32m = 8m · 4
56mp = 8m · 7p
32m + 56mp = 8m(4 + 7p)
The first thing we are going to do is find the area of the field. To do this we are going to use the area of a square formula:

Were

is the area in square kilometers

is one of the sides of the square
We know for our problem that the side lengths of the field are 0.9 kilometers, so

. Lets replace that value in our formula to find

:

Now, to find the population density of the filed, we are going to use the population density formula:

where

is the population density in <span>in burrows per square kilometer
</span>

is the number of burrows

is the are of the field
We know that

and

, so lets replace those values in our formula:


We can conclude that the <span>density of prairie dog burrows is approximately
2444 burrws per square kilometer.</span>
A] Given that the last years's sales was $144,600 and this years sales should increase by 1/3. Then:
i] Amount the sales should increased by will be:
(last year's sales)*(increase)
=144,600*(1/3)
=48,200
ii] The sales in the new year will be:
(last year's sales)+(increase)
=144600+48600
=$192, 800
2] Given that the sales of hifi which included 6% tax was 205,000. The actual sales was:
Actual percentage sales=100%
percentage sales after taxation=100-6=94%
thus the actual sales was:
(100)/(94)*205,000
=218, 085.1064
3]Given that the rate per $100 is $0.83, and the insurance was for 90000, the insurance premium will be:
(total insurance) *(unit rate)/(number of units)
plugging the values we obtain:
90000*0.83/100
$747
Let the initial point of the vector be (x,y). Then the magnitude of the vector v can be written as:

The magnitude of vecor v is given to be 10. So we can write:

Now from the given options, we have to check which one satisfies the above equation. That point will be the initial point of the vector.
The point in option d, satisfies the equation.
Thus, the answer to this question is option D
Hello,
A: roots: -1,-3
a point (-2,1)
Vertex=((-2,1)
y=k*(x+1)(x+3) using roots
but k*(-2+1)(-2+3)=1==>k*(-1)*1=1==>k=-1
eq: y=-(x+1)(x+3)
==>y=-(x²+3x+x+3)
==>y=-x²-4x-3
y=k(x+2)²+1 if x=-1,y=0 ==>k*1+1=0==>k=-1
==>y=-(x+2)²+1
Answer :A--> R,K
B)
y=k(x+4)²-2 and k=-1/2
y=-1/2(x+4)²-2
y=-1/2x²-4x-10
answer B--> I,≈W if it is written -1/2*x² (square has been forgotten)
C:
y=2x²-16x+30
y=2(x-4)²-2
answer : C-->S,J
D:
y=-(x+3)(x+1)
y=-x²-4x-3
=-(x+2)²+1
answer D--> V,L
E:
Here there is a problem: or the graph is wrong, or 2 equations are missing!
y=1(x+1)(x-3) using roots
y=x²-2x-3 ≈ T si it were -2x and not +2x.
y=(x-1)²-4 ≈H is it were -1 in place of +1 [H:y=(x+1)²-4]