Can somebody help pleaseIf a number is a whole number, then it cannot be an

The area of a rectangular pool was 120square meters. If the long side in 7 meters plus a number and the width is 6 meters. What

NeX [460]

Answer:

The value of the variable=13 meters

Step-by-step explanation:

Step 1

Express the dimensions of the pool

Length=7+x

width=6

Step 2

Express the area of a rectangle as shown;

A=L×W

where;

A=area of a rectangle

L=length of the rectangle

W=width of the rectangle

In our case;

A=120 m²

L=x+7

W=6

replacing in the equation;

120 m²=(x+7)×6

120=6 x+42

6 x+42=120

6 x=120-42

6 x=78

x=78/6

x=13

The value of the variable=x=13 meters

8 0

3 years ago

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

2y+18=42 what is y, and how would you find it?

Nesterboy [21]

Answer:

y=12

Step-by-step explanation:

Use order of operations. Subtract 18 from both sides and the divide by 2.

2y+18 -18=42-18

2y=24

y=12

3 0

3 years ago

WILL GIVE BRAINIEST TO THE PERSON THAT ANSWERS ALL THREE QUESTIONS ASAP!

Radda [10]

Answer:

1. Your answer would be C) 2√3-5√2.

2. Your answer would be -39.7i.

3. Your answer would be -1.

Step-by-step explanation:

8 0

3 years ago

Read 2 more answers

There is a direct path from Gayle's house to the store. Due to construction, the direct route is closed and Gayle must travel 6

Elina [12.6K]

Answer:

<em><u>c2=10</u></em>

Step-by-step explanation:

8^2+6^2=c2

64+36=120

c2=square root of 120..which is 10

3 0

4 years ago