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Sunny_sXe [5.5K]

2 years ago

11

Question 3 (Multiples and Factors] Three numbers are given below. Use prime factorisation to determine the HCF and LCM 1848 132

462

1 answer:

Ilia_Sergeevich [38]2 years ago

4 0

Prime factorization involves rewriting numbers as products

The HCF and the LCM of 1848, 132 and 462 are 66 and 1848 respectively

<h3>How to determine the HCF</h3>

The numbers are given as: 1848, 132 and 462

Using prime factorization, the numbers can be rewritten as:

$1848 = 2^3 * 3 * 7 * 11$

$132 = 2^2 * 3 * 11$

$462 = 2 * 3 * 7 * 11$

The HCF is the product of the highest factors

So, the HCF is:

$HCF = 2 * 3 * 11$

$HCF = 66$

<h3>How to determine the LCM</h3>

In (a), we have:

$1848 = 2^3 * 3 * 7 * 11$

$132 = 2^2 * 3 * 11$

$462 = 2 * 3 * 7 * 11$

So, the LCM is:

$LCM = 2^3 * 3 * 7 * 11$

$LCM = 1848$

Hence, the HCF and the LCM of 1848, 132 and 462 are 66 and 1848 respectively

Read more about prime factorization at:

brainly.com/question/9523814

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Help !! Please I can’t find the answer

SVETLANKA909090 [29]

Answer:

$\large\boxed{r^2=(x+5)^2+(y-4)^2}$

Step-by-step explanation:

The equation of a circle:

$(x-h)^2+(y-k)^2=r^2$

<em>(h, k)</em><em> - center</em>

<em>r</em><em> - radius</em>

<em />

We have diameter endpoints.

Half the length of the diameter is the length of the radius.

The center of the diameter is the center of the circle.

The formula of a distance between two points:

$d=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}$

Substitute the coordinates of the given points (-8, 2) and (-2, 6):

$d=\sqrt{(6-2)^2+(-2-(-8))^2}=\sqrt{4^2+6^2}=\sqrt{16+36}=\sqrt{52}$

The radius:

$r=\dfrac{d}{2}\to r=\dfrac{\sqrt{52}}{2}$

The formula of a midpoint:

$\left(\dfrac{x_1+x_2}{2},\ \dfrac{y_1+y_2}{2}\right)$

Substitute:

$x=\dfrac{-8+(-2)}{2}=\dfrac{-10}{2}=-5\\\\y=\dfrac{2+6}{2}=\dfrac{8}{2}=4$

$(-5,\ 4)\to h=-5,\ k=4$

Finally:

$(x-(-5))^2+(y-4)^2=\left(\dfrac{\sqrt{52}}{2}\right)^2\\\\(x+5)^2+(y-4)^2=\dfrac{52}{4}\\\\(x+5)^2+(y-4)^2=13$

5 0

3 years ago

A coin is tossed and a single 6 sided number cube is rolled. Find the probability of landing on the heads side of the coin and r

Delicious77 [7]

Answer:

The probability of landing on the heads side of the coin and rolling 3 on the number cube is 0.083.

Step-by-step explanation:

Given:

Coin is tossed and a single 6 sided number cube is rolled.

Let the event of tossing head be 'H' and event of rolling a 3 be 'R3'.

Now, we know that:

Probability of an event 'A' = Favorable outcomes of 'A' ÷ Total possible outcomes

Now, for tossing a coin, the total possible outcomes are head and tail. So, there are 2 possible outcomes.

So, probability of event 'H' is given as:

$P(H)=\dfrac{n(H)}{n(S)}\\\\P(H)=\frac{1}{2}=0.5$

Similarly, for rolling a 6 sided cube, the possible outcomes are numbers 1 to 6. So, the number of possible outcomes is, $n(S)=6$

Now, probability of rolling a 3 is given as:

$P(R3)=\frac{n(R3)}{n(S)}\\\\P(R3)=\frac{1}{6}$

Now, both the events 'H' and 'R3' occur together. So, the combined probability is the product of two individual probabilities as they are independent events. So,

$P(H\ and\ R3)=P(H)\times P(R3)\\\\P(H\ and\ R3)=\frac{1}{2}\times \frac{1}{6}\\\\P(H\ and\ R3)=\frac{1}{12}=0.083$

Note: Independent events are those events whose intersection is an empty set of events or the outcome of one event doesn't affect the outcome of the other event.

Therefore, the probability of landing on the heads side of the coin and rolling 3 on the number cube is 0.083.

4 0

3 years ago

Type the correct answers out, thank you - 20 POINTS

nadezda [96]

AB and BC form a right angle at their point of intersection. This means AB is perpendicular to BC.

We are given the coordinates of points A and B, using which we can find the equation of the line for AB.

Slope of AB will be:

$m= \frac{-1-1}{14-2}=-1/6$

Using this slope and the point (2,1) we can write the equation for AB as:

$y-1= \frac{-1}{6}(x-2) \\ \\ 
y= -\frac{1}{6}x+ \frac{1}{3}+1 \\ \\ 
y=-\frac{1}{6}x+ \frac{4}{3} 
$

The above equation is in slope intercept form. Thus the y-intercept of AB is 4/3.

Slope of AB is -1/6, so slope of BC would be 6. Using the slope 6 and coordinates of the point B, we can write the equation of BC as:

y - 1 = 6(x - 2)
y = 6x - 12 + 1
y = 6x - 11

Point C lies on the line y = 6x - 11. So if the y-coordinate of C is 13, we can write:

13 = 6x - 11
24 = 6x
x = 4

The x-coordinate of point C will be 4.

Therefore, the answers in correct order are:

4/3 , 6, -11, 4

8 0

2 years ago

Some number was divided into 104.13. This quotient was multiplied by 4, after which the resulting product was added to 5. Given

SVEN [57.7K]

Answer:

The initial number is <u>-1218.321</u>.

Step-by-step explanation:

Let the initial number be 'x'.

The number 'x' is divided into 104.13.

So, we divide the number 'x' by 104.13. This gives,

$\dfrac{x}{104.13}$

Now, the quotient is multiplied by 4. So, this means we need to multiply 4 to the fraction above. This gives,

$\dfrac{x}{104.13}\times 4\\\\\frac{4x}{104.13}$

Now, 5 is added to the result. This gives,

$\frac{4x}{104.13}+5$

Now, as per question:

$\frac{4x}{104.13}+5=-41.8$

Now, solving for 'x', we add -5 both sides. This gives,

$\frac{4x}{104.13}+5-5=-41.8-5\\\\\frac{4x}{104.13}=-46.8\\\\4x=-46.8\times 104.13\\\\4x=-4873.284\\\\x=\frac{-4873.284}{4}=-1218.321$

Therefore, the initial number is -1218.321.

6 0

3 years ago

Each of three bags A, B, C contains white balls and black balls. A has a1 white & b1 black, B has a2 white & b2 black an

Alex73 [517]

Answer:

See explanation ( Answers are too long)

Step-by-step explanation:

We will first compute a general probability for picking a white ball:

P (W) = a_1 / (a_1 + b_1) + a_2 / (a_2 + b_2) + a_3 / (a_3 + b_3)

part a)

We are asked to find the probability of white ball given that it pulled from bag A. So if we express it in notation we are asked for P ( A / W). We will use conditional probability to answer our question:

P ( A / W ) = P ( W & A ) / P (W)

P ( W & A ) = a_1 / (a_1 + b_1)

Hence,

P ( A / W ) = [a_1 / (a_1 + b_1)] / [a_1 / (a_1 + b_1) + a_2 / (a_2 + b_2) + a_3 / (a_3 + b_3)]

part b)

We are asked to find the probability of white ball given that it pulled from bag B. So if we express it in notation we are asked for P ( B / W). We will use conditional probability to answer our question:

P ( B / W ) = P ( W & B ) / P (W)

P ( W & B ) = a_2 / (a_2 + b_2)

Hence,

P ( A / W ) = [a_2 / (a_2 + b_2)] / [a_1 / (a_1 + b_1) + a_2 / (a_2 + b_2) + a_3 / (a_3 + b_3)]

part c)

We are asked to find the probability of white ball given that it pulled from bag C. So if we express it in notation we are asked for P ( C / W). We will use conditional probability to answer our question:

P ( C / W ) = P ( W & C ) / P (W)

P ( W & C ) = a_3 / (a_3 + b_3)

Hence,

P ( A / W ) = [a_3 / (a_3 + b_3)] / [a_1 / (a_1 + b_1) + a_2 / (a_2 + b_2) + a_3 / (a_3 + b_3)]

6 0

3 years ago