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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X^3 / X^5 = X^ -2
you have to subtract the exponents when you’re dividing
To find the root, replace y with 0
X^2-12x+35=0
A=1 B=-12 C=35
B^2-4ac=(-12)^2 -4(1)(35)
=144 -140
=4
x=(-b+/- square root of b^2 -4ac) /over/ (2a)
Plug in the numbers
x=-(-12) sqr (-12)^2 4(1)(35) / (2)(1)
X=12 +/-sqr 4 / 2
Positive outcome
x=12 + sqr 4 / 2
x=12+2/2
x=7 <— this one
Negative outcome
x=12-2i/2
x=6-i
Vertex: (6,-1)
Here’s the work - answer is 0.833 mi
The term that can be added to the list so the GCF is 12h3 would be 48h5.
The reason being is that 48 is first divisible by 12 and does not yield a fraction, and we can remove upon dividing 3 h's from this term as it contains a total of 5 h's.
