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3 years ago

What are the coordinates of the circumcenter of a triangle with vertices A(−1, 1) , B(5, 1) , and C(−1, −1) ?

1 answer:

7 0

Triangle ABC is a right triangle with vertex A containing the right angle.

You can tell because AB is a horizontal line since A and B have the same y-coordinate, and AC is a vertical line since A and C have the same x-coordinate.

A right triangle in a circumcircle is always in the semicircle, with the centre of the circle lying in the middle of the side opposite the right angle.

So, the circumcenter is the midpoint of BC.

The midpoint of BC is ((5-1)/2, (1-1)/2) = (2, 0)

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Help me on this problem

Wittaler [7]

Answer:

1 and 3 are both perpendicular to segment NY

Step-by-step explanation:

1. Find the slope of line NY

slope of NY = 5-(-7)/-11-5 = - 3/4

Any line that is perpendicular to NY should have a slope of the inverse of negative slope of NY.

2.Find the slope of perpendicular lines

the inverse of negative slope of NY = - (-4/3) = 4/3

3 0

3 years ago

Can you plz solve this

Lesechka [4]

Answer:

a) acute angle

b) obtuse angle

c) right angle

d) reflex angle

e) straight line angle

f) acute angle

Step-by-step explanation:

Acute angle

angle that is smaller than 90°
0° < θ < 90°

Right angle

angle that is 90°
shaped of a "L"

Obtuse angle

angle that is greater than 90° but smaller than 180°
90° < θ < 180°

Straight line angle

angle that is 180°
drawn in a straight line

Reflex angle

angle than is greater than 180°
θ > 180°

8 0

3 years ago

Solve only if you know the solution and show work.

SashulF [63]

$\displaystyle\int\frac{\cos x+3\sin x+7}{\cos x+\sin x+1}\,\mathrm dx=\int\mathrm dx+2\int\frac{\sin x+3}{\cos x+\sin x+1}\,\mathrm dx$

For the remaining integral, let $t=\tan\dfrac x2$ . Then

$\sin x=\sin\left(2\times\dfrac x2\right)=2\sin\dfrac x2\cos\dfrac x2=\dfrac{2t}{1+t^2}$
$\cos x=\cos\left(2\times\dfrac x2\right)=\cos^2\dfrac x2-\sin^2\dfrac x2=\dfrac{1-t^2}{1+t^2}$

and

$\mathrm dt=\dfrac12\sec^2\dfrac x2\,\mathrm dx\implies \mathrm dx=2\cos^2\dfrac x2\,\mathrm dt=\dfrac2{1+t^2}\,\mathrm dt$

Now the integral is

$\displaystyle\int\mathrm dx+2\int\frac{\dfrac{2t}{1+t^2}+3}{\dfrac{1-t^2}{1+t^2}+\dfrac{2t}{1+t^2}+1}\times\frac2{1+t^2}\,\mathrm dt$

The first integral is trivial, so we'll focus on the latter one. You have

$\displaystyle2\int\frac{2t+3(1+t^2)}{(1-t^2+2t+1+t^2)(1+t^2)}\,\mathrm dt=2\int\frac{3t^2+2t+3}{(1+t)(1+t^2)}\,\mathrm dt$

Decompose the integrand into partial fractions:

$\dfrac{3t^2+2t+3}{(1+t)(1+t^2)}=\dfrac2{1+t}+\dfrac{1+t}{1+t^2}$

so you have

$\displaystyle2\int\frac{3t^2+2t+3}{(1+t)(1+t^2)}\,\mathrm dt=4\int\frac{\mathrm dt}{1+t}+2\int\frac{\mathrm dt}{1+t^2}+\int\frac{2t}{1+t^2}\,\mathrm dt$

which are all standard integrals. You end up with

$\displaystyle\int\mathrm dx+4\int\frac{\mathrm dt}{1+t}+2\int\frac{\mathrm dt}{1+t^2}+\int\frac{2t}{1+t^2}\,\mathrm dt$
$=x+4\ln|1+t|+2\arctan t+\ln(1+t^2)+C$
$=x+4\ln\left|1+\tan\dfrac x2\right|+2\arctan\left(\arctan\dfrac x2\right)+\ln\left(1+\tan^2\dfrac x2\right)+C$
$=2x+4\ln\left|1+\tan\dfrac x2\right|+\ln\left(\sec^2\dfrac x2\right)+C$

To try to get the terms to match up with the available answers, let's add and subtract $\ln\left|1+\tan\dfrac x2\right|$ to get

$2x+5\ln\left|1+\tan\dfrac x2\right|+\ln\left(\sec^2\dfrac x2\right)-\ln\left|1+\tan\dfrac x2\right|+C$
$2x+5\ln\left|1+\tan\dfrac x2\right|+\ln\left|\dfrac{\sec^2\dfrac x2}{1+\tan\dfrac x2}\right|+C$

which suggests A may be the answer. To make sure this is the case, show that

$\dfrac{\sec^2\dfrac x2}{1+\tan\dfrac x2}=\sin x+\cos x+1$

You have

$\dfrac{\sec^2\dfrac x2}{1+\tan\dfrac x2}=\dfrac1{\cos^2\dfrac x2+\sin\dfrac x2\cos\dfrac x2}$
$\dfrac{\sec^2\dfrac x2}{1+\tan\dfrac x2}=\dfrac1{\dfrac{1+\cos x}2+\dfrac{\sin x}2}$
$\dfrac{\sec^2\dfrac x2}{1+\tan\dfrac x2}=\dfrac2{\cos x+\sin x+1}$

So in the corresponding term of the antiderivative, you get

$\ln\left|\dfrac{\sec^2\dfrac x2}{1+\tan\dfrac x2}\right|=\ln\left|\dfrac2{\cos x+\sin x+1}\right|$
$=\ln2-\ln|\cos x+\sin x+1|$

The $\ln2$ term gets absorbed into the general constant, and so the antiderivative is indeed given by A,

$\displaystyle\int\frac{\cos x+3\sin x+7}{\cos x+\sin x+1}\,\mathrm dx=2x+5\ln\left|1+\tan\dfrac x2\right|-\ln|\cos x+\sin x+1|+C$

5 0

3 years ago

Which number completes the square?<br> 2x^2 - 12x +

Fofino [41]

67-x-20 hope this helped!!

7 0

2 years ago

Given f(x) =4x-5,find f(3)

Crazy boy [7]

Answer:

f(3) = 7

Step-by-step explanation:

Substitute x = 3 into f(x) , that is

f(3) = 4(3) - 5 = 12 - 5 = 7

8 0

2 years ago