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leonid [27]
3 years ago
7

Find the least common multiple of 18, 24 *

Mathematics
2 answers:
emmainna [20.7K]3 years ago
7 0

Answer:

72

Step-by-step explanation:

Find the prime factorization of 18

18 = 2 × 3 × 3

Find the prime factorization of 24

24 = 2 × 2 × 2 × 3

Multiply each factor the greater number of times it occurs in steps above to find the LCM:

LCM = 2 × 2 × 2 × 3 × 3

LCM = 72

i hope this helps :)

grigory [225]3 years ago
4 0

the multiples of 18 are.. 18 36 54<u> 72</u>

the multiples of 24 are.. 24 48 <u>72</u>

LCM is 72

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3 years ago
Enter the number that belongs in the green box.
ale4655 [162]

Answer:

the green box is 0

Step-by-step explanation:

we are asked to find the <u>x-intercept</u>, which means y = 0

(x, y)

Another way to test this is plug it in 1 for x!  (1, y) where x = 1

y = \frac{2x^{2} -3x+1}{2x^{2} +8x+6} \\\\

y = \frac{2(1)^{2} -3(1)+1}{2(1)^{2} +8(1)+6}

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6 0
2 years ago
What is the first derivative of r with respect to t (i.e., differentiate r with respect to t)? r = 5/(t2)Note: Use ^ to show exp
adelina 88 [10]

Answer:

The first derivative of r(t) = 5\cdot t^{-2} (r(t)=5*t^{-2}) with respect to t is r'(t) = -10\cdot t^{-3} (r'(t) = -10*t^{-3}).

Step-by-step explanation:

Let be r(t) = \frac{5}{t^{2}}, which can be rewritten as r(t) = 5\cdot t^{-2}. The rule of differentiation for a potential function multiplied by a constant is:

\frac{d}{dt}(c \cdot t^{n}) = n\cdot c \cdot t^{n-1}, \forall \,n\neq 0

Then,

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The first derivative of r(t) = 5\cdot t^{-2} (r(t)=5*t^{-2}) with respect to t is r'(t) = -10\cdot t^{-3} (r'(t) = -10*t^{-3}).

5 0
3 years ago
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|\det J|=\left|\begin{vmatrix}\mathbf x_u&\mathbf x_v\\\mathbf y_u&\mathbf y_v\end{vmatrix}\right|=\dfrac1{32}

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4 0
3 years ago
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babunello [35]
The answer would be (-1,4)
7 0
3 years ago
Read 2 more answers
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