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2 years ago

Please help me out on this :(( struggling !

Mathematics

1 answer:

Yuki888 [10]2 years ago

3 0

Answer:

answer is B

Step-by-step explanation:

step by step explanation

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If the endpoints of the diameter of a circle are (−6, 6) and (6, −2), what is the standard form equation of the circle?

Sergeeva-Olga [200]

i Acehello :
let : A(-6,6) B(6,-2)
the center is w((-6+6)/2 , (6-2)/2)....(midel [ AB]
w(0 ,2)
the ridus is : r = AB/2
AB = √(-6-6)²+(6+2)² = √(144+64) =√208/2
an equation in standard form equation of the circle :
(x+0)²+(y-2)² = (√208/2)²=208/4 = 52

4 0

2 years ago

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The radius of a sphere is 12 units. Find the volume of the sphere. Write your answer in terms of pi.

stiv31 [10]

Answer:

V = 2304π units³

Step-by-step explanation:

The volume (V) of a sphere is calculated as

V = $\frac{4}{3}$ πr³ ( where r is the radius ) , then

V = $\frac{4}{3}$ π × 12³

= $\frac{4}{3}$ π × 1728

= 4π × 576

= 2304π units³

5 0

2 years ago

Write in simplest form 3m⋅4⋅n⋅n

Sedbober [7]

Hello from MrBillDoesMath!

Answer:

12 m n^2

Discussion:

3m * 4 * n * n = => 3* 4 = 12

12 m * n * n => as n*n = n^2

12 m n^2

Thank you,

MrB

7 0

3 years ago

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Pls I need help I will mark Brainly

Margarita [4]

Answer:

(10⁰x 10¹ x 1¹⁰)< (10⁰ + 10¹x 1¹⁰)<(10⁰ + 10¹ + 1¹⁰)< (10 x 10¹)

Step-by-step explanation:

increasing order-

(10⁰x 10¹ x 1¹⁰)

(10⁰ + 10¹x 1¹⁰)

(10⁰ + 10¹ + 1¹⁰)

(10 x 10¹)

7 0

3 years ago

1) Determine the discriminant of the 2nd degree equation below:

Aleksandr-060686 [28]

$\LARGE{ \boxed{ \mathbb{ \color{purple}{SOLUTION:}}}}$

We have, Discriminant formula for finding roots:

$\large{ \boxed{ \rm{x = \frac{ - b \pm \: \sqrt{ {b}^{2} - 4ac} }{2a} }}}$

Here,

x is the root of the equation.
a is the coefficient of x^2
b is the coefficient of x
c is the constant term

1) Given,

3x^2 - 2x - 1

Finding the discriminant,

➝ D = b^2 - 4ac

➝ D = (-2)^2 - 4 × 3 × (-1)

➝ D = 4 - (-12)

➝ D = 4 + 12

➝ D = 16

2) Solving by using Bhaskar formula,

❒ p(x) = x^2 + 5x + 6 = 0

$\large{ \rm{ \longrightarrow \: x = \dfrac{ - 5\pm \sqrt{( - 5) {}^{2} - 4 \times 1 \times 6 }} {2 \times 1}}}$

$\large{ \rm{ \longrightarrow \: x = \dfrac{ - 5 \pm \sqrt{25 - 24} }{2 \times 1} }}$

$\large{ \rm{ \longrightarrow \: x = \dfrac{ - 5 \pm 1}{2} }}$

So here,

$\large{\boxed{ \rm{ \longrightarrow \: x = - 2 \: or - 3}}}$

❒ p(x) = x^2 + 2x + 1 = 0

$\large{ \rm{ \longrightarrow \: x = \dfrac{ - 2 \pm \sqrt{ {2}^{2} - 4 \times 1 \times 1} }{2 \times 1} }}$

$\large{ \rm{ \longrightarrow \: x = \dfrac{ - 2 \pm \sqrt{4 - 4} }{2} }}$

$\large{ \rm{ \longrightarrow \: x = \dfrac{ - 2 \pm 0}{2} }}$

So here,

$\large{\boxed{ \rm{ \longrightarrow \: x = - 1 \: or \: - 1}}}$

❒ p(x) = x^2 - x - 20 = 0

$\large{ \rm{ \longrightarrow \: x = \dfrac{ - ( - 1) \pm \sqrt{( - 1) {}^{2} - 4 \times 1 \times ( - 20) } }{2 \times 1} }}$

$\large{ \rm{ \longrightarrow \: x = \dfrac{ 1 \pm \sqrt{1 + 80} }{2} }}$

$\large{ \rm{ \longrightarrow \: x = \dfrac{1 \pm 9}{2} }}$

So here,

$\large{\boxed{ \rm{ \longrightarrow \: x = 5 \: or \: - 4}}}$

❒ p(x) = x^2 - 3x - 4 = 0

$\large{ \rm{ \longrightarrow \: x = \dfrac{ - ( - 3) \pm \sqrt{( - 3) {}^{2} - 4 \times 1 \times ( - 4) } }{2 \times 1} }}$

$\large{ \rm{ \longrightarrow \: x = \dfrac{3 \pm \sqrt{9 + 16} }{2 \times 1} }}$

$\large{ \rm{ \longrightarrow \: x = \dfrac{3 \pm 5}{2} }}$

So here,

$\large{\boxed{ \rm{ \longrightarrow \: x = 4 \: or \: - 1}}}$

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5 0

2 years ago

Read 2 more answers