All categories

3 years ago

Angles are preserved in which of the following types of transformations? I. rotations II. reflections III. translations

Mathematics

2 answers:

Nuetrik [128]3 years ago

7 0

Angles are preserved in reflextions

monitta3 years ago

4 0

Answer:

I, II, and III

Step-by-step explanation:

i had this same question on study island

You might be interested in

Right to explain how this digit 6 changes value in number 666, 666, 666

zvonat [6]

To a higher number the six changes value in this number above

8 0

3 years ago

(a) Find a vector parallel to the line of intersection of the planes −4x+2y−z=1 and 3x−2y+2z=1.

valentinak56 [21]

Find the intersection of the two planes. Do this by solving for z in terms of x and y ; then solve for y in terms of x ; then again for z but only in terms of x.

-4x + 2y - z = 1 ==> z = -4x + 2y - 1

3x - 2y + 2z = 1 ==> z = (1 - 3x + 2y)/2

==> -4x + 2y - 1 = (1 - 3x + 2y)/2

==> -8x + 4y - 2 = 1 - 3x + 2y

==> -5x + 2y = 3

==> y = (3 + 5x)/2

==> z = -4x + 2 (3 + 5x)/2 - 1 = x + 2

So if we take x = t, the line of intersection is parameterized by

r(t) = ⟨t, (3 + 5t )/2, 2 + t⟩

Just to not have to work with fractions, scale this by a factor of 2, so that

r(t) = ⟨2t, 3 + 5t, 4 + 2t⟩

(a) The tangent vector to r(t) is parallel to this line, so you can use

v = dr/dt = d/dt ⟨2t, 3 + 5t, 4 + 2t⟩ = ⟨2, 5, 2⟩

or any scalar multiple of this.

(b) (-1, -1, 1) indeed lies in both planes. Plug in x = -1, y = 1, and z = 1 to both plane equations to see this for yourself. We already found the parameterization for the intersection,

r(t) = ⟨2t, 3 + 5t, 4 + 2t⟩

3 0

3 years ago

Find the equation of a line that contains the points (−3,−1) and (−4,−7). Write the equation in slope-intercept form

BARSIC [14]

Answer:

y =6x+17

Step-by-step explanation:

First we need to find the slope

m = (y2-y1)/(x2-x1)

= (-7- -1)/(-4 - -3)

= (-7+1)/(-4+3)

-6/-1

The slope intercept form of the equation is

y = mx+b where m is the slope and b is the y intercept

y = 6x+b

Substitute a point into the equation

-1 = 6(-3) +b

-1 = -18+b

Add 18 to each side

17 =b

The equation is

y =6x+17

8 0

3 years ago

Hey (; I'm taking this rigorous AP Stat course and I've mastered Binomial Sampling but this Geometric Sampling just has me stump

chubhunter [2.5K]

Answer:

1) It is geometric

a) In each trial you can obtain 11 or obtain something else (and fail)

b) Throw 2 dices and watch if the result is 11 or not

c) The probability of success is 1/18

2) It is not geometric, but binomal.

Step-by-step explanation:

1) This is effectively geometric. When you see the sum of 2 dices, you can separate the result in two different outcomes: when the sum is 11 and when the sum is different from 11.

A trial is constituted bu throwing 2 dices and watching if the sum of the dices is 11 or not.

In order to get 11 you need one 5 in one dice and 1 six in another. As a consecuence, you have 2 favourable outcomes (a 5 in the first dice and a 6 in the second one or the other way around). The total amount of outcomes is 6² = 36, and all of them have equal probability. This means that the probability of success is 2/36 = 1/18.

2) This is not geometric distribution. The geometric distribution meassures how many tries do you need for one success. The amount of success in 10 trias follows a binomial distribution.

4 0

3 years ago

Alexander Litvinenko was poisoned with 10 micrograms of the radioactive substance Polonium-210. Since radioactive decay follows

koban [17]

Answer:

The amount of Polonium-210 left in his body after 72 days is 6.937 μg.

Step-by-step explanation:

The decay rate of Polonium-210 is the following:

$N(t) = N_{0}e^{-\lambda t}$ (1)

Where:

N(t) is the quantity of Po-210 at time t =?

N₀ is the initial quantity of Po-210 = 10 μg

λ is the decay constant

t is the time = 72 d

The decay rate is 0.502%, hence the quantity that still remains in Alexander is 99.498%.

First, we need to find the decay constant:

$\lambda = \frac{ln(2)}{t_{1/2}}$ (2)

Where t(1/2) is the half-life of Po-210 = 138.376 days

By entering equation (2) into (1) we have:

$N(t) = N_{0}e^{-\frac{ln(2)}{t_{1/2}}*t}} = 10* \frac{99.498}{100}*e^{-\frac{ln(2)}{138.376}*72} = 6.937 \mu g$

Therefore, the amount of Polonium-210 left in his body after 72 days is 6.937 μg.

I hope it helps you!

8 0

3 years ago