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3 years ago

Which equation justifies why ten to the one third power equals the cube root of ten?

Mathematics

2 answers:

son4ous [18]3 years ago

6 0

Answer:

Step-by-step explanation:

$(a^{m})^{n}=a^{mn}.\ \text{Hence option b will be correct.}$

IgorC [24]3 years ago

6 0

Answer:

Step-by-step explanation:

AC and D are all wrong. The all add or subtract something from 3 which is not going to keep the two sides of the equation equal.

Only B is correct. To get the two sides equal, you need to multiply the powers by the same constant. B does this. The powers and the constant 1/3 * 3 are the same on both sides.

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We are given a sequence of five non-zero numbers, where the sum of each term and its neighboring terms is 15 or 25. Find the sum

PSYCHO15rus [73]

We have a sequence that meets the given criteria, and with that information, we want to get the sum of all the terms in the sequence.

We will see that the sum tends to infinity.

So we have 5 terms;

A, B, C, D, E.

We know that the sum of each term and its neighboring terms is 15 or 25.

then:

A + B + C = 15 or 25
B + C + D = 15 or 25
C + D + E = 15 or 25

Now, we want to find the sum of all the terms in the sequence (not only the 5 given).

Then let's assume we write the sum of infinite terms as:

$a_1 + a_2 + a_3 + a_4 + a_5 + a_6 + ...$

Now we group that sum in pairs of 3 consecutive terms, so we get:

$(a_1 + a_2 + a_3) + (a_4 + a_5 + a_6) + ...$

So we will have a sum of infinite of these, and each one of these is equal to 15 or 25 (both positive numbers). So when we sum that infinite times (even if we always have the smaller number, 15) the sum will tend to be infinite.

Then we have:

$(a_1 + a_2 + a_3) + (a_4 + a_5 + a_6) + ... \to \infty$

If you want to learn more, you can read:

brainly.com/question/21885715

3 0

2 years ago

Calculate AC if the perimeter of <br> ABC is 83 inches

Elza [17]

Since you already know BC is 23inches, you can minus it from 83 inches first.
Next, you have to realise that since angle ABC and angle ACB are both 65 degrees and equal, the triangle is an isosceles triangle it means that AC is equals to BC.
Thus, it means that you can divide (83-23) by 2, which is 30 inches

4 0

3 years ago

What’s the missing angle

charle [14.2K]

Answer:

the answer will be 28

Step-by-step explanation:

180-47-31=102

since vertical angles have the same value the top angles for both triangles will be 102 degrees

102+50=152

180-152=x

28=x

7 0

2 years ago

PLEASE ANSWER THESE! First person to do it gets brainliest!

True [87]

1. meters
2(a). mm
2(b). cm
3. Gram
4(a). g
4(b). kg
4(c). g
5. Liter

6 0

2 years 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

2 years ago