

A line is drawn so that it passes through the points (-3, -1) and (4, 2).

marishachu [46]

Answer:

(A) $\boxed{\bold{y=\frac{3}{7}x+\frac{2}{7}}}$

(B) $\boxed{\bold{2 \ = \ \frac{3}{7} * \ 4 \ + \ \frac{2}{7} }}$

Explanation:

(A) Slope: y = mx + b

m = 3/7

b = 2/7

Slope = $\bold{\frac{y_2-y_1}{x_2-x_1}}$

$\bold{\left(x_1,\:y_1\right)=\left(-3,\:-1\right),\:\left(x_2,\:y_2\right)=\left(4,\:2\right)}$

M = $\bold{\frac{2-\left(-1\right)}{4-\left(-3\right)}: \ \frac{3}{7} }$

Y intercept

$\bold{y=\frac{3}{7}x+b}$

Plug in $\bold{\left(-3,\:-1\right)\mathrm{:\:}\quad \:x=-3,\:y=-1}$

$\bold{-1=\frac{3}{7}\left(-3\right)+b}$

Isolate B

$\bold{-1=\frac{3}{7}\left(-3\right)+b}$

b = $\bold{\frac{2}{7} }$

(B) 4 = x, 2 = y

2 = 3/7 * 4 + 2/7

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3 years ago

Read 2 more answers

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Quadrilateral ABCD is reflected across the y-axis and translated 1 unit down to create quadrilateral EFGH .

sleet_krkn [62]

GH is parallel to FE

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