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| Give the standard form of the equation of the line that is perpendicular to y=6/5x+4 and contains (-8,4). |
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Definition
The given line has a slope of 6/5
so the line perpendicular must have a slope of -5/6
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form of the equation of the line is y=mx+b and you have y,x,and m.
4=(-5/6)(-8)+b
4=(20/3)+b
b=-8/3
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EQUATION:
y=(-5/6)x - 8/3
To put in standard form, multiply thru by 6 to get:
6y = -5x -16
5x+6y=-16 (standard form) |
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| 1.the length of a rectangular field is 11m more than it's width. if the length is decreased by 12m and the width is increased by 10m, the area decreases by 24m2.find the length and the width of the field. |
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Definition
width =x
length = x+11
length decreased by 12
x+11-12
=x-1
width = x+10
area dcreases by 24
Original area - changed area = 24
x(x+11)-(x-1)(x+10)=24
x^2+11x-(x^2+9x-10)=24
x^2+11x-x^2-9x+10=24
2x+10=24
2x=24-10
2x=14
/2
x=7 m the width
Length = x+11= 18m
..
126-102=24
x=14/20
x=7/10 |
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| The second angle in a triangle is 24 degrees larger then the first. And the third angle is 69 degrees larger than the second. What are the measures of all three angles? |
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Definition
Let 1st be "x":
Then 2nd is "x+24":
And 3rd is "x+24+69"
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Equation:
x + x+24 + x + 93 = 180
3x = 180-117
3x = 63
x = 21 degrees
x+24 = 45 degrees
x+93 = 114 degrees |
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| The equation x^2 + k = 6 x has two distinct real roots. Find the range of values of k |
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Definition
Ans: 2 distict Real roots implies b^2-4ac > 0
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Rearrange the equation:
x^2-6x+k = 0
Solve:; 6^2-4*1*k > 0
36 -4k > 0
36 > 4k
k < 9
That is the range of the values of "k". |
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| A 12 centimeter cube (i.e. 12x12x12) was painted red on the outside. The cube was then cut into one centimeter cubes. How many of these smaller cubes have paint on one side? Two sides? Three sides? No sides? (hint: examine smaller cubes first) show/explain the procedure used. |
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Definition
a cube each side has 12 rows of 12 squares, when you subtract the outside squares (they have paint on 2 and 3 sides, you get 10 rows of 10 squares
by 6 sides there are 600 squares with one side painted
each corner of the square would have 3 sides painted, as there are 8 corners, 8 cubes have 3 sides painted.
now all we have left are the outside painted cubes, because the corners are missing, there is 10 per row, (12 per row subtracting 2 corners in a row=10 per row)
there are 12 rows on a cube, (4 on the top, 4 on the bottom, and 4 on the sides, because they all share sides we can not state 4X6 sides we have to logically subtract the sides.
because there are 12 rows with 10 in a row, there is 120 cubes with 2 sides painted.
last, there are 10 rows, 10 columns, and 10 stacks left, hence there is 1000 unpainted cubes. here is the completed list.
8=cubes with 3 painted sides
120=cubes with 2 painted sides
600=cubes with 1 painted side
1000=cubes with no painted sides.
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1728 cubes in total
check, 12x12x12=1728, so we have accounted for all the squares.
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