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loris [4]
3 years ago
6

What is the rule for reflection across the y axis

Mathematics
1 answer:
maw [93]3 years ago
3 0

Answer:

(x, y) ----> (-x, y).

Step-by-step explanation:

(x, y) ----> (-x, y).

The y value stays the same but the x-value changes sign.

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If cos a>0 and cot a is undefined then where does the terminal side of a located?
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Answer:C

Therefore: In Quadrant I, cos(θ) > 0, sin(θ) > 0 and tan(θ) > 0 (All positive). For an angle in the second quadrant the point P has negative x

Step-by-step explanation:

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Find the mean for each set of data. Round to the nearest tenth if necessary.
valkas [14]

Answer:

15

Step-by-step explanation:

yes.

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25−3x=40 what does x equal
Anna007 [38]

Answer:

-5

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25-3x=40

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4 0
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Read 2 more answers
Find the missing side length BM. Round answer to the nearest tenth.
xenn [34]

Answer: BM = 21.4

Step-by-step explanation:

Considering the given triangle BMS, to determine angle BM, we would apply the sine rule. It is expressed as

m/SinM = s/SinS = b/SinB

Where m, s and b are the length of each side of the triangle and angle M, Angle S and angle B are the corresponding angles of the triangle.

From the information given,

Angle M = 102°

Angle B = 35°

Angle S = 180 - (102 + 35) = 43°

b = MS = 18

s = BM

Therefore

18/Sin 35 = BM/Sin 43

Cross multiplying, it becomes

BMSin35 = 18 × Sin 43

BM × 0.5736 = 18 × 0.6820

0.5736BM = 12.276

Dividing the left hand side and the right hand side of the equation by 0.5736, it becomes

0.5736BM/0.5736 = 12.276/0.5736

BM = 21.4

6 0
3 years ago
Restless Leg Syndrome and Fibromyalgia
Anna11 [10]

Answer:

The probability that a person with restless leg syndrome has fibromyalgia is 0.183.

Step-by-step explanation:

Denote the events as follows:

<em>F</em> = a person with fibromyalgia

<em>R</em> = a person having restless leg syndrome

The information provided is as follows:

P (R | F) = 0.33

P (R | F') = 0.03

P (F) = 0.02

Consider the tree diagram attached below.

Compute the probability that a person with restless leg syndrome has fibromyalgia as follows:

P(F|R)=\frac{P(R|F)P(F)}{P(R|F)P(F)+P(R|F')P(F')}

            =\frac{(0.33\times 0.02)}{(0.33\times 0.02)+(0.03\times 0.98)}\\\\=\frac{0.0066}{0.0066+0.0294}\\\\=0.183333\\\\\approx 0.183

Thus, the probability that a person with restless leg syndrome has fibromyalgia is 0.183.

3 0
3 years ago
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