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gogolik [260]
3 years ago
5

The statement cot theta = 12/5, sec(theta) = - 13/5 and the terminal point determined by is in quadrant 2 ?

Mathematics
2 answers:
melomori [17]3 years ago
8 0

Answer:

answer: A, cannot be true because cot ∅ is less than zero in quadrant 2.

Step-by-step explanation:

slavikrds [6]3 years ago
8 0
Anwser is a don’t know how to explain tho
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Please help answer these questions
vfiekz [6]

<u>Step</u><u> </u><u>1</u>

given

<u>Step</u><u> </u><u>2</u>

\angle HKM \cong \angle OKM, \angle HMK \cong \angle OMK

<u>Step</u><u> </u><u>3</u>

\overline{KM} \cong \overline{KM}

Reason: Reflexive property

<u>Step</u><u> </u><u>4</u>

ASA

8 0
2 years ago
What is the 1st term when −5z2 + 7z4 + 11z − 8z3 is arranged in descending order
patriot [66]
Descending order...largest to smallest
the largest one is the one with the biggest exponent
so the first term is : 7z^4
3 0
3 years ago
Read 2 more answers
Describe parallelograms in 5 sentence
mixer [17]

Answer:

Their lines never touch. They have 2 different sides. They are like a square but differen. They always have even number.

Step-by-step explanation:

3 0
3 years ago
Read 2 more answers
What are the coordinates of P' when you translate P 4 units to the right and 3 unit down?
OLga [1]

Answer:

(3,2)

Step-by-step explanation:

The coordinates of the point P on the given graph is (-1,5).

So, when we translate point P on the graph by 4 units to the right its x-coordinate will change to (- 1 + 4) = 3.

Again, when we translate point P on the graph by 3 units down then its y-coordinate will change to (5 - 3) = 2.

Therefore, the new coordinates of the point P will be (3,2) (Answer)

4 0
3 years ago
Scores on a university exam are Normally distributed with a mean of 78 and a standard deviation of 8. The professor teaching the
mafiozo [28]

Answer:

Porcentage of students score below 62 is close to 0,08%

Step-by-step explanation:

The rule

68-95-99.7

establishes:

The intervals:

[ μ₀ - 0,5σ ,  μ₀ + 0,5σ] contains 68.3 % of all the values of the population

[ μ₀ - σ ,  μ₀ + σ]   contains 95.4 % of all the values of the population

[ μ₀ - 1,5σ ,  μ₀ + 1,5σ] contains 99.7 % of all the values of the population

In our case such intervals become

[ μ₀ - 0,5σ ,  μ₀ + 0,5σ]   ⇒  [ 78 - (0,5)*8 , 78 + (0,5)*8 ]  ⇒[ 74 , 82]

[ μ₀ - σ ,  μ₀ + σ]  ⇒ [ 78 - 8 , 78 +8 ]   ⇒  [ 70 , 86 ]

[ μ₀ - 1,5σ ,  μ₀ + 1,5 σ]  ⇒ [ 78 - 12 , 78 + 12 ]  ⇒ [ 66 , 90 ]

Therefore the last interval

[ μ₀ - 1,5σ ,  μ₀ + 1,5 σ]    ⇒  [ 66 , 90 ]

has as lower limit 66 and contains 99.7 % of population, according to that the porcentage of students score below 62 is very small, minor than 0,15 %

100 - 99,7  = 0,3 %

Only 0,3 % of population is out of   μ₀ ± 1,5 σ, and by symmetry 0,3 /2 = 0,15 % is below the lower limit, 62 is even far from 66 so we can estimate, that the porcentage of students score below 62 is under 0,08 %

6 0
3 years ago
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