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antiseptic1488 [7]
3 years ago
12

What is the factored form of this expression? x2 + 17x + 60

Mathematics
2 answers:
mel-nik [20]3 years ago
8 0

Answer:

( x + 5 ) ( x + 12 )

Step-by-step explanation:

hope it helps

charle [14.2K]3 years ago
3 0
(x+5) (x+12)
I just took it
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PLS HELP
Katena32 [7]

Answer:

-30x^2-9x+12  all real numbers

Step-by-step explanation:

f(x) = -6x + 3 and g(x) = 5x + 4

f(x) * g(x) = (-6x + 3) * ( 5x + 4)

FOIL

            = -30x^2 -24x+15x +12

Combine like terms

           =-30x^2-9x+12

The domain is what numbers x can take

There are no restrictions so all real numbers

3 0
3 years ago
CAN SOMEONE HELP ME PLEASE ASAP!?
VikaD [51]

Answer:

40°

Step-by-step explanation:

∠AOE and ∠COD are vertical angles meaning they have the same value.

8 0
3 years ago
Graph the ordered pair (-4,-1)
Bingel [31]

Answer:

-4 would be in the x axes and -1 would be in the y axes

Step-by-step explanation:

I hop i explained it well

The drawing i took dosent want to post

3 0
2 years ago
Read 2 more answers
Cal has a sports trading card collection with 30 baseball cards, 25 basketball cards, and 45 football cards. What percent of his
olga_2 [115]

total cards = 30 + 25 + 45 = 100

25 are basket ball

25/100 = 0.25

0.25 = 25%

 25% are basketball

5 0
3 years ago
Read 2 more answers
Show that (2 + √2i)^20+ (2 − √2i)^20is an integer.
Tju [1.3M]

Answer:

Yes, it is, we need to use the Moivre theorem and we get

Step-by-step explanation:

Hi, first, let´s introduce Moivre theorem to find the nth power of a complex number.

z^{n} =r(Cos(n\alpha )+iSin(n\alpha ))

Where:

r = module of the complex number

n= power

alpha= inclination angle

to find the module of the complex number, we need to use the following formula.

r=\sqrt{a^{2} +b^{2} }

Where:

z= a+bi

a= real part of the coplex number

b=imaginary part of the complex number

Finally, in order to find the angle (alpha), we have to use the following.

\alpha =tan^{-1} (\frac{b}{a} )

But, using Moivre for a complex number to the 20th power is not very practical, so we are going to assume some things first

z_{1} =(2+\sqrt{2} i)

z_{2} =(2-\sqrt{2} i)

So, first we are going to find the value of z_{1} ^{2} and elevate it to the 10th power in order to get (2+\sqrt{2} i)^{20}

First, lets find the module of z1

r_{1} =\sqrt{2^{2} +(\sqrt{2} )^{2} }=\sqrt{4+2} =\sqrt{6}

and its angle is:

\alpha =tan^{-1} (\frac{\sqrt{2} }{2} )=45

we are all set, now let´s find the value of z_{1} ^{2}

z_{1} ^{2} =\sqrt{6} (Cos(2*45 )+iSin(2*45 ))

z_{1} ^{2} =\sqrt{6} (Cos(90 )+iSin(90 ))}

z_{1} ^{2} =\sqrt{6} (0+i(1))

z_{1} ^{2}=\sqrt{6} i

Now, let´s find the value of z_{1} ^{20}

(z_{1} ^{2})^{10} =(\sqrt{6} i)^{10} =7,776(i)^{4} (i)^{4} (i)^{2} =7,776(1)(1)(-1)=-7,776

therefore:

(2 + \sqrt{2} i)^{20} =-7,776

We do the same for (2 − √2i)^20, this time:

z_{2} =(2-\sqrt{2})

r_{2} =\sqrt{2^{2} +(-\sqrt{2} )^{2} }=\sqrt{4+2} =\sqrt{6}

And the angle is

\alpha =tan^{-1} (\frac{-\sqrt{2} }{2} )=-45

Therefore, we get:

z_{2} ^{2} =\sqrt{6} (Cos(2*(-45) )+iSin(2*(-45) ))

z_{2} ^{2} =\sqrt{6} (Cos(-90 )+iSin(-90 ))

z_{2} ^{2} =\sqrt{6} (0+i(-1))

z_{2} ^{2}=-\sqrt{6} i

Now, let´s find the value of z_{2} ^{20}

(z_{2} ^{2})^{10} =(-\sqrt{6} i)^{10} =7,776(i)^{4} (i)^{4} (i)^{2} =7,776(1)(1)(-1)=-7,776

therefore:

(2 - \sqrt{2} i)^{20} =-7,776

And then, we add them up

(2+\sqrt{2} i)^{20}+(2-\sqrt{2} i)^{20}=-7,776+(-7,776)=-15,552

So, yes, the result is an integer, -15,552

7 0
3 years ago
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