Can anyone help me with this question please:

Mathematics

1 answer:

ladessa [460]3 years ago

3 0

The problem is asking us to isolate B. The given equation is solved for P, and we need to rearrange it for B.

First we need to square both sides. This will cancel out the square root on the right side.

P^2 = E + A^2/B^2

Next, subtract E from both sides.

P^2 - E = A^2/B^2

Next we need to get the B^2 out of the denominator. Multiply both sides by B^2.

B^2(P^2 - E) = A^2

Next divide both sides by (P^2 - E).

B^2 = A^2/(P^2 - E)

Lastly, take the square root of both sides.

B = sqrt(A^2/(P^2 - E))

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What is the equation of a circle with center (−8, 3) and radius 8?

Genrish500 [490]

Answer:

x^2 + y^2 + 16x + 6y + 9 = 0

Step-by-step explanation:

Using the formula for equation of a circle

(x - a)^2 + (y + b)^2 = r^2

(a, b) - the center

r - radius of the circle

Inserting the values given in the question

(-8,3) and r = 8

a - -8

b - 3

r - 8

[ x -(-8)]^2 + (y+3)^2 = 8^2

(x + 8)^2 + (y + 3)^2 = 8^2

Solving the brackets

( x + 8)(x + 8) + (y +3)(y+3) = 64

x^2 + 16x + 64 + y^2 + 6y + 9 = 64

Rearranging algebrally,.

x^2 + y^2 + 16x + 6y + 9+64 - 64 = 0

Bringing in 64, thereby changing the + sign to -

Therefore, the equation of the circle =

x^2 + y^2 + 16x + 6y + 9 = 0

7 0

3 years ago

At what point on the curve y = 19 2x is the tangent line perpendicular to the line 20x 4y = 1?

Gennadij [26K]

The points are; (7/2, -1/2).

<h3>What is the given point?</h3>

Now the tangent line is given as;20x 4y = 1. When we rewrite it in the slope intercept form, we have the equation as; y = 1 - 20x/4 or y = 1/4 - 5x.

Then to obtain the slope of the curve we have; y = 19 - 2x

dy/dx = 2

Using the relation;

m1m2 = -1

m2 = -1/2

Hence;

y = 1/4 - 5(-1/2)

y = 1/4 + 5/2

y = 7/2

Thus the points are; (7/2, -1/2)

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