Answer:
One plane has a speed of 450 km/h and the other has a speed of 900 km/h.
Step-by-step explanation:
I am going to say that:
The speed of the first plane is x.
The speed of the second plane is y.
One plane is flying at twice the speed of the other.
I will say that y = 2x. We could also say that x = 2y.
Two airplanes leave an airport at the same time, flying in the same direction
They fly in the same direction, so their relative speed(difference) at the end of each hour is y - x = 2x - x = x.
If after 4 hours they are 1800 km apart, find the speed of each plane
After 1 hour, they will be x km apart. After 4, 1800. So
1 hour - x km apart
4 hours - 1800 km apart
4x = 1800
x = 1800/4
x = 450
2x = 2*450 = 900
One plane has a speed of 450 km/h and the other has a speed of 900 km/h.
Answer:
a) 1/2; reduction
b) 5/4; enlargement
Step-by-step explanation:
In each case, the scale factor is CP'/CP. When it is more than 1, the dilation is an enlargement.
Even before you run the numbers, you can tell if it is an enlargement or not. If the dilated figure is larger, P is closer to C than is P'. If P' is closer to C, then it is a reduction.
a) CP'/CP = (4-2)/4 = 2/4 = 1/2 . . . . a reduction
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b) CP'/CP = 25/20 = 5/4 . . . . an enlargement
g(x)= -|x+2| -1
The domain is the input values, or x values
it is all real number
The range is the output values, or g(x)
the absolute value at is smallest is 0
0-1 is -1 this is the max since the absolute value is multiplied by -1
g(x) <=-1
the output is less than or equal to -1
Step-by-step explanation:
solution.
Let S represent side of the equilateral triangle.
perimeter of equilateral ∆ =3×S
Therefore,if we let width to be represented by W
Width of rectangle=W
Length of rectangle=2W
one side of equilateral triangle= W+8
Therefore after analysing the question, the true statement is; The length of the rectangle is 2W
