Let f(x) = x² + 6x²-x+ 5 then ,
number to be added be P
then,
f(x) = x² + 6x²-x+ 5 +P
According to the qn,
(x+3) is exactly divisible by zero then,
R=0
comparing .. we get a= -3
now by remainder theorm
R=f(a)
0=f(-3)
0=(-3)² + 6(-3)²-(-3)+ 5 + P
0= 9 + 54 + 3 + 5 + P
-71=P
therefore, -71 should be added.
Hope you understand
9514 1404 393
Answer:
C. 12
Step-by-step explanation:
Corresponding sides are proportional. Sides with the same mark are the same length.
WX/XY = LM/MN = LM/KJ
WX/10 = 18/15
WX = 180/15 . . . . multiply by 10
WX = 12
Answer:
1) The terminal side lies in quadrant 2
2) cosФ = -15/17
3) tanФ = -8/15
Step-by-step explanation:
1) First, we should realize that since sinФ = 8/17, it is positive. Thus, Ф must be in quadrants 1 or 2. And since cosФ is negative, Ф must be in quadrant 2 because cos, sin, and tan are all positive in quadrant 1
2) Next, we must realize that the sides given are a 8-15-17 right triangle. And in this case, since sinФ = 8/17, cosФ should be 15/17. But since cosФ is negative, cosФ = -15/17
3) Lastly, tanФ is opposite/adjacent. Since opposite of Ф is 8 and adjacent of Ф is 15, then tanФ should be 8/15, but since Ф is in quadrant 2, then tanФ is -8/15
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D. Pyramid
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Answer:
and 
Step-by-step explanation:
Given
See attachment for complete question
Required
Determine the equilibrium solutions
We have:


To solve this, we first equate
and
to 0.
So, we have:


Factor out R in 

Split
or 
or 
Factor out W in 

Split
or 
Solve for R


Make R the subject


When
, we have:




Collect like terms

Solve for W




When
, we have:



Collect like terms

Solve for R


So, we have:

When
, we have:





So, we have:

Hence, the points of equilibrium are:
and 