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Galina-37 [17]
3 years ago
12

Please Answer Correctly - Will Give Branliest & Extra Points. (links or unidentified/wrong answers will be reported and auto

banned)

Mathematics
2 answers:
jonny [76]3 years ago
6 0

Answer:

Step-by-step explanation:

11 no

46+4x=90 degree(being perpendicular)

4x=90-46

x=44/4

x=11

12 no

They are complementary angles

6 no

angle B+27=180 degree(being straight line)

angle B=180-27

angle B=153

10 no

angle 38 +angle DFE=180 degree(being supplementary)

angle DFE=180-38

angle DFE=142 degree

5 no

They are vertically opposite angles.

7 no

angle AGF and angle CGD (because they have common arm and vertex)

Tatiana [17]3 years ago
4 0

Answer:

Solution given:

11.4x+46=90[being right angle]

4x=90-46

x=44/4

x=11

12.

<u>Complementary</u><u> </u><u>angles</u>

<u>5</u><u>.</u>

vertical

<u>6</u><u>.</u>

<ABD+27=180°[linear pair]

<ABD=180-27

<ABD=153° degree

10.

<DFE+38=180[being supplementary]

<DFE=180-38

<u><</u><u>DFE</u><u>=</u><u>1</u><u>4</u><u>2</u><u> </u><u>degrees</u><u>.</u>

<u>7</u><u>.</u>

<u>angle</u><u> </u><u>DGE</u><u> </u><u>and</u><u> </u><u>angle</u><u> </u><u>DGC</u>

<u>8</u><u>.</u>

<u><</u><u>A</u><u>G</u><u>B</u><u> </u><u>and</u><u> </u><u><</u><u>AGE</u>

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Use the given zero to find the remaining zeros of each function f(x)=2x^4+5x^3+5x^2+20x-12 zero:-2i
frutty [35]

Answer:

1/2, 3

Step-by-step explanation:

This is a pretty involved problem, so I'm going to start by laying out two facts that our going to help us get there.

  1. The Fundamental Theorem of Algebra tells us that any polynomial has <em>as many zeroes as its degree</em>. Our function f(x) has a degree of 4, so we'll have 4 zeroes. Also,
  2. Complex zeroes come in pairs. Specifically, they come in <em>conjugate pairs</em>. If -2i is a zero, 2i must be a zero, too. The "why" is beyond the scope of this response, but this result is called the "complex conjugate root theorem".

In 2., I mentioned that both -2i and 2i must be zeroes of f(x). This means that both x-2i and x+2i are factors of f(x), and furthermore, their product, x^2+4, is <em>also</em> a factor. To see what's left after we factor out that product, we can use polynomial long division to find that

2x^4+5x^3+5x^2+20x-12=(x^2+4)(2x^2+5x-3)

I'll go through to steps to factor that second expression below:

2x^2+5x-3=2x^2+6x-x-3\\=2x(x+3)-(x+3)\\=(2x-1)(x+3)

Solving both of the expressions when f(x) = 0 gets us our final two zeroes:

2x-1=0\\2x=1\\x=1/2

x+3=0\\x=-3

So, the remaining zeroes are 1/2 and 3.

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