If 2 triangles are similar, the ratio of one side of the triangle to the corresponding side of the other is always same for all 3 sides.
We can make use of this rule to calculate the length of sides.
From the image,
3 / 15 = 6 / x
Simplify,
1 / 5 = 6 / x
Use cross method to solve this.
5 x 6 = 1 x x
X = 30
Therefore the length of x is 30ft., A.
To answer this question, you need to know that a quarter is worth $0.25 and dimes worth $0.10. In this case, the total number of money that made from quarters and dimes is $2.80
With Q=quarter and D=dime, then the equation should be:
2.80= 0.25 Q + 0.10 D
The condition that needs to be fulfilled in this case is that the quarter should be 8 more than dimes. Then you need to find out the highest possible number of the quarter and lowest possible number of the dimes.
If you divide 2.8 with 0.25 you will found it will be 11.2 but 11 quarter will result with 0.05 value which you cannot remove, so the highest possible of the quarter should be 10.
If the highest possible of the quarter is 10, then the lowest possible of dimes should be:
2.8= 0.25(10) + 0.10D
0.10D= 0.3
D=3
Since 10-3 is 7, then it is not possible for the quarter to be 8 more than the dimes.
Answer:
x = -7 2/3, y = 1 1/3 and z = 5 1/3.
Step-by-step explanation:
2x+4y+3z=6 ..... 1
x-2y+z=-5 ...... 2
-x-3y-2z=-7 .......3
Add equations 2 and 3 to eliminate x:
-5y - z = -12 .....4
Multiply equation 2 by - 2:
-2x + 4y - 2z = 10
Add this to equation 1:
8y + z = 16 ........ 5
Now add equation 4 to equation 5:
3y = 4
y = 4/3 = 1 1/3.
Now find z by substituting for y in equation 4:
-5(4/3) - z = -12
z = 12 - 20/3
z = 36/3 - 20/3 = 16/3 = 5 1/3.
Finally, we find x by substituting for y and z in equation 1:
2x + 4*4/3 + 3*16/3 = 6
2x = 6 - 16/3 - 16
2x = 18/3 - 16/3 - 48/3 = -46/3
x = 23/3 = 7 2/3.
Domain: (-∞,∞)
Range: (3,∞)
x-intercepts: none
y-intercepts: (0,7)
Interval positive: (3,∞)
Interval negative: none
Interval increasing: (7,∞)
Interval decreasing: (-∞,7)
I'm not sure what the average rate of change over is though.
