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IrinaK [193]
3 years ago
14

angle 3 and angle 4 form a linear pair. the measure of angle 3 is four more than three times the measure of angle 4. find the me

asure of each angle
Mathematics
1 answer:
lisabon 2012 [21]3 years ago
4 0
Since angles 3 and 4 form a linear pair, the sum of the measures of their angles must add up to 180. 

We can define angle 3 as x, and angle 4 as (180-x). 

x = 3 ( 180 - x ) + 4.

Solving for x, we get x=138. This is angle 3. Angle 4 is 180-138, so angle 4 is 42.
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Answer:

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Step-by-step explanation:

\dfrac{7x}{3}

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The line with equation a + 4b = 0 coincides with the terminal side of an angle θ in standard position and cos θ>0 . What is t
Nikitich [7]

Answer:

\frac{-1}{\sqrt{17} }

Step-by-step explanation

We have given that equation of line a+4b=0

⇒4b= -a

⇒b= \frac{-a}{4}

slope of line = tan\theta =\frac{P}{B}= \frac{-1}{4}

⇒ tan\theta is negative therefore it lies into 4th quadrant

By, Pythagoras theorem H^2=P^2+B^2

by putting value of P= -1 & B=4

the value of H=\sqrt{17}

with the given condition cos\theta>0

i.e. \theta∈ (3π/2, 2π)

now, sin\theta= \frac{P}{H}= \frac{-1}{\sqrt{17}}



6 0
3 years ago
Mrs. Purdue needs to pick three random students in her class to be representatives on the student council. Explain how Mrs. Purd
Xelga [282]

Answer:

See below

Step-by-step explanation:

She could have each student in the class write his or her name on a slip of paper and put it in a bowl. Then, she could shake the bowl and pull out any slip of paper with a student's name on it. After 3 times, the random sample would be over, and Mrs. Purdue would have her 3 representatives.

5 0
2 years ago
ive males with an​ X-linked genetic disorder have one child each. The random variable x is the number of children among the five
Nonamiya [84]

Answer:

a   c    d  

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5 0
3 years ago
A biology test is worth 100 points and has 36 questions.
Ad libitum [116K]

Answer:

(a) 7 essays and 29 multiple questions

(b) Your friend is incorrect

Step-by-step explanation:

Represent multiple choice with M and essay with E.

So:

M + E= 36 --- Number of questions

2M + 6E = 100 --- Points

Solving (a): Number of question of each type.

Make E the subject of formula in M + E= 36

E = 36 - M

Substitute 36 - M for E in 2M + 6E = 100

2M + 6(36 - M) = 100

2M + 216 - 6M = 100

Collect Like Terms

2M - 6M = 100 - 216

-4M = - 116

Divide both sides by -4

M = \frac{-116}{-4}

M = 29

Substitute 29 for M in E = 36 - M

E = 36 - 29

E = 7

Solving (b): Can the multiple questions worth 4 points each?

It is not possible.

See explanation.

If multiple question worth 4 points each, then

2M + 6E = 100 would be:

4M + xE = 100

Where x represents the number of points for essay questions.

Substitute 7 for E and 29 for M.

4 * 29 + x * 7 = 100

116 + 7x = 100

Subtract 116 from both sides

116-116 + 7x = 100 -116

7x = 100-116

7x = -16

Make x the subject

x = -\frac{16}{7}

Since the essay question can not have worth negative points.

Then, it is impossible to have the multiple questions worth 4 points

<em>Your friend is incorrect.</em>

6 0
3 years ago
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