All categories

3 years ago

The dotted line represents the mid segment of the trapezoid. Find the value of X

Mathematics

1 answer:

Ganezh [65]3 years ago

6 0

<u>Given</u>:

Given that the bases of the trapezoid are 21 and 27.

The midsegment of the trapezoid is 5x - 1.

We need to determine the value of x.

<u>Value of x:</u>

The value of x can be determined using the trapezoid midsegment theorem.

Applying the theorem, we have;

$Midsegment = \frac{b_1+b_2}{2}$

where b₁ and b₂ are the bases of the trapezoid.

Substituting Midsegment = 5x - 1, b₁ = 21 and b₂ = 27, we get;

$5x-1=\frac{21+27}{2}$

Multiplying both sides of the equation by 2, we have;

$2(5x-1)=(21+27)$

Simplifying, we have;

$10x-2=48$

Adding both sides of the equation by 2, we get;

$10x=50$

Dividing both sides of the equation by 10, we have;

$x=5$

Thus, the value of x is 5.

You might be interested in

Ali receives a 10% pay

DiKsa [7]

Answer:

95.04 pounds

122.54 US dollars

Step-by-step explanation:

I had to do the math in US dollars first then turn it into pounds.

All you do is this

136.16 * -.10 (US dollars)

13.616(US) 10.56(pounds)

Subtract 13.616 from 136.16 to get

122.544 or 122.54 (US)

95.04(pounds)

It might be a bit confusing, but the answers are up top.

8 0

4 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

Help a girl out pleaseeee!!

fiasKO [112]

Spinal Cord

hope this helps

7 0

3 years ago

Read 2 more answers

A golfer bought 24 golf balls for $10.08. Which is the price per golf ball?

Vlada [557]

Answer: 42 cents

Step-by-step explanation: Take 10.08 divide it by 24 and there's your answer!