The distance from Humood's house to the school is 500m
<h3>How to determine the distance from Humood's house to the school?</h3>
The given parameters are:
Humood's house is (-5,7)
The school is (3,1)
The distance between both points is calculated using
Substitute the known values in the above equation
Evaluate
d = 10
Each unit in the graph is 50m.
So, we have
Distance =10 * 50m
Evaluate the product
Distance = 500m
Hence, the distance from Humood's house to the school is 500m
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130 - 40 = 90 (because you need to have soil to have flowers to grow)
90/6 = 15
So 15 flowers she can buy
To answer your question, this could be the possible answer and i hope you understand and interpret it correctly:
<span>[Integrate [0, 1/2] xcos(pi*x
let u=x so that du=dx
and v=intgral cos (xpi)dx
v=(1/pi)sin(pi*x)
integration by parts
uv-itgral[0,1/2]vdu just plug ins
(1/pi)sinpi*x]-(1/pi)itgrlsin(pi*x)dx from 0 to 1/2
(1/pi)x sinpi*x - (1/pi)[-(1/pi) cos pi*x] from 0 to 1/2
=(1/2pi)+(1/pi^2)[cos pi*x/2-cos 0]
=1/2pi - 1/2pi^2
=(pi-2)/2pi^2 ans</span>
Answer:
the sides are 5ft long
Step-by-step explanation:
since it is a cube we know each side is the same and the cube root of 125 is 5
<span>The vertex of the parabola is the highest or lowest point of the graph.
</span><span>y=-4x^2+8x-12 = -4 (x^2 -2x +3)
Lets work with this now: </span>x^2 -2x +3
x^2 -2x +3 -> what is the closeset perfect square?
x^2 -2x +1 = (x-1)^2
So
x^2 -2x +3 = (x-1)^2 +2
Replacing to original
y=-4x^2+8x-12 = -4 (x^2 -2x +3) = -4 ((x-1)^2 +2) = -4 (x-1)^2 - 8
The min or max point is where the squared part = 0
So when x=1 , y= -4*0-8=-8
This will be the max of the parabola as there is - for the highest factor (-4x^2)
The max: x=1, y= -8