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boyakko [2]
3 years ago
10

What is the lcm of 5 And 10

Mathematics
2 answers:
Maksim231197 [3]3 years ago
8 0

If you want to know what is the LCM of 5 and 10 it is 10 Usually, this is written as

<span>LCM(5,10) = 10</span>
Kisachek [45]3 years ago
4 0

Answer:

50

Step-by-step explanation:

just multiply the two

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18. Find the length of the segment indicated. Round your answer to the
Usimov [2.4K]

Answer:

b. 6.2

Step-by-step explanation:

x =√18.2²-17.1² =√331.24- 292.41

= √38.83 = 6.2

8 0
3 years ago
Please help will give brainliest it’s for a test<br><br> Round 29.20 to 30 <br> Show work please
djverab [1.8K]

Answer:

you cant round that to 30

Step-by-step explanation

29.20

29.2 2 is less than 5

so you cant round you can only round if the .tens place is 5 or more

4 0
3 years ago
Thank you in advance for those who provided support to my question!
alekssr [168]

GCF of 10 and 12 is 2

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5 0
3 years ago
Read 2 more answers
What are the coordinates of point B on \overline{AC} such that AB = 1/2 BC
marta [7]

Answer:

C. B(x,y) = \left(3,-\frac{16}{3} \right)

Step-by-step explanation:

Let A(x,y) = (6,-7), B(x,y) = (x,y) and C(x,y) = (-3,-2). From statement we know that AB = \frac{1}{2}\cdot BC, which is equivalent to the following linear algebraic formula:

B(x,y) -A(x,y) = \frac{1}{2}\cdot [C(x,y)-B(x,y)] (1)

B(x,y)-A(x,y) = \frac{1}{2}\cdot C(x,y)-\frac{1}{2}\cdot B(x,y)

\frac{3}{2}\cdot B(x,y) = \frac{1}{2}\cdot C(x,y)+A(x,y)

B(x,y) = \frac{1}{3}\cdot C(x,y) +\frac{2}{3}\cdot A(x,y) (2)

Then, the coordinates of point B on AC are:

B(x,y) = \frac{1}{3}\cdot (-3,-2)+\frac{2}{3}\cdot (6,-7)

B(x,y) = \left(-1, -\frac{2}{3}\right)+\left(4, -\frac{14}{3} \right)

B(x,y) = \left(3,-\frac{16}{3} \right)

Which means that correct answer is C.

5 0
3 years ago
THIS IS DUE TODAY 1. A circle has a mass of 3 grams and a triangle has a mass of 5 grams.
horsena [70]

Answer:

3+2=5

5+2=7

Step-by-step explanation:

the square is 7 mass because from 3 to 5 the numbers increase by 2 so I add 5 and 2

7 0
3 years ago
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