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Roman55 [17]
3 years ago
6

Find the smallest possible interger value n for which 99n is a multiple of 24

Mathematics
2 answers:
anastassius [24]3 years ago
8 0
Thats going to be 3 cos 3 * 8 and 3 times 23 is both of the answers
motikmotik3 years ago
3 0
I would guess that you would need to find the Least Common Multiple (LCM) of 99 and 24 first. 

After listing out all of the multiples, you would have 792 be the LCM of 99 and 24. But since we need to find the integer value of n, in 99n (which means 99 x n), then we need to divide 792 by 99, which would be 8.

n = 8 
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3 years ago
MO bisects angle LMN , angle LMO = 6x-22 and angle NMO = 2x +34. Solve for x and find angle LMN.
Mkey [24]

Answer:

  • x=14
  • \angle{LMN}=124\textdegree

Step-by-step explanation:

<u><em>To Determine:</em></u>

Solve for x and find angle LMN.

<u><em>Fetching Information and Solution Steps:</em></u>

Considering the angle \angle{LMN}

As MO bisects the angle \angle{LMN} into two equal angle parts. These equal angles are:

\angle{LMO}=6x-22

\angle{NMO}=2x+34

As these angles are equal. i.e.

\angle{LMO}=\angle{NMO}

6x-22=2x+34

\mathrm{Add\:}22\mathrm{\:to\:both\:sides}

6x-22+22=2x+34+22

6x=2x+56

\mathrm{Subtract\:}2x\mathrm{\:from\:both\:sides}

6x-2x=2x+56-2x

4x=56

\mathrm{Divide\:both\:sides\:by\:}4

\frac{4x}{4}=\frac{56}{4}

\mathrm{Simplify}

x=14

Hence, x=14

As  \angle{LMN} was cut into two equal parts \angle{LMO} and \angle{NMO}

So,

\angle{LMN} = \angle{LMO} + \angle{NMO}

            = 6x-22+2x+34

            = 8x + 12

            = 8(14) + 12     ∵ x=14

            = 124\textdegree

Therefore, \angle{LMN}=124\textdegree

Keywords: angle bisector, congruent angles

Learn more about angle bisector and congruent angles from brainly.com/question/711370

#learnwithBrainly

3 0
3 years ago
Can someone give an explaination please<br><br>​
iren [92.7K]

Answer:

x ≥ -3

x ≤ 3

Step-by-step explanation:

For the first inequality, just add 3 to both sides.

For the second inequality, add 4 to both sides then divide both sides by 3.

3 0
3 years ago
Fundamental theorem of calculus<br> <img src="https://tex.z-dn.net/?f=g%28s%29%3D%5Cint%5Climits%5Es_6%20%7B%28t-t%5E4%29%5E6%7D
mr_godi [17]

Answer:

\displaystyle g'(s) = (s-s^4)^6

Step-by-step explanation:

The Fundamental Theorem of Calculus states that:
\displaystyle \frac{d}{dx}\left[ \int_a^x f(t)\, dt  \right] = f(x)

Where <em>a</em> is some constant.

We can let:
\displaystyle g(t) = (t-t^4)^6

By substitution:

\displaystyle g(s) = \int_6^s g(t)\, dt

Taking the derivative of both sides results in:
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Hence, by the Fundamental Theorem:

\displaystyle \begin{aligned} g'(s) & = g(s) \\ \\  & = (s-s^4)^6\end{aligned}

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