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3 years ago

State the domain of the function y=3/x

1 answer:

8 0

$Notice \ the \ equation \ contains \ x \ in \ the \ denominator.$

$Since \ any \ denominator \ must \ not \ equal \ zero,$

$our \ domain \ is \ restricted \ to \$ x<span>≠0</span>

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Will give brainliest

Galina-37 [17]

The percentage of the radioactive isotope left is 25.0000%.

The half life of a radioactive isotope is the time that will elapse before only half of the number of original isotopes will remain.

Now;

We are told that;

Half life (t1/2) = 55 days

Time taken (t) = 115 days

Ratio of isotope remaining = N/No

Thus;

N/No = (1/2)^t/t1/2

N/No = (1/2)^115/55

N/No = (1/2)^2

N/No =1/4

Hence, the percentage = 1/4 * 100/1 = 25.0000%

Learn more about half life of an isotope:brainly.com/question/16387602

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7 0

2 years ago

How would I do this?

Georgia [21]

Work shown above! Answer would be 110 degrees!

3 0

3 years ago

The population of a town has grown at an annual rate of 2.7%. How long does it take for it’s population of 18,450 people to doub

charle [14.2K]

It would take 26 years.

The equation will be in the form y = a(1+r)ˣ, where a is the initial population, r is the growth rate as a decimal number, and x is the amount of time. Using our information we have

y = 18450(1+0.027)ˣ = 18450(1.027)ˣ

We want the population to be doubled; 18450*2 = 36900:
36900 = 18450(1.027)ˣ

Divide both sides by 18450:
36900/18450 = 18450(1.027)ˣ/18450
2 = 1.027ˣ

Using logarithms, we have
$\log_{1.027}2=x
\\
\\x=26.017$

3 0

3 years ago

4. Use the process outlined in the lesson to approximate the number 2√3. Use the approximation √3 ≈ 1.732 050 8.

jeka57 [31]

Answer:

sequence of five intervals

(1) 2 < $2^{\sqrt{3} }$ < $2^{2}$

(2) $2^{1.7}$ < $2^{\sqrt{3} }$ < $2^{1.8}$

(3) $2^{1.73}$ < $2^{\sqrt{3} }$ < $2^{1.74}$

(4) $2^{1.732}$ < $2^{\sqrt{3} }$ < $2^{1.733}$

(5) $2^{1.7320}$ < $2^{\sqrt{3} }$ < $2^{1.7321}$

Step-by-step explanation:

as per question given data

√3 ≈ 1.732 050 8

to find out

sequence of five intervals

solution

as we have given that √3 value that is here

√3 ≈ 1.732 050 8 ........................1

when we find $2^{\sqrt{3} }$ ................2

put here √3 value in equation number 2

we get $2^{\sqrt{3} }$ that is 3.322

sequence of five intervals

(1) 2 < $2^{\sqrt{3} }$ < $2^{2}$

(2) $2^{1.7}$ < $2^{\sqrt{3} }$ < $2^{1.8}$

(3) $2^{1.73}$ < $2^{\sqrt{3} }$ < $2^{1.74}$

(4) $2^{1.732}$ < $2^{\sqrt{3} }$ < $2^{1.733}$

(5) $2^{1.7320}$ < $2^{\sqrt{3} }$ < $2^{1.7321}$

8 0

3 years ago

Dy/dx = (cos x) e^(y+sinx) and y = 0 when x = 0. find the original equation

Korvikt [17]

Find the general solution by separating the variables then integrating:
dy / dx = cosx℮^(y + sinx)
dy / dx = cosx℮ʸ℮^(sinx)
℮^(-y) dy = cosx℮^(sinx) dx
∫ ℮^(-y) dy = ∫ cosx℮^(sinx) dx
-℮^(-y) = ℮^(sinx) + C
℮^(-y) = C - ℮^(sinx)
-y = ln[C - ℮^(sinx)]
y = -ln[C - ℮^(sinx)]

Find the particular solution by solving for the constant:
When x = 0, y = 0
-ln(C - 1) = 0
ln(C - 1) = 0
C - 1 = 1
C = 2
<span>y = -ln[2 - ℮^(sinx)]

I hope my answer has come to your help. Thank you for posting your question here in Brainly. We hope to answer more of your questions and inquiries soon. Have a nice day ahead!</span>

3 0

3 years ago