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3 years ago

3. Tom, Sam and Matt are counting drum beats.

Mathematics

1 answer:

just olya [345]3 years ago

6 0

Answer:

<em>When 60 beats are heard, Tom hits 15 snare drums, Sam hits 6 kettle drums, and Matt hits 5 bass drums.</em>

Step-by-step explanation:

The Least Common Multiple ( LCM )

The LCM of two integers a,b is the smallest positive integer that is evenly divisible by both a and b.

For example:

LCM(20,8)=40

LCM(35,18)=630

Since Tom, Sam, and Matt are counting drum beats at their own frequency, we must find the least common multiple of all their beats frequency.

Find the LCM of 4,10,12. Follow this procedure:

List prime factorization of all the numbers:

4 = 2*2

10 = 2*5

12 = 2*2*3

Multiply all the factors the greatest times they occur:

LCM=2*2*3*5=60

Thus, when 60 beats are heard, Tom hits 15 snare drums, Sam hits 6 kettle drums, and Matt hits 5 bass drums.

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2) How many sides does a regular polygon have if the interior angles each measure 156 degrees ?

spayn [35]

Answer:

B; 15 sides

Step-by-step explanation:

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3 years ago

What facts are required in order to conclude that m∠ADE = m∠ABC?

mariarad [96]

Answer:

The correct option is;

DE = 2·(BC), AD = 2·(AB), and AE = 2·(AC)

Step-by-step explanation:

Given that we have;

1) The side AD of the angle m∠ADE corresponds to the side AB of the angle m∠ABC

2) The side DE of the angle m∠ADE corresponds to the side BC of the angle m∠ABC

3) The side AE of the angle m∠ADE corresponds to the side AC of the angle m∠ABC

Then when we have DE = 2·(BC), AD = 2·(AB), and AE = 2·(AC), we have by sin rule;

AE/(sin(m∠ADE)) = 2·(AC)/(sin(m∠ABC)) = AE/(sin(m∠ABC))

∴ (sin(m∠ADE)) = (sin(m∠ABC))

m∠ADE) = m∠ABC).

3 0

3 years ago

Please help!!!!!!!!!!!!!!!!!!!!

Valentin [98]

Answer:

I am not sure but think that the answer is 1/20

7 0

3 years ago

Read 2 more answers

If mLABC = (2x + 4)º, m_DEF= (3x – 5)°, and ABC and

Harrizon [31]

Answer:

-5x-1

Step-by-step explanation:

4 0

3 years ago

Two problems here I need solved! I need every step, so please have that with your answers!!

Free_Kalibri [48]

QUESTION 1

We want to solve,

$\frac{1}{(x-4)}+\frac{x}{(x-2)}=\frac{2}{x^{2}-6x+8}$

We factor the denominator of the fraction on the right hand side to get,

$\frac{1}{(x-4)}+\frac{x}{(x-2)}=\frac{2}{x^{2}-4x - 2x+8}.$

This implies
$\frac{1}{(x-4)}+\frac{x}{(x-2)}=\frac{2}{x(x-4) - 2(x - 4)}.$

$\frac{1}{(x-4)}+\frac{x}{(x-2)}=\frac{2}{(x-4)(x - 2)}$

We multiply through by LCM of
$(x-4)(x - 2)$

$(x - 2) + x(x-4) = 2$

We expand to get,

$x - 2 + {x}^{2} - 4x= 2$

We group like terms and equate everything to zero,

${x}^{2} + x - 4x - 2 - 2 = 0$

We split the middle term,

${x}^{2} + - 3x - 4 = 0$

We factor to get,

${x}^{2} + x - 4x- 4 = 0$

$x(x + 1) - 4(x + 1) = 0$

$(x + 1)(x - 4) = 0$

$x + 1 = 0 \: or \: x - 4 = 0$

$x = - 1 \: or \: x = 4$

But
$x = 4$
is not in the domain of the given equation.

It is an extraneous solution.

$\therefore \: x = - 1$
is the only solution.

QUESTION 2

$\sqrt{x+11} -x=-1$

We add x to both sides,

$\sqrt{x+11} =x-1$

We square both sides,

$x + 11 = (x - 1)^{2}$

We expand to get,

$x + 11 = {x}^{2} - 2x + 1$

This implies,

${x}^{2} - 3x - 10 = 0$

We solve this quadratic equation by factorization,

${x}^{2} - 5x + 2x - 10 = 0$

$x(x - 5) + 2(x - 5) = 0$

$(x + 2)(x - 5) = 0$

$x + 2 = 0 \: or \: x - 5 = 0$

$x = - 2 \: or \: x = 5$

But
$x = - 2$
is an extraneous solution

$\therefore \: x = 5$

7 0

4 years ago