What is the equation of the line that is parallel to the line y − 1 = 4(x + 3) and passes through the point (4, 32)?

Find the measure of line segment MK. K 8 I 7 J M 5 2x+1

mrs_skeptik [129]

Hello!

This is a problem about relating circle theorems to line lengths.

We can first see that both line segment MK and CM are secants within the circle that come from a common point K.

This means that the Intersecting Secant Theorem applies here.

The Intersecting Secant Theorem states that if two secants are formed from a common point outside the circle, the length of each secant multiplied by the length of its corresponding external secant are equivalent.

We can set up the following equation.

$8(8+x+5)=7(7+2x+1)$

$8(x+13)=7(2x+8)$

$8x+104=14x+56$

$6x=48$

$x=8$

Using this value, we can find the length of line segment MK.

$MK=x+5+8$

$MK=8+5+8$

$MK=21$

Hope this helps!

5 0

3 years ago

Helppppppppplppppppp

irinina [24]

Answer:

number 1 is 2 number 2 is 15 number 4 is 30

Step-by-step explanation:

6 0

3 years ago

Solve the equation : 2x + 7 = 21 and find the value of x. Is 2x + 7 = 21 a first degree equation or not?

Nady [450]

2x + 7 = 21
first you subtract the 7 from the 7 and 21 which will leave you with 2x=14
then see what multiples with 2 to get 14 which is 7!

7 0

3 years ago

Assume that the number of messages input to a communication channel in an Exponential distribution with 7 messages arriving in a

NARA [144]

The probability that more than 3 messages will arrive during a 30-second interval is $P(X=n)=\dfrac{e^{-0.3 \times 20} (0.3 \times 20)^n}{n!}$ .

According to the statement

we have given that the an Exponential distribution with 7 messages arriving in a 10 second period and we have to find the probability that more than 3 messages will arrive during a 30-second interval.

So, For this purpose, we know that the

The probability is the measure of the likelihood of an event to happen. It measures the certainty of the event.

And the given information is that :

3 messages will arrive during a 30-second interval.

Then

Probability = P(X=1) + P(X=2) + P(X=3).Then

The probability become according to the exponential distribution:

$P(X=1)=\dfrac{e^{-0.3 \times 20} (0.3 \times 20)^1}{1!}\\P(X=2)=\dfrac{e^{-0.3 \times 20} (0.3 \times 20)^2}{2!}\\P(X=3)=\dfrac{e^{-0.3 \times 20} (0.3 \times 20)^3}{3!}\\$

And then substitute the values in it then

$Probability = \dfrac{e^{-0.3 \times 20} (0.3 \times 20)^1}{1!}\ +\dfrac{e^{-0.3 \times 20} (0.3 \times 20)^2}{2!}\ + \dfrac{e^{-0.3 \times 20} (0.3 \times 20)^3}{3!}\\$

This is the probability.

So, The probability that more than 3 messages will arrive during a 30-second interval is $P(X=n)=\dfrac{e^{-0.3 \times 20} (0.3 \times 20)^n}{n!}$ .

Learn more about probability here

brainly.com/question/24756209

#SPJ4

7 0

2 years ago

Which statement describes the following expression? 7 x (2,304 x 681)

pantera1 [17]

7 times the value of 2,304 times 681 ...

This is your answer...

Hope it helped you... pls mark brainliest also

7 0

3 years ago