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3 years ago

What is the least common factor of 30 and 42

Mathematics

1 answer:

shutvik [7]3 years ago

4 0

Answer:

210

Step-by-step explanation:

Least common multiple can be found by multiplying the highest exponent prime factors of 30 and 42

GCF(30,42) = 6

LCM(30,42) = ( 30 × 42) / 6

LCM(30,42) = 1260 / 6

LCM(30,42) = 210

hope this helps :)

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IgorC [24]

Answer:

Step-by-step explanation:

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3 years ago

There are 3x-2 trees planted in each row of a rectangular Parcel of land. If there are a total of 24x-16 trees planted and the p

m_a_m_a [10]

If one raw contains 3x-2 trees & if the total trees in the rectangle are 24x-16, that means Horizontal Rows x Vertical Rows =Total number of trees:

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3 years ago

Answer asap which equation could have been used to create this function table?

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3 years ago

Need help asap. Simplify the expression 6/c+4/c^2 show work

Zina [86]

By taking a common denominator, we can find out that the simplified expression of the given expression is $\frac{6c+4}{c^{2} }$ .

The given expression is -

$\frac{6}{c}+\frac{4}{c^{2} }$ ---- (1)

We have to simplify the expression.

From equation (1), we have the given expression as -

$\frac{6}{c}+\frac{4}{c^{2} }$

Establishing common denominator $c^{2}$ , we have

$\frac{6c}{c^{2} }+\frac{4}{c^{2}}$

Combining the numerator and common denominator, we have

$\frac{6c+4}{c^{2} }$

Thus, the simplified expression of the given expression is $\frac{6c+4}{c^{2} }$ which is obtained by taking a common denominator.

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1 year ago

The logistic equation for the population (in thousands) of a certain species is given by:

Eva8 [605]

Answer:

b. 1.5

c. 1.5

d. No

Step-by-step explanation:

a. First, let's solve the differential equation:

$\frac{dp}{dt} =3p-2p^2$

Divide both sides by $3p-2p^2$ and multiply both sides by dt:

$\frac{dp}{3p-2p^2}=dt$

Integrate both sides:

$\int\ \frac{1}{3p-2p^2} dp =\int\ dt$

Evaluate the integrals and simplify:

$p(t)=\frac{3e^{3t} }{C_1+2e^{3t}}$

Where C1 is an arbitrary constant

I sketched the direction field using a computer software. You can see it in the picture that I attached you.

b. First let's find the constant C1 for the initial condition given:

$p(0)=3=\frac{3e^{0} }{C_1+2e^{0} } =\frac{3}{C_1+2}$

Solving for C1:

$C_1=-1$

Now, let's evaluate the limit:

$\lim_{t \to \infty} \frac{3e^{3t} }{2e^{3t}-1 } \\\\Divide\hspace{3}the\hspace{3}numerator\hspace{3}and\hspace{3}denominator\hspace{3}by\hspace{3}e^{3t} \\\\ \lim_{t \to \infty} \frac{3 }{2-e^{-3x} }$

The expression $-e^{-3x}$ tends to zero as x approaches ∞ . Hence:

$\lim_{t \to \infty} \frac{3e^{3t} }{2e^{3t}-1 } =\frac{3}{2} =1.5$

c. As we did before, let's find the constant C1 for the initial condition given:

$p(0)=0.8=\frac{3e^{0} }{C_1+2e^{0} } =\frac{3}{C_1+2}$

Solving for C1:

$C_1=1.75$

Now, let's evaluate the limit:

$\lim_{t \to \infty} \frac{3e^{3t} }{2e^{3t}+1.75 } \\\\Divide\hspace{3}the\hspace{3}numerator\hspace{3}and\hspace{3}denominator\hspace{3}by\hspace{3}e^{3t} \\\\ \lim_{t \to \infty} \frac{3 }{2+1.75e^{-3x} }$

The expression $-e^{-3x}$ tends to zero as x approaches ∞ . Hence:

$\lim_{t \to \infty} \frac{3e^{3t} }{2e^{3t}+1.75 } =\frac{3}{2} =1.5$

d. To figure out that, we need to do the same procedure as we did before. So, let's find the constant C1 for the initial condition given:

$p(0)=2=\frac{3e^{0} }{C_1+2e^{0} } =\frac{3}{C_1+2}$

Solving for C1:

$C_1=-\frac{1}{2} =-0.5$

Can a population of 2000 ever decline to 800? well, let's find the limit of the function when it approaches to ∞:

$\lim_{t \to \infty} \frac{3e^{3t} }{2e^{3t}-0.5 } \\\\Divide\hspace{3}the\hspace{3}numerator\hspace{3}and\hspace{3}denominator\hspace{3}by\hspace{3}e^{3t} \\\\ \lim_{t \to \infty} \frac{3 }{2-0.5e^{-3x} }$

The expression $-e^{-3x}$ tends to zero as x approaches ∞ . Hence:

$\lim_{t \to \infty} \frac{3e^{3t} }{2e^{3t}-0.5 } =\frac{3}{2} =1.5$

Therefore, a population of 2000 never will decline to 800.

6 0

3 years ago