All categories

3 years ago

HELP ME PLEASEEEEEEEEEEEEE

Mathematics

1 answer:

kakasveta [241]3 years ago

3 0

Bro we cant help you, there must be a kind of image before that question. You have to attach it since we dont know what “Organism A and B” are.. we dont know what its referring to so you might have to repost this question again

You might be interested in

Hi w do you find c for standard normal distribution

siniylev [52]

You need to a table of the standard normal cumulative distribution
Here is one:
https://math.ucalgary.ca/files/math/normal_cdf.pdf

the closest value I see is 0.85

5 0

3 years ago

If 20% of a school was made up of out-of-district students last year, but the number of out-of-district students increased by 50

Nadya [2.5K]

Answer:

<h2>30% is enrolled out-of-district.</h2>

Step-by-step explanation:

According to the word problem, the number is increased by 50%.

To increase a number by a percent, multiply the number and the percent.

We can do this by changing both percents to decimals.

To change a percent to a decimal, divide it by 100.

$20/100 = 0.2$

$50/100=0.5$

The number is INCREASED by 50% so 0.5 would become 1.5.

$0.2 * 1.5=0.3$

4 0

3 years ago

What is the multiplicative inverse of y? y 1 y -y 0<br><br><br>Ill give brainly if correct :D

kondor19780726 [428]

Answer:

The multiplicative inverse of y is - 1/y

Step-by-step explanation:

hope this helps :)

4 0

3 years ago

Describe a relationship in which it would be reasonable to have a negative number in the domain or

Aliun [14]

Answer:

Relationship between temperature at different times of the day during the Winter period in places like Alaska and Minnesota in the USA.

Step-by-step explanation:

An example of a relationship that would have a negative number in the domain or range would be the relationship between the temperature at different times of the day during the snow period in the winter. It's well known that in places like Alaska and Minnesota in the USA, that during the winter when there's very heavy snow, the average temperature is usually below 0°C. The temperature is the dependent variable and is most likely negative in those 2 states and thus the range would have negative numbers.

7 0

3 years ago

Evaluate the integral, show all steps please!

Aloiza [94]

Answer:

$\displaystyle \int \dfrac{1}{(9-x^2)^{\frac{3}{2}}}\:\:\text{d}x=\dfrac{x}{9\sqrt{9-x^2}} +\text{C}$

Step-by-step explanation:

<u>Fundamental Theorem of Calculus</u>

$\displaystyle \int \text{f}(x)\:\text{d}x=\text{F}(x)+\text{C} \iff \text{f}(x)=\dfrac{\text{d}}{\text{d}x}(\text{F}(x))$

If differentiating takes you from one function to another, then integrating the second function will take you back to the first with a constant of integration.

Given indefinite integral:

$\displaystyle \int \dfrac{1}{(9-x^2)^{\frac{3}{2}}}\:\:\text{d}x$

Rewrite 9 as 3² and rewrite the 3/2 exponent as square root to the power of 3:

$\implies \displaystyle \int \dfrac{1}{\left(\sqrt{3^2-x^2}\right)^3}\:\:\text{d}x$

<u>Integration by substitution</u>

<u />

<u /> $\boxed{\textsf{For }\sqrt{a^2-x^2} \textsf{ use the substitution }x=a \sin \theta}$

$\textsf{Let }x=3 \sin \theta$

$\begin{aligned}\implies \sqrt{3^2-x^2} & =\sqrt{3^2-(3 \sin \theta)^2}\\ & = \sqrt{9-9 \sin^2 \theta}\\ & = \sqrt{9(1-\sin^2 \theta)}\\ & = \sqrt{9 \cos^2 \theta}\\ & = 3 \cos \theta\end{aligned}$

Find the derivative of x and rewrite it so that dx is on its own:

$\implies \dfrac{\text{d}x}{\text{d}\theta}=3 \cos \theta$

$\implies \text{d}x=3 \cos \theta\:\:\text{d}\theta$

<u>Substitute</u> everything into the original integral:

$\begin{aligned}\displaystyle \int \dfrac{1}{(9-x^2)^{\frac{3}{2}}}\:\:\text{d}x & = \int \dfrac{1}{\left(\sqrt{3^2-x^2}\right)^3}\:\:\text{d}x\\\\& = \int \dfrac{1}{\left(3 \cos \theta\right)^3}\:\:3 \cos \theta\:\:\text{d}\theta \\\\ & = \int \dfrac{1}{\left(3 \cos \theta\right)^2}\:\:\text{d}\theta \\\\ & = \int \dfrac{1}{9 \cos^2 \theta} \:\: \text{d}\theta\end{aligned}$

Take out the constant:

$\implies \displaystyle \dfrac{1}{9} \int \dfrac{1}{\cos^2 \theta}\:\:\text{d}\theta$

$\textsf{Use the trigonometric identity}: \quad\sec^2 \theta=\dfrac{1}{\cos^2 \theta}$

$\implies \displaystyle \dfrac{1}{9} \int \sec^2 \theta\:\:\text{d}\theta$

$\boxed{\begin{minipage}{5 cm}\underline{Integrating $\sec^2 kx$}\\\\$\displaystyle \int \sec^2 kx\:\text{d}x=\dfrac{1}{k} \tan kx\:\:(+\text{C})$\end{minipage}}$