Standard form: 7z^4-6z^2-6z
This IS a polynomial, it's degree is 4 and has 3 terms
 
        
             
        
        
        
Answer:
4y=x+33
y=x/4+33/4 (slope-intercept form)
Step-by-step explanation:
y-3 = -4(x+2)
y-3 -4x-8
y= -4x-8+3
y= -4x-5
m1 = -4
For perpendicularity
m2= -1/-4 = 1/4
The equation is
y-y1 = m2(x-x1)
y-7 = 1/4(x-(-5))
y-7 = x/4+5/4
Multiply through by 4
4y-28=x+5
4y=x+5+28
4y=x+33
Divide through by 4
y=x/4+33/4 (slope-intercept form)
 
        
             
        
        
        
Answer:
71
Step-by-step explanation:
<u>refer</u><u> </u><u>the</u><u> attachment</u>
to solve the question we need to recall one of the most important theorem of circle known as two tangent theorem which states that <u>tangents </u><u>which</u><u> </u><u>meet </u><u>at</u><u> the</u><u> </u><u>same</u><u> </u><u>point</u><u> </u><u>are </u><u>equal</u><u> </u><u> </u>that is being said
since  and it's given that FA and BA are 17 and 29 FB should be
 and it's given that FA and BA are 17 and 29 FB should be
therefore,
once again by two tangent theorem we acquire:
As BC=BH+CH,BC is
- 12+2.5
 
likewise,AD=AI+DI so,
- 21=17+DI [AD=21(given) and AI=17 (by the theorem)]
thus,
- DI=21-17= 
By the theorem we obtain:
Similarly,DC=DG+CH therefore,
- DC=4+2.5= 
Now <u>finding</u><u> </u><u>the</u><u> </u><u>Perimeter</u><u> </u><u>of </u><u>ABCD</u>
substitute what we have and got
simplify addition:
hence,
the Perimeter of ABCD is <u>7</u><u>1</u>