Someone please help me with angle relationships in circles with algebra.

Mathematics

1 answer:

Mrac [35]3 years ago

6 0

If angle TSV=6x+17 then arc it intercept (SV)would be double it

6x+17=1/2 (15x+13)
12x+34=15x+13
21=3x
x=7

Arc SV=15(7)+13
=118

And angle TSV=6(7)+17
=59

There are 360 degrees in a circle and the measure of the arc would be equal to the central point so the sum of all the arcs equals 360

Arc SV+ arc SUV=360
118+SUV=360
Arc SUV=242

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The circle given by : x^2 + y^2 - 6y - 12 = 0 can be written as:

deff fn [24]

Step-by-step explanation:

Given equation of circle is,

${x}^{2} + {y}^{2} - 6y - 12 = 0 \\ by \: compairing \: it \: with \: \\ {x}^{2} + {y}^{2} + 2gx + 2fy = 0 \: we \: get \\2 g = 0 \: or \: g = 0 \\ 2f= - 6 \\ or \: f= - 3 \\ c = - 12$

again the another form of circle is,

${x}^{2} + {(y - k)}^{2} = 21 \\ or \: {x}^{2} + {y}^{2} - 2ky + {k}^{2} - 21 = 0 \\ by \: compairing \:it \: with \: \\ {x}^{2} + {y}^{2} + 2gx + 2fy = 0 \: we \: get \\ g = 0 \\f = - k \\ c = {k}^{2} - 21$

now equating the values of f in both equations,

-k=-3

i.e. k=3

therefore k=3

4 0

4 years ago

a slitter assembly contains 48 blades five blades are selected at random and evaluated each day for sharpness if any dull blade

12345 [234]

Answer:

P(at least 1 dull blade)=0.7068

Step-by-step explanation:

I hope this helps.

This is what it's called dependent event probability, with the added condition that at least 1 out of 5 blades picked is dull, because from your selection of 5, you only need one defective to decide on replacing all.

So if you look at this from another perspective, you have only one event that makes it so you don't change the blades: that 5 out 5 blades picked are sharp. You also know that the probability of changing the blades plus the probability of not changing them is equal to 100%, because that involves all the events possible.

P(at least 1 dull blade out of 5)+Probability(no dull blades out of 5)=1

P(at least 1 dull blade)=1-P(no dull blades)

But the event of picking one blade is dependent of the previous picking, meaning there is no chance of picking the same blade twice.

So you have 38/48 on getting a sharp one on your first pick, then 37/47 (since you remove 1 sharp from the possibilities, and 1 from the whole lot), and so on.

Also since are consecutive events, you need to multiply the events.

The probability that the assembly is replaced the first day is:

P(at least 1 dull blade)=1-P(no dull blades)

$P(at least 1 dull blade)=1-(\frac{38}{48}* \frac{37}{47} *\frac{36}{46}*\frac{35}{45}*\frac{34}{44})$