my brother did that too but I'm sorry i cant help you but I'm gonna say just keep up the good work
<h2><em>To enlarge or reduce any shape you must begin by working out the scale factor, this is calculated by using the following formula:
</em></h2><h2><em>
</em></h2><h2><em>For an enlargement = large number ÷ small number
</em></h2><h2><em>
</em></h2><h2><em>For a reduction = small number ÷ large number</em></h2>
For
ax+by=c
slope=-a/b
perpendiculare lines have slopes that multiply to -1
so
2x+3y=9
slope=-2/3
-2/3 times what=-1
what=3/2
point slope form
the equation of a line that passes through the point (x1,y1) and the slope is m is
y-y1=m(x-x1)
slope=3/2
point i (8,5)
equation is
y-5=3/2(x-8)
converted to standard form
2y-10=3x-24
-3x+2y=-14
3x-2y=14
Answer:
Step-by-step explanation:
Part A:
To find the total length of sides 1, 2, and 3 of the quadrilateral add all three sides together.
Side 1: 3y2 + 2y − 6
Side 2: 3y − 7 + 4y2
Side 3: −8 + 5y2 + 4y
3y^2+2y-6+3y-7+4y^2-8+5y^2+4y
Combine the like terms:
12y^2+9y-21
Part B:
To find the length of the fourth side of the quadrilateral we have the total perimeter and the sum of three of the sides, so you just need that fourth side value. Let the fourth side be d.
P= 4y3 + 18y2 + 16y − 26.
Sides = 12y^2+9y-21
12y² + 10y – 21 + d = 4y³ + 18y² + 16y – 26
Solve for d
Separate d from rest of the terms.
d = -12y^2-10y+21+4y^3+18y^2+16y-26
d = 4y^3+6y^2+6y-5
Part C:
If closed means that the degree that these polynomials are at stay that way, then yes, this is true in these cases because you will notice that each side had a y^2, y and no coefficient value except for the fourth one. This didn't change, because you only add and subtract like terms....
Answer:
idk i have to answer 5\
Step-by-step explanation:
