PLZ HELP!! WILL GIVE BRAINLIST IF CORECT!!

What is the slope of the line through the points (-3, -5) and (-4, -2)

Ede4ka [16]

Answer:

slope= -3

Step-by-step explanation:

the slope formula is $\frac{y_{2} -y_{1} }{x_{2} -x_{1} }$

plug in your numbers: $\frac{(-2)-(-5)}{(-4)-(-3)}$

solve: $\frac{3}{-1}$

simplify: -3

5 0

3 years ago

Marco goal is to eat only 2,000 calories each day one day for breakfast he consumed in 310 calories for lunch you consume 200 mo

spayn [35]

Calories in (Breakfast + Lunch +Dinner) = (310 +(310+200)+800)

=310+500+800= 1610>2000.

So, he was unable to do that.

8 0

3 years ago

Read 2 more answers

Choose the correct fraction and decimal equivalent for 78%.

vazorg [7]

Answer: 0.78 and 39/50

<h3>Explanation:</h3>

What we know:

We need to find 78% as a fraction and decimal

How to solve:

The decimal is easy enough, and then the fraction can be found from there. D will represent decimal, p will represent percentage, and F will represent fraction.

<h3>Process:</h3>

Decimal

Set up equation D = p / 100

Substitute D = 78% / 100

Simplify D = 0.78%

Remove % sign D = 0.78

Fraction

Set up equation F = p / 100

Substitute F = 78% / 100

Remove % sign F = 78 / 100

Simplify if needed F = /2 /2

F = 39 / 50

Solution: 0.78 and 39/50

Check:

Multiply the decimal by 100 to check the decimal

0.78 x 100 = 78; 78%

Divide the fraction to check the fraction

39/50 = 0.78; see above

8 0

2 years ago

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The theoretical probability of selecting a consonant at random from a list of letters in the alphabet is 21/26. Wayne opens a bo

marysya [2.9K]

Answer:

You need a corpus of text. It usually gathers text from passages, chapters, or sections of a book.

6 0

3 years ago

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Find $\int \dfrac{x}{\sqrt{1-x^4}}$ Please, help

ki77a [65]

If you're using the app, try seeing this answer through your browser: brainly.com/question/2867785

_______________

Evaluate the indefinite integral:

$\mathsf{\displaystyle\int\! \frac{x}{\sqrt{1-x^4}}\,dx}\\\\\\ \mathsf{=\displaystyle\int\! \frac{1}{2}\cdot 2\cdot \frac{1}{\sqrt{1-(x^2)^2}}\,dx}\\\\\\ \mathsf{=\displaystyle \frac{1}{2}\int\! \frac{1}{\sqrt{1-(x^2)^2}}\cdot 2x\,dx\qquad\quad(i)}$

Make a trigonometric substitution:

$\begin{array}{lcl}
\mathsf{x^2=sin\,t}&\quad\Rightarrow\quad&\mathsf{2x\,dx=cos\,t\,dt}\\\\
&&\mathsf{t=arcsin(x^2)\,,\qquad 0\ \textless \ x\ \textless \ \frac{\pi}{2}}\end{array}$

so the integral (i) becomes

$\mathsf{=\displaystyle\frac{1}{2}\int\!\frac{1}{\sqrt{1-sin^2\,t}}\cdot cos\,t\,dt\qquad\quad (but~1-sin^2\,t=cos^2\,t)}\\\\\\
\mathsf{=\displaystyle\frac{1}{2}\int\!\frac{1}{\sqrt{cos^2\,t}}\cdot cos\,t\,dt}$

$\mathsf{=\displaystyle\frac{1}{2}\int\!\frac{1}{cos\,t}\cdot cos\,t\,dt}\\\\\\
\mathsf{=\displaystyle\frac{1}{2}\int\!\f dt}\\\\\\
\mathsf{=\displaystyle\frac{1}{2}\,t+C}$

Now, substitute back for t = arcsin(x²), and you finally get the result:

$\mathsf{\displaystyle\int\! \frac{x}{\sqrt{1-(x^2)^2}}\,dx=\frac{1}{2}\,arcsin(x^2)+C}$ ✔

________

You could also make

x² = cos t

and you would get this expression for the integral:

$\mathsf{\displaystyle\int\! \frac{x}{\sqrt{1-(x^2)^2}}\,dx=-\,\frac{1}{2}\,arccos(x^2)+C_2}$ ✔

which is fine, because those two functions have the same derivative, as the difference between them is a constant:

$\mathsf{\dfrac{1}{2}\,arcsin(x^2)-\left(-\dfrac{1}{2}\,arccos(x^2)\right)}\\\\\\
=\mathsf{\dfrac{1}{2}\,arcsin(x^2)+\dfrac{1}{2}\,arccos(x^2)}\\\\\\
=\mathsf{\dfrac{1}{2}\cdot \left[\,arcsin(x^2)+arccos(x^2)\right]}\\\\\\
=\mathsf{\dfrac{1}{2}\cdot \dfrac{\pi}{2}}$

$\mathsf{=\dfrac{\pi}{4}}$ ✔

and that constant does not interfer in the differentiation process, because the derivative of a constant is zero.

I hope this helps. =)

6 0

3 years ago