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3 years ago

Quadrilateral ABCD is inscribed in this circle.

Mathematics

2 answers:

Alecsey [184]3 years ago

8 0

The sum of all interior angles in a polygon is

180(n-2), where n = sides, well, this is a QUADrilateral, so it has 4 sides, so the total is 360°.

now, let's find what angle C is first,

$\bf \stackrel{A}{(2x-40)}+\stackrel{B}{(116)}+C+\stackrel{D}{(x)}=360\implies C+3x+76=360
\\\\\\
C+3x=284\implies C=284-3x$

now, recall the "inscribed quadrilateral conjecture", where opposite angles are "supplementary angles", thus

$\bf \stackrel{\measuredangle A}{(2x-40)}+\stackrel{\measuredangle C}{(284-3x)}=180\implies -x+244=180
\\\\\\
64=x\\\\
-------------------------------\\\\
\measuredangle A=2(64)-40$

victus00 [196]3 years ago

6 0

The answer should be 88

Hope this helps and have good day

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Solve the system of linear equations y= 2x -3Y= x + 6

Alja [10]

Answer:

x=9, y=15. (9, 15).

Step-by-step explanation:

y=2x-3

y=x+6

----------

2x-3=x+6

2x-x-3=6

x-3=6

x=6+3

x=9

y=9+6=15

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2 years ago

What shapes can you obtain by taking a cross section of a cone?

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3 years ago

Using the quadratic formula to solve 5x = 6x- - 3, what are the values of x?

goldfiish [28.3K]

Answer:

x= -3

Step-by-step explanation:

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8 0

2 years ago

Read 2 more answers

Set up the integral that represents the arc length of the curve f(x) = ln(x) + 5 on [1, 3], and then use Simpson's Rule with n =

marta [7]

Answer:

The integral for the arc of length is:

$\displaystyle\int_1^3\sqrt{1+\frac{1}{x^2}}dx$

By using Simpon’s rule we get: 1.5355453

And using technology we get: 2.3020

The approximation is about 33% smaller than the exact result.

Explanation:

The formula for the length of arc of the function f(x) in the interval [a,b] is:

$\displaystyle\int_a^b \sqrt{1+[f'(x)]^2}dx$

We need the derivative of the function:

$f'(x)=\frac{1}{x}$

And we need it squared:

$[f'(x)]^2=\frac{1}{x^2}$

Then the integral is:

$\displaystyle\int_1^3\sqrt{1+\frac{1}{x^2}}dx$

Now, the Simposn’s rule with n=4 is:

$\displaystyle\int_a^b g(x)}dx\approx\frac{\Delta x}{3}\left( g(a)+4g(a+\Delta x)+2g(a+2\Delta x) +4g(a+3\Delta x)+g(b) \right)$

In this problem:

$a=1,b=3,n=4, \displaystyle\Delta x=\frac{b-a}{n}=\frac{2}{4}=\frac{1}{2},g(x)= \sqrt{1+\frac{1}{x^2}}$

So, the Simposn’s rule formula becomes:

$\displaystyle\int_1^3\sqrt{1+\frac{1}{x^2}}dx\\\approx \frac{\frac{1}{3}}{3}\left( \sqrt{1+\frac{1}{1^2}} +4\sqrt{1+\frac{1}{\left(1+\frac{1}{2}\right)^2}} +2\sqrt{1+\frac{1}{\left(1+\frac{2}{2}\right)^2}} +4\sqrt{1+\frac{1}{\left(1+\frac{3}{2}\right)^2}} +\sqrt{1+\frac{1}{3^2}} \right)$

Then simplifying a bit:

$\displaystyle\int_1^3\sqrt{1+\frac{1}{x^2}}dx \approx \frac{1}{9}\left( \sqrt{1+\frac{1}{1^2}} +4\sqrt{1+\frac{1}{\left(\frac{3}{2}\right)^2}} +2\sqrt{1+\frac{1}{\left(2\right)^2}} +4\sqrt{1+\frac{1}{\left(\frac{5}{2}\right)^2}} +\sqrt{1+\frac{1}{3^2}} \right)$

Then we just do those computations and we finally get the approximation via Simposn's rule:

$\displaystyle\int_1^3\sqrt{1+\frac{1}{x^2}}dx\approx 1.5355453$

While when we do the integral by using technology we get: 2.3020.

The approximation with Simpon’s rule is close but about 33% smaller:

$\displaystyle\frac{2.3020-1.5355453}{2.3020}\cdot100\%\approx 33\%$

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3 years ago

There is a line whose slope is 1/6 and whose y-intercept is 4. what is its equation in slope intetecept form?

Stolb23 [73]

Answer:
y = $\frac{1}{6}$ x + 4

Explanation:
The equation of the lone has a general formula:
y = mx + c
where:
m is the slope of the line = 1/6
c is the y-intercept = 4

Substitute with the givens in the above formula to get the equation as follows:
y = mx + c
y = $\frac{1}{6}$ x + 4

Hope this helps :)

6 0

3 years ago

Read 2 more answers

Quadrilateral ABCD ​ is inscribed in this circle.

Quadrilateral ABCD is inscribed in this circle.