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3 years ago

2b-5/b-2 = 3/b+2 SOLVE

1 answer:

7 0

Cross multiply:-

3(b - 2) = (2b - 5)(b + 2)

3b - 6 = 2b^2 + 4b - 5b - 10

2b^2 - 4b - 4 = 0

b^2 - 2b - 2 = 0

b = 2.73, -0.73 Answer

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Here is a scale drawing of a window frame that uses a scale of 1 cm to 6 inches.Create another scale drawing of the window frame

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Answer:

Step-by-step explanation:

Below is the rectangle in the attachment.

Current scale:

1 cm : 6 inches

If the dimensions of the rectangle is:

Length = a cm

Width = b cm

Using the scale:

Length = a × 6 inches

Width = b × 6 inches

Using the same dimensions of the rectangle is:

Length = a cm

Width = b cm

Using the scale:

Length = a × 12 inches

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Note that there is an enlargement of the rectangle to form the new rectangle. The length and width of new rectangle drawn will be 2 × the length and width of the rectangle seen below.

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3 years ago

A developer sold an apartment complex for $606,150. She had purchased the complex years earlier for $449,000. What percent of th

Juli2301 [7.4K]

She marked down the price of the apartment by 26%. This is because of this equation (Percent change=Amount of change divided by the original number and then multiply by 100)

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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

Solve the equation. Check your answer. -11=5+8x x=?

zloy xaker [14]

Answer:

x=-2

Step-by-step explanation:

Consider equation -11=5+8x

Collecting the like terms

-8x=11+5, -8x=16

Dividing through by -8 ob both sides of the equation

(-8x)-/8=16/-8