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raketka [301]
4 years ago
7

Find the least common multiple of 8b^2 and 5n^3

Mathematics
2 answers:
liq [111]4 years ago
8 0

For this case, we have that by definition, the LCM of two or more natural numbers, is the smallest natural number that is a common multiple of all of them.

Then, the LCM of 8 and 5 is the smallest positive integer that divides both numbers without leaving residues.

Multiplos:

8: 8,16,24,32,40,48

5: 1,10,15,20,25,30,35,40,45

Thus, the LCM of 8 and 5 is 40.

Therefore the LCM of the given expressions is:

40b ^ 2n ^ 3

Answer:

40b ^ 2n ^ 3

muminat4 years ago
5 0

Answer:

The lcm is; 40b^2n^3

Step-by-step explanation:

There are no common factors between the two expressions and as such the lcm will be found by obtaining the product of he two;

8b^2*5n^3 = 40b^2n^3

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A triangle has sides that measure 5cm, 12 cm and 13cm. What type of triangle is it?
kykrilka [37]

Answer:

i think its an isoceles

Step-by-step explanation:

6 0
3 years ago
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Given m||n, find the value of x.<br> m<br> t<br> (5x-5)<br> (6x-27)
almond37 [142]

Step-by-step explanation:

due to the laws of symmetry of the angles of 2 crossing lines (and parallel lines are simply repeating and mirroring these angles), both angles must be equal.

5x - 5 = 6x - 27

-5 = x - 27

x = 22

4 0
2 years ago
1
Lina20 [59]

Answer:

\textbf{The slope of the parallel line is given by $ 2 $}\\

Step-by-step explanation:

\textup{The slope of any two parallel lines are equal and they only differ in their $y- intercepts$. }\\\textup{Given $y = 2x + 4 $}\\\textup{This is  comparable to $y = mx + c$, where $m$ is the slope of that line.}\\\textup{Therefore, the slope of the given line and any line parallel to it is $ 2$.}

7 0
3 years ago
HELP HELP I NEED HELP ASAP
tia_tia [17]
I am pretty sure the answer is 180
5 0
3 years ago
Please help answer these questions. My teacher said they were really easy but I just don't understand. Will mark brainliest !!!
kodGreya [7K]

Answer:

1.  A = 59

2.  A = 43

Step-by-step explanation:

If we have a right triangle  we can use sin, cos and tan.

sin = opp/ hypotenuse

cos= adjacent/ hypotenuse

tan = opposite/ adjacent


For the first problem, we know the opposite and adjacent sides to angle A

tan A = opposite/ adjacent

tan A = 8.8 / 5.2

Take the inverse of each side

tan ^-1 tan A = tan ^-1 (8.8/5.2)

A = 59.42077313

To the nearest degree

A = 59 degrees


For the second problem, we know the  adjacent side and the hypotenuse to angle A

cos A = adjacent/hypotenuse

cos A = 15.3/21

Take the inverse of each side

cos ^-1 cos A = cos ^-1 (15.3/21)

A = 43.23323481

To the nearest degree

A = 43 degrees


6 0
4 years ago
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