Answer:
2
Step-by-step explanation:
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]
Answer: The required solution is
Step-by-step explanation: We are given to solve the following differential equation :
Let us consider that
be an auxiliary solution of equation (i).
Then, we have
Substituting these values in equation (i), we get
![m^2e^{mt}+10me^{mt}+25e^{mt}=0\\\\\Rightarrow (m^2+10y+25)e^{mt}=0\\\\\Rightarrow m^2+10m+25=0~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~[\textup{since }e^{mt}\neq0]\\\\\Rightarrow m^2+2\times m\times5+5^2=0\\\\\Rightarrow (m+5)^2=0\\\\\Rightarrow m=-5,-5.](https://tex.z-dn.net/?f=m%5E2e%5E%7Bmt%7D%2B10me%5E%7Bmt%7D%2B25e%5E%7Bmt%7D%3D0%5C%5C%5C%5C%5CRightarrow%20%28m%5E2%2B10y%2B25%29e%5E%7Bmt%7D%3D0%5C%5C%5C%5C%5CRightarrow%20m%5E2%2B10m%2B25%3D0~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~%5B%5Ctextup%7Bsince%20%7De%5E%7Bmt%7D%5Cneq0%5D%5C%5C%5C%5C%5CRightarrow%20m%5E2%2B2%5Ctimes%20m%5Ctimes5%2B5%5E2%3D0%5C%5C%5C%5C%5CRightarrow%20%28m%2B5%29%5E2%3D0%5C%5C%5C%5C%5CRightarrow%20m%3D-5%2C-5.)
So, the general solution of the given equation is
Differentiating with respect to t, we get
According to the given conditions, we have
and
Thus, the required solution is
2040 = n/2(20+(n-1)4)
4080 = n(20+4n-4)
4080= 20n +4n^2 -4n
1020 = 4n + n^2
n^2 +4n -1020 =0
use common formula (can't write out so just look at answers. sorry)
which gives answers of n=-34 and n=30. since n can only be positive, n=30 so there are 30 rows. I liked that challenge
The area of the sector which is the white triangle adding the shaded region is 68.9/360*π*9.28^2=51.78005(rounding to 5th digit after decimal point for accuracy before we do final round for answer)
The area of the white triangle in the sector has area 1/2*9.28^2*sin(68.9)= 40.17223(rounding to 5 digits again for some accuracy.
Now we take out the white triangle from the sector.
51.78005-40.17223=11.60782
rounding to the nearest tenth we get 11.6 cm^2
Problem done!
Hope this helped and if you have any questions about my explanation just ask in the comment and I will answer.
