Either perpendicular lines or linear pairs I️t depends on which one you use. Perpendicular lines are lines in which one line goes straight through another and forms right angles. Transversals can be slanted which would form linear pairs
Answer:
Step-by-step explanation:
Keep in mind that the tangent function is negative in Quadrants II and IV.
Of the three possible answers, the 2nd is invalid because the tangent is positive in Q III.
If the reference angle were in Quadrant II and had the value 30° (second possible answer), then the opposite side would be +1, the adjacent side -√3 and the hypotenuse 2. Based on this, the tangent of theta would be opp / adj, or +1 / (-√3) or -√3/3. This is correct.
48=4% loaded on truck and they have 96% left to load
So set up a proportion and it should look like this
48 over x = 4 over 100
48/x=4/100
Do cross multiplication 100×48= 4,800÷ 4 = 1,200 total on truck
Answer:
The absolute value of the quadratic term, 
is less than 1
Step-by-step explanation:
The given function is y = (-1/2)·x² - 7
The parent function is y = x²
The vertical compression or stretching of a quadratic function is given by the value of the coefficient, <em>a</em>, of the quadratic term, x² of a quadratic function, a·x²
A quadratic function is vertically compressed if the coefficient, 
< 1.
In the given function, y = (-1/2)·x² - 7, the absolute value of the coefficient of the quadratic term, < 1, therefore, the equation, y = (-1/2)·x² - 7, will be vertically compressed compared to the parent function, y = x².
Answer:
3 patients had all the three complaints
Step-by-step explanation:
Let U be the set of patients who reported at the hospital on that day
Let F be the set of patients who complained of fever
Let S be the set of patients who had stomach troubles
Let I be the set of injured patients
Then the given data can be written as:
- n(U) = n(F∪S∪I) = 100
- n(F) = 70
- n(S) = 50
- n(I) = 30
- n(F∩S) + n(S∩I) + n(I∩F) - 3×n(F∩S∩I) = 44
n(F∩S∩I) = ?
Using the formula for the cardinal number of union of three sets:
n(F∪S∪I) = n(F) + n(S) + n(I) - n(F∩S) - n(S∩I) - n(I∩F) + n(F∩S∩I)
100 = 70 + 50 + 30 - (44 + 3×n(F∩S∩I)) + n(F∩S∩I)
100 = 150 - 44 - 2×n(F∩S∩I)
2×n(F∩S∩I) = 106 - 100 = 6
<u>n(F∩S∩I) = 3</u>
