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-BARSIC- [3]
3 years ago
5

What is the least common multiple of 26 and 39?

Mathematics
1 answer:
GuDViN [60]3 years ago
5 0

Answer:

78

Step-by-step explanation:

Find the prime factorizations of the two numbers:

26 = 2 * 13

39 = 3 * 13

The LCM is the product of common and not common factors with the higher exponent.

LCM = 2 * 3 * 13 = 78

Answer: 78

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What is the explicit formula for this sequence?
Zarrin [17]

Step-by-step explanation:

The above sequence is a geometric sequence

For an nth term in a geometric sequence

a(n) = a ({r})^{n - 1}

where

n is the number of terms

a is the first term

r is the common ratio

From the question

a = 5

r = 10/5 = 2

Therefore the explicit formula for this sequence is

a(n) = 5( {2})^{n - 1}

Hope this helps you

8 0
3 years ago
If ay = 9 and an = -5an-1 + 5 then find the value of az.<br> Answer:<br> Submit Answer
Butoxors [25]
The answer would be 10 because 9+-5=4+-1=5+5=10
5 0
3 years ago
Can someone help with an explanation? This is finding Surface Area.​
Y_Kistochka [10]
The answer should be 95in^2

Explanation:

You’re essentially just finding the area of all the shapes on the triangle and adding them together.

Area of a triangle = 1/2•b•h
Area of a square = l•w or just s2

The area of the triangles given what we have in the diagram can be found with
1/2•5•7 = 17.5

There are 4 triangles so you multiply 17.5 by 4 which gives 70

The base of the triangle (the square) = 25

70+25=95
7 0
2 years ago
(a) Find the three-digit number that increases by 30% when each of its
Elden [556K]

Step-by-step explanation:

So a three digit number can be expressed as: 100a + 10b + c where a is the third digit, b is the second digit, and c is the first digit. Or in other words a is the hundreds place, b is the tens place, and c is the ones place. When something "increases" by 30%, it is 130% it's original value, and to calculate how much that is, you simply convert 130% to a decimal by dividing by 100, which gives you 1.30. And since all the digits are increased by 1, you have the equation:

100(a+1) + 10(b+1) + 1(c+1) = 1.30(100a+10b+c)

Distribute the multiplication on the left side:

100a+100+10b+10+c+1=1.30(100a+10b+c)

Distribute the multiplication on the right side:

100a+100+10b+10+c+1=130a+13b+1.3c

Add like terms on the left side:

100a+10b+c+111=130a+13b+1.3c

Subtract 100a, 10b, and c from both sides

111=30a+3b+0.3c

So "technically" you can just plug in any two values, and then solve for the last value, but since you have a 3 digit number, you have the restriction of a < 10, b < 10, and c < 10, and also a, b, and c, should only be integers and all have the same sign or it wouldn't be a 3 digit number.

So let's start with a, since it has the highest coefficient, well you can fit 30 into 111, 3 times without going over so that's the first value

111 = 30(3) + 3b + 0.3c

111 = 90 + 3b + 0.3c

Now subtract the 90 from both sides

21=3b+0.3c

Well 3 can fit into 21, 7 times!

21 = 3(7) + 0.3c

21 = 21 + 0.3c

subtract 21 from both sides

0 = 0.3c

and now obviously c is 0, if you want you can divide both sides by 0.3 but it's a bit redundant

c = 0

This gives you the three values, a=3, b=7, c=0. which is the number 370. Now let's double check. Adding 1 to each digit would give you 481 and 481/370 = 1.3, so it is correct!

part b:

So to prove there is no three digit number, is to realize there is no solution, given the restriction or integers, greater than or equal to 0, and less than 10, and all of them must have the same sign.

So let's start with the same equation except this time instead of 1.3 it's 1.4

100(a+1) + 10(b+1) + 1(c+1) = 1.40(100a+10b+c)

Distribute on the left side;

100a+100+10b+10+c+1=1.40(100a+10b+c)

Distribute on the right side:

100a+100+10b+10+c+1=140a + 14b + 1.4c

Add like terms on left side:

100a + 10b + c + 111 = 140a + 14b + 1.4c

Subtract 100a, 10b, and c from both sides:

111 = 40a + 4b + 0.4c

Now to do the same process, let's start by finding how many times we can fit 40 into 111, and if you're wondering why we start with 40, it's because let's say for example I just say, I can fit another 40 into it, but I decide not to, and let b do that, well even if it's just 40, b will have be at least 10, which does not fit our restrictions, so you have to fit as many 40's into the number first then go the other numbers.

So only 2 40's can fit in 111 without going over the value

111 = 40(2) + 4b + 0.4c

Subtract 80 from both sides

31=4b+0.4c

4 can fit into 31, 7 times

31 = 4(7) + 0.4c

31 = 28 + 0.4c

subtract 28 from both sides

3 = 0.4c

divide both sides by 0.4

7.5 = c.

Since c is not an integer there is no 3 digit number that exists that increases by 40% whenever you increase it by 1.

7 0
2 years ago
What is the highest fractions 2/4 5/6 3/2 1/9 4/5 8/2 ​
Akimi4 [234]

Answer:

8/2

eight halves

Step-by-step explanation:

please mark this answer as brainliest

6 0
2 years ago
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