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3 years ago

Single power What’s 7^2 x 7^3 written as a single power?

Mathematics

2 answers:

Katarina [22]3 years ago

7 0

Answer:

7^5

Step-by-step explanation:

When multiplying powers in a binomial, you simply add the powers togehter and leave the same base.

irina1246 [14]3 years ago

6 0

7^5
Explanation:
2+3=5 (times them all at once)

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-0.38 (8 repeating) written as a fraction is _____.

natulia [17]

Answer:

$\frac{7}{18}$

Step-by-step explanation:

We first let 0.38 (8 being repeated) be T.

Since z is recurring in 1 decimal places, we multiply it by 10. 10z = 3.88

Next, we subtract them. 10 r T 3.88 0.38 9x 3.5

Lastly, we divide both sides by 9 to get IC as a fraction. 3.5 T 9 35 90 7 18

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4 years ago

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a missing fraction on a number line is located exactly halfway between 3/6 and 5/6. what is the missing fraction

FinnZ [79.3K]

The missing number is 4/6

the number line goes 1/6 2/6 3/6 4/6 5/6 6/6

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3 years ago

What is the answer to 1/5(5x-15)-2x

Contact [7]

Answer:

-1x - 3 or -x - 3

Step-by-step explanation:

1/5 (5x - 15) - 2x =

1x - 3 - 2x =

1x - 2x - 3 =

-1x - 3

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3 years ago

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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 ∞:

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