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2 years ago

Given: Circle X with a Radius r and circle Y with radius s Prove: Circle x is similar to circle y

Mathematics

2 answers:

lana [24]2 years ago

8 0

Two figures are similar if one is the scaled version of the other.

This is always the case for circles, because their geometry is fixed, and you can't modify it in anyway, otherwise it wouldn't be a circle anymore.

To be more precise, you only need two steps to prove that every two circles are similar:

Translate one of the two circles so that they have the same center
Scale the inner circle (for example) unit it has the same radius of the outer one. You can obviously shrink the outer one as well

Now the two circles have the same center and the same radius, and thus they are the same. We just proved that any two circles can be reduced to be the same circle using only translations and scaling, which generate similar shapes.

Recapping, we have:

Start with circle X and radius r
Translate it so that it has the same center as circle Y. This new circle, say X', is similar to the first one, because you only translated it.
Scale the radius of circle X' until it becomes $s$ . This new circle, say X'', is similar to X' because you only scaled it

So, we passed from X to X' to X'', and they are all similar to each other, and in the end we have X''=Y, which ends the proof.

stealth61 [152]2 years ago

5 0

Answer:

Step-by-step explanation:

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Two window washers start at the heights shown. (A: 21 ft high rising 8 in per second. The other is 50 feet high descending 11inc

babunello [35]

For this case, the first thing we are going to do is write the generic equation of motion for the vertical axis.

We have then:

$h = \frac {1} {2} gt ^ 2 + vo * t + h0$

Where,

g: acceleration of gravity
vo: initial speed
h0: initial height

For the first body:

$h1 = \frac {1} {2} gt ^ 2 + \frac {8} {12} * t + 21$

For the second body:

$h2 = \frac {1} {2} gt ^ 2 - \frac {11} {12} * t + 50$

By the time both bodies have the same height we have:

$h1 = h2\\$

$\frac {1} {2} gt ^ 2 + \frac {8} {12} * t + 21 = \frac {1} {2} gt ^ 2 - \frac {11} {12} * t + 50$

Rewriting we have:

$\frac {8} {12} * t + 21 = - \frac {11} {12} * t + 50$

$\frac {8} {12} * t + \frac {11} {12} * t = 50 - 21$

$\frac {19} {12} * t = 29$

Clearing time:

$t = 29 (\frac {12} {19})\\t = 18.31s$

Answer:

it takes 18.31s for the two window washers to reach the same height

6 0

2 years ago

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Seventy percent of a town's population voted in an election. If 1,589 people voted, what is the population of the town?

Crank

The total population of the town is 2270 people.

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2 years ago

If 25% of 150 is equal to 700 percent of n, then n is equal to what number?

adelina 88 [10]

25% of 150 = 37.5
700% of x = 37.5
x = 5.35714285

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2 years ago

If a certain number of Grade IV students share 28 or 39 skill books in Mathematics,there are remainders of 4 and 3 books. What i

nadya68 [22]

Answer:

12 students

Step-by-step explanation:

Below numbers are divisible by the number of students:

28 - 4 = 24
39 - 3 = 36

Lets find GCF of 24 and 36:

24 = 2*2*2*3
36 = 2*2*3*3
GCF(24,36) = 2*2*3 = 12

The largest possible number of students is 12

7 0

1 year ago

Find the real numbers x and y if -3+ix^2y and x^2+y+4i are conjugate of each other. Pls solve with the steps

Firdavs [7]

ANSWER
x = ±1 and y = -4.
Either x = +1 or x = -1 will work

EXPLANATION
If -3 + ix²y and x² + y + 4i are complex conjugates, then one of them can be written in the form a + bi and the other in the form a - bi. In other words, between conjugates, the imaginary parts are same in absolute value but different in sign (b and -b). The real parts are the same

For -3 + ix²y
⇒ real part: -3
⇒ imaginary part: x²y

For x² + y + 4i
⇒ real part: x² + y (since x, y are real numbers)
⇒ imaginary part: 4

Therefore, for the two expressions to be conjugates, we must satisfy the two conditions.

Condition 1: Imaginary parts are same in absolute value but different in sign. We can set the imaginary part of -3 + ix²y to be the negative imaginary part of x² + y + 4i so that the

x²y = -4 ... (I)

Condition 2: Real parts are the same

x² + y = -3 ... (II)

We have a system of equations since both conditions must be satisfied

x²y = -4 ... (I)
x² + y = -3 ... (II)

We can rearrange equation (II) so that we have

y = -3 - x² ... (II)

Substituting into equation (I)

x²y = -4 ... (I)
x²(-3 - x²) = -4
-3x² - x⁴ = -4
x⁴ + 3x² - 4 = 0
(x² + 4)(x² - 1) = 0
(x² + 4)(x-1)(x+1) = 0

Therefore, x = ±1.
Leave alone (x² + 4) as it gives no real solutions.

Solve for y:

y = -3 - x² ... (II)
y = -3 - (±1)²
y = -3 - 1
y = -4

So x = ±1 and y = -4. We can confirm this results in conjugates by substituting into the expressions:

-3 + ix²y
= -3 + i(±1)²(-4)
= -3 - 4i

x² + y + 4i
= (±1)² - 4 + 4i
= 1 - 4 + 4i
= -3 + 4i

They result in conjugates

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2 years ago

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