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slavikrds [6]
3 years ago
14

12x + 2y = 6A. solve for yB. What is the slope and what is the y-intercept?​

Mathematics
1 answer:
VashaNatasha [74]3 years ago
8 0

Answer:

y=3

B.Slope is -6 Y intercept is 0.5

Step-by-step explanation:

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Find the modulus of -2 - 6i.<br><br> 40<br> -40<br> 6.3<br> -6.3
olga2289 [7]
The modulus of complex number can be computed with this equation:
|z|=sqrt(Re(z)^2+Im(z)^2)
z = -2-6i
Re(z)=-2, so Re(z)^2=4
Im(z)=-6, so Im(z)^2=36
|z|=sqrt(36+4)=sqrt(40)=6.3
6 0
3 years ago
Use Newton’s Method to find the solution to x^3+1=2x+3 use x_1=2 and find x_4 accurate to six decimal places. Hint use x^3-2x-2=
luda_lava [24]

Let f(x) = x^3 - 2x - 2. Then differentiating, we get

f'(x) = 3x^2 - 2

We approximate f(x) at x_1=2 with the tangent line,

f(x) \approx f(x_1) + f'(x_1) (x - x_1) = 10x - 18

The x-intercept for this approximation will be our next approximation for the root,

10x - 18 = 0 \implies x_2 = \dfrac95

Repeat this process. Approximate f(x) at x_2 = \frac95.

f(x) \approx f(x_2) + f'(x_2) (x-x_2) = \dfrac{193}{25}x - \dfrac{1708}{125}

Then

\dfrac{193}{25}x - \dfrac{1708}{125} = 0 \implies x_3 = \dfrac{1708}{965}

Once more. Approximate f(x) at x_3.

f(x) \approx f(x_3) + f'(x_3) (x - x_3) = \dfrac{6,889,342}{931,225}x - \dfrac{11,762,638,074}{898,632,125}

Then

\dfrac{6,889,342}{931,225}x - \dfrac{11,762,638,074}{898,632,125} = 0 \\\\ \implies x_4 = \dfrac{5,881,319,037}{3,324,107,515} \approx 1.769292663 \approx \boxed{1.769293}

Compare this to the actual root of f(x), which is approximately <u>1.76929</u>2354, matching up to the first 5 digits after the decimal place.

4 0
2 years ago
Lines c and d are perpendicular. if the slope of line c is -4, what is the slope of line d?
torisob [31]
The slope of line d is 1/4.
8 0
3 years ago
(100 points)
LekaFEV [45]

Answer:

30/10 and 5/11

Step-by-step explanation:

8 0
3 years ago
I don’t know the answer
shutvik [7]

Answer:

HJ is 8

JE is 4

Step-by-step explanation:

Nice song in the other tab by the way:)

7 0
3 years ago
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