2 years ago

Can you help me with my math

Mathematics

1 answer:

romanna [79]2 years ago

5 0

I can help if you want me to

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There are two coins in a bag. for coin 1, p(h) = ½ and for coin 2, p(h) = 1/3. your friend chooses one of the coins at random an

lana [24]

The answer is 3 to this question

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3 years ago

Solve 3x+k= c for x.

Debora [2.8K]

Answer:

x = $\frac{c-k}{3}$

Step-by-step explanation:

Given

3x + k = c ( isolate the term 3x by subtracting k from both sides )

3x = c - k ( divide both sides by 3 )

x = $\frac{c-k}{3}$

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2 years ago

Let $r$ and $s$ be the roots of $3x^2 + 4x + 12 = 0.$ Find $r^2 + s^2.$ Pls help.

nirvana33 [79]

Answer:

-56/9

Step-by-step explanation:

By Vieta's formulas,

$r + s = -\frac{4}{3}$ and $rs = \frac{12}{3} = 4.$ Squaring the equation $r + s = -\frac{4}{3},$ we get

$r^2 + 2rs + s^2 = \frac{16}{9}.$ Therefore,

$r^2 + s^2 = \frac{16}{9} - 2rs = \frac{16}{9} - 2 \cdot 4 = -\frac{56}{9}}$

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2 years ago

The answer to the problem

kotegsom [21]

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3 years ago

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Determine the amplitude or period as requested. Period of a y = - 1/3 * sin x - 1/3 pi/3 1/3 d 2pi

zysi [14]

Answer:

$Period = 2\pi$

Step-by-step explanation:

Given

$y = -\frac{1}{3} * \sin(x - \frac{1}{3})$

Required

Determine the period

A sine function is represented as:

$y =A \sin(Bx + C) + D$

Where

$Period = \frac{2\pi}{B}$

By comparing:

$y =A \sin(Bx + C) + D$ and $y = -\frac{1}{3} * \sin(x - \frac{1}{3})$

$Bx = x$

So:

$B = 1$

So, we have:

$Period = \frac{2\pi}{B}$

$Period = \frac{2\pi}{1}$

$Period = 2\pi$

Hence, the period of the function is $2\pi$

6 0

2 years ago