# Find lcm and gcd of the following pairs of integers and verify that lcm is gcd = product of the two (2023)

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Sol. (1) Prime factors of 26 = 2 x 13
Prime factors of 91 = 7 x 13
Therefore HCF = common factors between 26 and 91 = 13 and LCM = 13x2x7 = 182

Now product of numbers 26 and 91
= 26 x 91 = 2366 en product van HCF en LCM = 13 x 182 = 2366

So check if the product of two numbers is the product of HCF and LCM.

(2) Prime factors of 510 = 2 x 3 x 5 x 17
Prime factors of 92 = 2 x 2 x 23
Therefore HCF = 2 and LCM = 2 × 2 × 3 × 5 × 17 × 23 = 23460

Now the product of the numbers 510 and 92 = 46920 and the product of HCF and LCM = 2 x 23460 = 46920

So you verified that the product of two numbers with the number 18 is equal to the product of their HCF and their LCM.

(3) Priemfactoren van 336 = 2 x 2 x 2 x 2 x 3 x 7

Prime factors of 54 = 2 x 3 x 3 x 3
Hence HCF (product of the common factors of 336 and 54)
=2x3=6

And LCM (product of all common factors with the other factors)
=(2x3)x2x2x2x3x3x7=3024

Now the product of the numbers 336 and 54 = 336 x 54 = 18144
en product van HCF en LCM = 6 x 3024 = 18144
Hence product of two numbers: product of HCF and LCM.

## Related Questions

Four bells toll at 8, 12, 15 and 18 seconds apart. The four bells begin to ring simultaneously. How many fit together in an hour without counting the start time?

The nesting time when they play together is indicated by the LCM of 8,12,15 and 18

8= 2*2*2
12=2*2*3
15 = 3*5
18=2*3*3

mm = 2*2*2*3*3*5 = 360

So they sound together every 360 seconds

So in an hour they go to the toll 3600/360 = 10 times the answer

Sol. LCM go 8,12,15 in 18 = 360

So see every 360 seconds. Four bells ring at once every 6 minutes, so keep going

1 hour that in 60 minutes these four bells ring simultaneously =
= 10 Mal

Explain why 7 × 11 × 13 +13 and 7 × 6 × 5 × 4 × 3 × 2 × 1 × 5 are composite numbers.

G. We have 7 x 11 x 13 + 13 = 1001 + 13 = 1014
1014 = 2 x 3 x 13 x 13

So it is the product of more than two prime numbers. 2, 3 and 13.

So it is a composite number. 7x6x5x4x3x2x1x3x2x1 + 5 = 5040+5=5045
5045 = 5x1009

It is the product of prime factors 5 and 1009.

So it is a composite number.

11∗13∗15∗17+17=17(11∗13∗15∗1+1)
11

13

15

17
+
17
=
17
(
11

13

15

1
+
1
)

Non, 11∗13∗15∗1
11

13

15

1
is a product of odd numbers that is always odd.

So we have 17∗(odd+1)
17

(
O
D
D
+
1
)

= impar∗(impar+impar)
=
O
D
D

(
O
D
D
+
O
D
D
)

= oneven∗par
=
O
D
D

mi
v
mi
norte

= par
=
mi
v
mi
norte

Now 2 is the only prime number that is even. Therefore, a given number cannot be prime and is therefore composite

Assume that the z-scores are normally distributed with a mean of 0 and a standard deviation of 1. If p(z > c. = 0.1093, find
C.

Provided that μ=0 and σ=1, P(z>c)=0.1093
therefore
P(z

THE SMARTEST ANSWER!!!! A number is three times as large as another number. The sum of two numbers is 48. Find the two numbers. Show your work.

To leave
x -------> largest number
y ------> secondary number

we know
x=3y -------> equation 1
x+y=48 ----> equation 2
Replace 1 with 2

[3d]+d=48------> 4d=48-----> d=48/4-----> d=12
x=3j----> x=3*12-----> x=36

the two numbers are 36 and 12

Find the largest number that divides 224,250 and 302, leaving a remainder of 3 in each.

To get a remainder of 3, we first subtract 3 from each number.

wear 3:
224 - 3 = 221
250 - 3 = 247
302 - 3 = 299

Prime factorization of each of the numbers:
221 = 13 x 17
247 = 13x19
299 = 13x23

Find HCF:
HCF = 13

Answer: The largest number that can be divided is 13.

Number Required = HCF of (224-3), (250-3) and (302-3)
The number required is HCF of 221, 247 and 299. Number required = HCF of 221 and 247 then HCF of (221 and 247) and 299.

Step (1) HCF of 221 and 247 = 13 [using Euclid's division algorithm]

Step (2) HCF of 13 and 299 = 13 [using Euclid's division algorithm]

Therefore, the HCF of 221, 247, and 299 is 13. Therefore, the required number is 13.

Find the largest number by dividing 628.3129 and 15630 to leave memories 3,4 and 5 respectively.

number required
= HCF of (628 3), (3129 -4), and (15630 5), which is the number required
= HCF van 625, 3125, 15625

To find HCF of 625, 3125 and 15625, we first find HCF of 625 and 3125 and then HCF of HCF of (625 and 3125) and 15625.

Step (1): HCF of 625 and 3125 using Euclid's division algorithm is 3125 = 625x5+0
=:> HCF of 625 and 3125 is 625.

Step (2): The HCF of 625 and 15625 is now: 15625 = 625 x 25 + O
=:> HCF of 625 and 15625 is 625. . Therefore, the HCF of 625, 3125, and 15625 is 625. Therefore, the required number is 625.

Prove using Euclid's division algorithm that: 847.2160 are relatively prime

Definition of EO primes or relative primes: Two numbers are called coprimes or relative primes if their HCF is 1. Therefore, to prove that 847 and 2160 are coprimes, we find their HCF and what 1 must be

The new steps to find HCF are as follows

2160 = 847 x 2+ 466
847 = 466 × 1 +381
466 = 381x1 + 85
381 =85 x 4+ 41
85 =41x2+3
41 =3 x 13+ 2
3 =2x1+1
2 = 1x2+0

Hence HCF=1. Therefore, the numbers are relatively prime (relatively prime).

How much does it cost to deposit into a bank account earning 12% compound interest per year to have \$4,614 after 18 years?

A = P*(1 +r)^t
4614 = P*1,12^18
4614/7,689966 = 600,00

To have \$4,614 after 18 years, a \$600 deposit is required.

For the solid described below, determine the dimensions that result in the minimum surface area with a volume of 64 cm3. a closed cylinder with radius r cm and height h cm.

The contents of the cylinder are:
V = pi * r ^ 2 * h = 64
The area averages:
A = 2 * pi * r^2 + 2 * pi * r * h
We write the area as a function of r:
A(r) = 2 * pi * r ^ 2 + 2 * pi * r * (64 / (pi * r ^ 2))
rewrite:
A(r) = 2 * pi * r ^ 2 + 2 * (64 / r)
A(r) = 2 * pi * r ^ 2 + 128 / r
derivative:
A'(r) = 4*pi*r - 128/r^2
We set it equal to zero and solve for r:
0 = 4 * pi * r - 128 / r^2
128 / r^2 = 4 * pi * r
r ^ 3 = 128 / (4 * pi)
r = (128 / (4 * pi)) ^ (1/3)
r = 2,17cm
The height is:
h = 64 / (pi * r ^ 2)
h = 64 / (pi * (2.17) ^ 2)
Height = 4.33 cm
The dimensions that make up the minimum surface are:
r = 2,17cm
Height = 4.33 cm

ASAP The sixth grade is organizing a canning campaign. The number of doses collected each day was 45, 21, 3, 15, 32, 97, 68, 27, 29, and 52. What is the interquartile range of doses collected each day?
Find the range of the lower quartile, upper quartile, and interquartile:

Upper quartile = 52

Inter-Kwartiel area =31

Step-by-step explanation:

The given dates: 45, 21, 3, 15, 32, 97, 68, 27, 29, 52.

put in order,

3, 15, 21, 27, 29, 32, 45, 52, 68, 97

Bottom half = 3, 15, 21, 27, 29

Top half = 32, 45, 52, 68, 97

Lower quartile = median of the lower half of the data

[The median is the medium term.]

Top quartile = median of the bottom half of the data

[The median is the medium term.]

Interkwartiel area =

Bottom quartile - 21st
Oberes Quartile-52
Inter-Kwartiel area – 31

I hope it helps!!

Si (2 − 3i) + (x + yi) = 6, ¿cuál es x + yi? 4 + 3i 4 − 3i -4 − 3i -4 + 3i 4x + 3i

If (2 − 3i) + (x + yi) = 6, we subtract (2-3i) from both sides of the equation to get x + yi.

(2 − 3i) + (x + yi) - (2 − 3i)= 6 - (2 − 3i)
(x + yi) = 6 - (2 - 3i)
Open the bracket and note the characters
(x + yi) = 6 - (2 - 3i)
= 6 - 2 + 3i
= 4 + 3i

If a simple machine reduces the intensity of a force, what needs to be increased? A. The speed of input force.
B. The size of the simple machine
C. The distance over which the force is applied.
D. The work done by the simple machine
OMG, just realized this is the wrong topic for math guys, but you can still comment if you want :D

I think the answer is C. The distance over which the force is applied. I hope it helps!

What is the standard form of (2-4i)(3+5i) divided by (3+i)

First we need to simplify the numerator of the fraction

(2 - 4i)(3 + 5i)
= 6 +10i - 12i - 20i²
= 6 - 2i + 20
= 26 - 2i

The fraction now becomes:

To convert the fraction to its standard form, we need to multiply and divide the fraction by the conjugate of the denominator, as shown below:

What is the best estimate for the total weight of these sausages? 1 7/8 pounds bologna, 1 1/2 pounds ham, and 7/8 pounds roast beef

Well, it just seems like the sum of all shapes and sizes. Now let's first convert the mixed numbers to "improper" and then add them.

If the median score in this pop quiz is more than 170, the teacher will cancel the next pop quiz. Pop quiz results followed this distribution: {180, 175, 163, 186, 153, 194, 198, 183, 187, 174, 177, 196, 162, 185, 174, 195, 164, 152, 144, 138, 125, 110}. What is the median of the distribution? 168.86 170.78 183.23 174.50 180.56

Arr sorts the 22 scores in ascending order. The median value is the two "middle" values, i.e. the average value of the 11th and 12th points.

The middle two values ​​are 174 and 175, so the median is 174.50.

What is the equation of the line passing through (1,3) and (-2,-3)?

m = y2-y1/x2-x1 -> -3-3/-2-1 = -6/-3 -> +2

y=mx+b of 3=2 × 1+b -> b = 1 (y-intercept)

y = 2x+1

34 HELP POINTS CAN YOU HELP ME SOLUTION AND SHOW HOW YOU DONE IT? 1.4X+5=6X

2. 4X+2(X+2)=4(X-4)-10

SOLVE FOR AND
3. 2X+3Y=12

4.X=3Y+4

SOLUTION FOR X

5. 2X^2+5=25

Solve for R

6. I=PRT

1. Let's solve your equation step by step.
4x+5=6x
Step 1: Subtract 6x from both sides.
4x+5−6x=6x−6x
−2x+5=0
Step 2: Subtract 5 from both sides.
−2x+5−5=0−5
−2x=−5
Step 3: Divide both sides by -2.
−2x/−2=−5/−2

What is the maximum error when measuring to the nearest quarter inch? 1/8 in. 1/4 in. 1/2 inch.

The maximum possible error is half the unit of measurement you are using.

Multiply the unit used by 1/2:

The largest possible error is 1/8 inch.

The furthest distance from the correct 1/4 inch mark is
1/8 inch.

What are the possible values ​​of x and y for two different points, (5, -2) and (x, y) to represent a function?

The possible values ​​of x and y can be:

x can be any value except 5

and y can be any value

Step-by-step explanation:

A function f is a mapping from a non-empty set A to B if each element of A maps to a unique element of B.

Here we get the following:

f(5)= -2

for two different points such that f(x)=y

x can be any value except 5

(since 5 has an image of -2 and for a function each element's image is unique)

and y can be any value

The possible values ​​of x and y can be:

x can be any value except 5 and y can be any value

X can be any value except 5.
and can have any value.

What is the equation of the line containing points (5,2), (10,4) and (15,6)?

All y/x values ​​have the same ratio, 2/5. the line is

y = (2/5)x

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