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2 years ago

Write the equation in standard form for a circle that has a diameter with endpoints (8,0) and (-8,0)

Mathematics

1 answer:

riadik2000 [5.3K]2 years ago

6 0

Answer:

x²+y² = 64

Step-by-step explanation:

The standard form of writing the equation of a circle is expressed as;

(x-a)²+(y-b)² = r² where;

(a, b) is the centre of the circle

r is the radius of the circle

Given the diameter of the circle with endpoints (8,0) and (-8, 0)

d = √(x2-x1)²+(y2-y1)²

d = √(-8-8)²+(0-0)²

d = √(-16)²

d = √256

d = 16

radius r = 16/2

r = 8

The centre of the circle will be the midpoint of the coordinates

M = (8-8/2, 0+0/2)

M = (0/2, 0/2)

M = (0,0)

Hence the centre is at (0,0)

Get the required equation

Recall that (x-a)²+(y-b)² = r²

Substitute the centre and the radius

(x-0)²+(y-0)² = 8²

x²+y² = 64

This gives the required equation

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SIZIF [17.4K]

Answer:

an exponential function

Step-by-step explanation:

An exponential function has the characteristic that it increases by a constant factor when the independent variable increases by a constant increment. The function you describe is an exponential function.

It can be written in the form ...

f(t) = f(0)·2^(t/d) . . . . . where d is the doubling time

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3 years ago

Findℒ{f(t)}by first using a trigonometric identity. (Write your answer as a function of s.)f(t) = 12 cost −π6

allsm [11]

Answer:

$L(f(t)) = \dfrac{6}{S^2+1} [\sqrt{3} \ S +1 ]$

Step-by-step explanation:

Given that:

$f(t) = 12 cos (t- \dfrac{\pi}{6})$

recall that:

cos (A-B) = cos AcosB + sin A sin B

∴

$f(t) = 12 [cos\ t \ cos \dfrac{\pi}{6}+ sin \ t \ sin \dfrac{\pi}{6}]$

$f(t) = 12 [cos \ t \ \dfrac{3}{2}+ sin \ t \ sin \dfrac{1}{2}]$

$f(t) = 6 \sqrt{3} \ cos \ (t) + 6 \ sin \ (t)$

$L(f(t)) = L ( 6 \sqrt{3} \ cos \ (t) + 6 \ sin \ (t) ]$

$L(f(t)) = 6 \sqrt{3} \ L [cos \ (t) ] + 6\ L [ sin \ (t) ]$

$L(f(t)) = 6 \sqrt{3} \dfrac{S}{S^2 + 1^2}+ 6 \dfrac{1}{S^2 +1^2}$

$L(f(t)) = \dfrac{6 \sqrt{3} +6 }{S^2+1}$

$L(f(t)) = \dfrac{6( \sqrt{3} \ S +1 }{S^2+1}$

$L(f(t)) = \dfrac{6}{S^2+1} [\sqrt{3} \ S +1 ]$

7 0

2 years ago

Someone please help me

Maurinko [17]

The correct answer would be 13:4. 
When reducing, you can divide both 600 and 1950 by 150. 1950/ 150= 13 and 600/150=4.

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2 years ago

Please help! (Numbers 1 and 2)

Svetradugi [14.3K]

1. is 8
2. is 4

Subtract them by the percentage

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2 years ago

SOMEBODY HELPPP PLEASEEEE. I don’t understand

gulaghasi [49]

Answer:

a) 18

b) 10

c) x^2 - 4x + 5

d) 4x^2 + 2

Step-by-step explanation:

f(x) = 4x - 2

g(x) = x^2 + 1

a) f(5) = 4(5) - 2 = 20 - 2 = 18.

b) g(-3) = (-3)^2 + 1 = 9 + 1 = 10.

c) g(x - 2)...

g(x - 2) = (x - 2)^2 + 1 = x^2 - 2x - 2x + 4 + 1 = x^2 - 4x + 5.

d) f(g(x)) = f(x^2 + 1) = 4(x^2 + 1) - 2 = 4x^2 + 4 - 2 = 4x^2 + 2.

Hope this helps!

8 0

3 years ago