Answer:
18:162
Step-by-step explanation:
1:9
1+9=10
(1×180)÷10= 18
(9×180)÷10=162
A straight line passing through the point (-2,1) and having a gradient of -3 yields the equation y = -3x - 5.
We know that a straight line is an infinitely long line with no curves on it. A straight line's equation is
y = mx + c...(1), where m is the gradient of the straight line and c is a constant.
Given that the gradient of the given straight line = m = -3.
Putting this value in (1), we get
y = -3x + c ...(2)
Again, the given straight line passes through the point (-2,1). So, we can put x = -2 and y = 1 to get the value of the constant c.
So, 1 = (-3)(-2) + c
i.e. 6 + c = 1
i.e. c = 1 - 6 = -5
(2) can be written as
y = -3x - 5
Therefore the equation of the given straight line is
y = -3x - 5
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Answer:
t=2Q-sr/r
Step-by-step explanation:
We are to make t the subject of formula from the given question
Q=r/2 (s+t)
Q=Sr+tr/2
Cross multiply
2Q=Sr+tr
2Q-sr=tr
t=2Q-sr/r
Yes. If the diagonals bisect the angles, the quadrilateral is always a parallelogram, specifically, a rhombus.
Consider quadrilateral ABCD. If diagonal AC bisects angles A and C, then ΔACB is congruent to ΔACD (ASA). Hence AB=AD and BC=CD (CPCTC).
Likewise, if diagonal BD bisects angles B and D, triangles BDA and BDC are congruent, thus AB=BC and AD=CD. (CPCTC again). Now, we have AB=BC=CD=AD, so the figure is a rhombus, hence a parallelogram.
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