Which of the following functions shows an initial amount of $15 and an increase of 35%
1 answer:
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Answer:
units
Step-by-step explanation:
If we draw a perpendicular from point (6,2) on the line 6x - y = - 3, then we have to find the length of the perpendicular.
We know the formula of length of perpendicular from a point to the straight line ax + by + c = 0 is given by
Therefore, in our case the perpendicular distance is
units. (Answer)
Consider the picture.
Let MN be the midsegment of the trapezoid.
That is M is the midpoint of AD, N is the midpoint of BC.
Being the midsegment of the trapezoid, MN is parallel to the bases.
Let O and K be the intersections of the diagonals with the midsegment.
MN//AB, so MO//AB, and since M is the midpoint of DA, O must be the midpoint of DB,
Similarly we prove that K is the midpoint of CA.
Thus O is F and K is E.
O and K lie on the midsegment MN, so F and E lie on the midsegment.
MO is a midsegment of triangle ABD so |MO|=1/2 |AB|=1/2 * 10=5
MK is a midsegment of triangle ADC, so |MK|=1/2 * |DC|=1/2 * 22=11
|OK|=|MK|-|MO|=11-5=6 (units)
Let MN be the midsegment of the trapezoid.
That is M is the midpoint of AD, N is the midpoint of BC.
Being the midsegment of the trapezoid, MN is parallel to the bases.
Let O and K be the intersections of the diagonals with the midsegment.
MN//AB, so MO//AB, and since M is the midpoint of DA, O must be the midpoint of DB,
Similarly we prove that K is the midpoint of CA.
Thus O is F and K is E.
O and K lie on the midsegment MN, so F and E lie on the midsegment.
MO is a midsegment of triangle ABD so |MO|=1/2 |AB|=1/2 * 10=5
MK is a midsegment of triangle ADC, so |MK|=1/2 * |DC|=1/2 * 22=11
|OK|=|MK|-|MO|=11-5=6 (units)