Whereby circle $\bigodot$ P can be obtained from circle $\bigodot$ O by applying the transformations of a translation of T₍₁₄, ₋₈₎ followed by a dilation by a scale factor of 2.4, $\bigodot$ O is similar to $\bigodot$ P

Step-by-step explanation:

The given center of the circle $\bigodot$ O = (-2, 7)

The radius of $\bigodot$ O, r₁ = 5

The given center of the circle $\bigodot$ P = (12, -1)

The radius of $\bigodot$ P, r₂ = 12

The similarity transformation to prove that $\bigodot$ O and $\bigodot$ P are similar are;

a) Move circle $\bigodot$ O 14 units to the right and 8 units down to the point (12, -1)

b) Apply a scale of S.F. = r₂/r₁ = 12/5 = 2.4

Therefore, the radius of circle $\bigodot$ O is increased by 2.4

We then obtain $\bigodot$ O' with center at (12, -1) and radius r₃ = 2.4×5 = 12 which has the same center and radius as circle $\bigodot$ P

∴ Circle $\bigodot$ P can be obtained from circle $\bigodot$ O by applying similarity transformation of translation of T₍₁₄, ₋₈₎ followed by a dilation by a scale factor of 2.4, $\bigodot$ O is similar to $\bigodot$ P.

7 0

3 years ago

The rent of a room is increased from rupees 2500 to rupees 3000 . Find the percent increase in rent

evablogger [386]

3,000 - 2,500 = 500

500/2,500 = 20%

The percent increase is 20%.

Hope this helps!

If it does it would be really helpful if you could make me brainliest.

3 0

3 years ago

Read 2 more answers

A fair coin is tossed three times and the events A, B, and C are defined as follows: A: \{ At least one head is observed \} B: \

Yanka [14]

Answer:

a) $P(A)=0.875$

b) $\text{P(A or B)}=0.875$

c) $\text{P((not A) or B or (not C))}=0.625$

Step-by-step explanation:

Given : A fair coin is tossed three times and the events A, B, and C are defined as follows: A: At least one head is observed, B: At least two heads are observed, C: The number of heads observed is odd.

To find : The following probabilities by summing the probabilities of the appropriate sample points ?

Solution :

The sample space is

S={HHH,HHT,HTT,HTH,TTT,TTH,THH,THT}

n(S)=8

A: At least one head is observed

i.e. A={HHH,HHT,HTT,HTH,TTH,TTH,THH,THT}

n(A)=7

B: At least two heads are observed

i.e. B={HHH,HTT,TTH,THT}

n(B)=4

C: The number of heads observed is odd.

i.e. C={HHH,HTT,THT,TTH}

n(c)=4

a) Probability of A, P(A)

$P(A)=\frac{n(A)}{n(S)}$

$P(A)=\frac{7}{8}$

$P(A)=0.875$

b) P(A or B)

Using formula,

$\text{P(A or B)}=P(A)+P(B)-\text{P(A and B)}$

$\text{P(A or B)}=\frac{n(A)}{n(S)}+\frac{n(B)}{n(S)}-\frac{\text{n(A and B)}}{n(S)}$

$\text{P(A or B)}=\frac{7}{8}+\frac{4}{8}-\frac{4}{8}$

$\text{P(A or B)}=\frac{7}{8}$

$\text{P(A or B)}=0.875$

(c) P((not A) or B or (not C))

A={HHH,HHT,HTT,HTH,TTH,TTH,THH,THT}

not A = {TTT} = 1

B={HHH,HTT,TTH,THT}

C={HHH,HTT,THT,TTH}

not C = {HHT,HTH,THH,TTT} = 4

So, not A or B or not C = {HHH,HHT,HTH,THH,TTT}=5

$\text{P((not A) or B or (not C))}=\frac{5}{8}$

$\text{P((not A) or B or (not C))}=0.625$

4 0

3 years ago