All categories

2 years ago

Can anybody help me ?????????????????????? please geometry

Mathematics

2 answers:

AVprozaik [17]2 years ago

7 0

Where the angle bisectors intersect determines the center of the incircle.
Where the side perpendicular bisectors intersect determines the center of the center of the circumcircle. From these definitions, you should be able to determine the points you want.

Ulleksa [173]2 years ago

7 0

See, pls, the suggested explanation:
S - the centre of inscribed circle, P - the centre of circumscribed circle.
Details are in the attached picture: red lines for small circle (inscribed), green ones for large (circumscribed).

You might be interested in

Solve for x going step by step .

OleMash [197]

Answer:

x=11

Step-by-step explanation:

We know that the angle is equal to 1/2 the difference of the long arc length between where the line segments touch the circle and the small arc length where the line segments touch the circle

They long distance is 360 - (86+74) = 200

The small distance is 86

4x+13 = 1/2 ( 200 - 86)

4x+13 = 1/2 (114)

4x+13 = 57

Subtract 13 from each side

4x+13-13 = 57-13

4x= 44

Divide by 4

4x/4 = 44/4

x=11

6 0

2 years ago

To the nearest cubic centimeter, what is the volume of the regular hexagonal prism? 3 cm 8 cm 2.6 cm The volume of the regular

blagie [28]

Divide multiple 478 by c to max

4 0

3 years ago

Find the roots of h(t) = (139kt)^2 − 69t + 80

Sonbull [250]

Answer:

The positive value of $k$ will result in exactly one real root is approximately 0.028.

Step-by-step explanation:

Let $h(t) = 19321\cdot k^{2}\cdot t^{2}-69\cdot t +80$ , roots are those values of $t$ so that $h(t) = 0$ . That is:

$19321\cdot k^{2}\cdot t^{2}-69\cdot t + 80=0$ (1)

Roots are determined analytically by the Quadratic Formula:

$t = \frac{69\pm \sqrt{4761-6182720\cdot k^{2} }}{38642}$

$t = \frac{69}{38642} \pm \sqrt{\frac{4761}{1493204164}-\frac{80\cdot k^{2}}{19321} }$

The smaller root is $t = \frac{69}{38642} - \sqrt{\frac{4761}{1493204164}-\frac{80\cdot k^{2}}{19321} }$ , and the larger root is $t = \frac{69}{38642} + \sqrt{\frac{4761}{1493204164}-\frac{80\cdot k^{2}}{19321} }$ .

$h(t) = 19321\cdot k^{2}\cdot t^{2}-69\cdot t +80$ has one real root when $\frac{4761}{1493204164}-\frac{80\cdot k^{2}}{19321} = 0$ . Then, we solve the discriminant for $k$ :

$\frac{80\cdot k^{2}}{19321} = \frac{4761}{1493204164}$

$k \approx \pm 0.028$

The positive value of $k$ will result in exactly one real root is approximately 0.028.

7 0

2 years ago

The probability of A is 1/6 and the probability of B is 1/2. what is the probability of A or B or both happening?

Damm [24]

Answer:

$p(a \: or \: b) = p(a) + p(b) - p(a \: and \: b) \\ p(a \: and \: b) = p(a)p(b)$

Write the above equation in capital in your case, all you need to do is plug in your fractions...

5 0

2 years ago

Why is it necessary to check that n modifyingabove p with caret greater than or equals 5np≥5 and n modifyingabove q with caret g

Nookie1986 [14]

Both of these conditions must be true in order for the assumption that the binomial distribution is approximately normal. In other words, if $np \ge 5$ and $nq \ge 5$ then we can use a normal distribution to get a good estimate of the binomial distribution. If either np or nq is smaller than 5, then a normal distribution wouldn't be a good model to use.

side note: q = 1-p is the complement of probability p

4 0

2 years ago