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Simora [160]
3 years ago
5

If point A is between B and C then BA + AC = BC. always sometimes never

Mathematics
2 answers:
Romashka [77]3 years ago
7 0
Ur right it is always that
vladimir2022 [97]3 years ago
4 0

Answer:

Always.

Step-by-step explanation:

If a point lies  in between  two end points of a line then the sum of distance between the points in between is always equal to the whole line.

If A is in between two points B and C then the sum of line segments BA and AC will always equal to the length of BC.

Or BA+AC=BC in all cases whatever be the position of point A between the points Band C.

Always is the right option.


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How do I verify #15 using fundamental trig identities?
expeople1 [14]

We start with the more complicated side which is the left side, and show that, on using some trigonometric identities, we will get the term on the right side .

\frac{sin \theta + tan \theta}{1+cos \theta}

Using Quotient identity for tangent function, we will get

\frac{sin \theta+ \frac{sin \theta}{cos \theta}}{1+cos \theta}

\frac{sin \theta cos \theta + sin \theta}{cos \theta(1+cos \theta)}

Taking out sine function from the numerator

=\frac{sin \theta(1+cos \theta)}{cos \theta(1+cos \theta)}

Cancelling the common term of numerator and denominator

=\frac{sin \theta}{cos \theta} = tan \theta

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3 years ago
A) Se, com 10 quilogramas de laranja, é possível fazer 6 litros de suco,
julia-pushkina [17]

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2 years ago
Write the equation of a line, in slope-intercept form, that has a slope of m=9 and passes through the point (4, 31)
Darina [25.2K]

Answer:

9x-y = 5

Step-by-step explanation:

Equation of line in slope intercept form:

y-y1 = m*(x-x1), m=9, (x1,y1)=(4,31)

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4 0
3 years ago
Someone help with this
elena-14-01-66 [18.8K]

The quadrants are I, II, III, and IV, (meaning 1, 2, 3, and 4). The first quadrant is in the upper right, which has both positive x and positive y values. The second quadrant is the upper left, which has negative x and positive y values. The third quadrant is the lower left, which has both negative x and y values. The fourth quadrant is the lower right, which has positive x and negative y values. Using this knowledge and our positive and negative signs, the following are the answers to questions 1-6.

1) (-4, -2) - Quadrant III, both x and y are negative

2) (0, -7) - This point is actually on the y-axis. The x value is 0 and the y value is -7, so the graph is 7 units down from the origin on the y-axis.

3) (0,0) - This point is the origin, or where the x and y axes cross (the middle of the graph).

4) (6, -9) - This point is in Quadrant IV, because it has a positive x value and a negative y value

5) (3,5) - This point is in Quadrant I, because both the x and y values are positive.

6) (8,0) - This point is on the x-axis. The y-value is zero, so this point is 8 units to the right of the origin between Quadrant I and Quadrant IV.

Using the knowledge presented above, to graph the points given to you in the second part of the problem, first you can figure out what quadrant or part of the graph the point is on. Then, you can count the number of units (squares on the graph) in the right direction (remember that up is positive on the y-axis and down is negative, and to the right is positive on the x-axis and to the left is negative) in order to plot the points. Then, you must connect the points that correspond to the same figure in order to create the figures.

Please comment if you have any questions!

Hope this helps!

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On Halloween, Mr. McCoy and two of his children ate a 6 pound bag of
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They would have each had around 2 pounds each
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2 years ago
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