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4 years ago

6

Please please please answer my question! It's the one before this one.

2 answers:

Alina [70]4 years ago

6 0

Answer:

Ok..

Step-by-step explanation:

tia_tia [17]4 years ago

3 0

Oh sure thing.........

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Calculate an estimate of a square root of 119 + 120i

uysha [10]

Answer:

Step-by-step explanation:

Given

z=119+120 i

Let $\sqrt{119+120 i}=p+iq$

Squaring both sides

$119+120 i=p^2-q^2+2ipq$

Comparing real and imaginary part

Re(LHS)=Re(RHS)

$119=p^2-q^2$ -----------1

comparing Im(LHS)=Im(RHS)

120=2pq

$q=\frac{60}{p}$

Substitute q in 1

$119=p^2-(\frac{60}{p})^2$

$p^4-119p^2-(68)^2=0$

Let $x=p^2$

$x^2-119x-4624=0$

$x=frac{119\pm \sqrt{119^2+4\times 4624}}{2}$

$x=\frac{119\pm 180.71}{2}$

we take only Positive value because $p^2=x$

x=149.85

$p^2=149.85$

thus $p=\pm 12.24$

$q=\mp 4.90$

thus $\sqrt{119+120 i}=\pm (12.24+i 4.90)$

6 0

4 years ago

If triangle XYZ’s points are X (-2, 1), Y (-4,-3), and Z (0,-2) what are the new coordinates of X’, Y’, and Z’ if the triangle i

8090 [49]

Answer:

The coordinates of image are X'(-3,4), Y'(-5,0) and Z'(-1,1).

Step-by-step explanation:

The vertices of triangle XYZ are X(-2,1), Y(-4,-3) and Z(0,-2).

It is given that the triangle translated by rule

$(x,y)\rightarrow (x-1,y+3)$

The coordinates of image of XYZ are

$X(-2,1)\rightarrow X'(-2-1,1+3)$

$X(-2,1)\rightarrow X'(-3,4)$

$Y(-4,-3)\rightarrow Y'(-4-1,-3+3)$

$Y(-4,-3)\rightarrow Y'(-5,0)$

$Z(0,-2)\rightarrow Z'(0-1,-2+3)$

$Z(0,-2)\rightarrow Z'(-1,1)$

Therefore coordinates of image are X'(-3,4), Y'(-5,0) and Z'(-1,1).

5 0

3 years ago

What is the solution to the equations x^2-2x+y=8 and x-y=-2

kherson [118]

Hello!!

For the first equation:

$-2x + x^2 + y = 8$ (Solve for y)

$y = -x^2 + 2x +8$

For the second equation:

x - y = -2 (Solve for y)

-y = -x - 2

-y = x + 2

Good luck :)

6 0

4 years ago

I know how to work out all but the third row can you help

labwork [276]

Here I have attached a photo to help you with this:

4 0

4 years ago

Find the 4th term of the expansion<br><br> A<br> B<br> C<br> D

Marysya12 [62]

Answer:

option-A

Step-by-step explanation:

We can use rth term binomial expansion formula

$(x+y)^n$

$T_r=(n,r-1)x^{n-(r-1)}y^{r-1}$

we are given

$(a-\sqrt{2})^8$

we can compare

$x=a,y=-\sqrt{2}$

n=8

r=4

now, we can find 4th term

$T_r=(8,4-1)a^{8-(4-1)}(-\sqrt{2})^{4-1}$

now, we can simplify it

$T_4=\frac{8!}{5!3!}\times -2\sqrt{2}a^5$

$T_4=56\times -2\sqrt{2}a^5$

$T_4=-112\sqrt{2}a^5$

5 0

4 years ago

Read 2 more answers