You're told there is a 10th degree polynomial. There are four roots of multiplicity 1 and one of multiplicity 2 (a double root).
Anytime there is a root of multiplicity 1 of a polynomial, its graph crosses the x-axis at that root. Anytime there is one of multiplicity 2, it means that we count the root twice and the graph has a tangency point.
The degree of a polynomial tells you how how many roots it has. Ours is degree 10, so it has ten roots. We have the four roots of multiplicity 1 and the one of multiplicity two, for a total of 6. (four and two).
So there are six real roots.
The rest of the roots are imaginary and non-real, and 10 - 6 = 4. So there are four imaginary roots.
Thus choice C is best.
Simplify: [{y^(2/7)}/{y^(1/2)}]
Since, [{a^(p/q)}/{a^(r/s)}] = a^{(p/q)-(r/s)}
Where,
- a = p
- p/q = 2/7 and
- r/s = 1/2
so,
= y^{(2/7)-(1/2)}
Take the LCM of denominator i.e.,2 & 7 is 14.
= y^{(2*2 - 1*7)/14}
= y^{(4-7)/14}
= y^(-3/14) Ans.
<u>read</u><u> </u><u>more similar</u><u> questions</u><u>:</u> Which equation can be simplified to find the inverse of y = x2 – 7? a: x=y ^ 2 - 1/7 b: 1/x = y^2 - 7 c: x = y^2 – 7 d: –x = y^2 – 7..
brainly.com/question/2396514?referrer
Surface area is taking the area of each face and adding it together. The equation is 2lw + 2hw + 2hl. So, for example, you'd take the length and the width to the find the area of the top. Multiply this by 2 to find the top AND the bottom. You need to end with 6 sides and all these together to find your surface area.
Answer: see proof below
<u>Step-by-step explanation:</u>
Given: A + B + C = 90° → A + B = 90° - C
→ C = 90° - (A + B)
Use the Double Angle Identity: cos 2A = 1 - 2 sin² A
→ sin² A = (1 - cos 2A)/2
Use Sum to Product Identity: cos A + cos B = 2 cos [(A + B)/2] · cos [(A - B)/2]
Use the Product to Sum Identity: cos (A - B) - cos (A + B) = 2 sin A · sin B
Use the Cofunction Identities: cos (90° - A) = sin A
sin (90° - A) = cos A
<u>Proof LHS → RHS:</u>
LHS: sin² A + sin² B + sin² C

![\text{Sum to Product:}\quad 1-\dfrac{1}{2}\bigg[2\cos \bigg(\dfrac{2A+2B}{2}\bigg)\cdot \cos \bigg(\dfrac{2A-2B}{2}\bigg)\bigg]+\sin^2 C\\\\\\.\qquad \qquad \qquad =1-\cos (A+B)\cdot \cos (A-B)+\sin^2 C](https://tex.z-dn.net/?f=%5Ctext%7BSum%20to%20Product%3A%7D%5Cquad%201-%5Cdfrac%7B1%7D%7B2%7D%5Cbigg%5B2%5Ccos%20%5Cbigg%28%5Cdfrac%7B2A%2B2B%7D%7B2%7D%5Cbigg%29%5Ccdot%20%5Ccos%20%5Cbigg%28%5Cdfrac%7B2A-2B%7D%7B2%7D%5Cbigg%29%5Cbigg%5D%2B%5Csin%5E2%20C%5C%5C%5C%5C%5C%5C.%5Cqquad%20%5Cqquad%20%5Cqquad%20%3D1-%5Ccos%20%28A%2BB%29%5Ccdot%20%5Ccos%20%28A-B%29%2B%5Csin%5E2%20C)
Given: 1 - cos (90° - C) · cos (A - B) + sin² C
Cofunction: 1 - sin C · cos (A - B) + sin² C
Factor: 1 - sin C [cos (A - B) + sin C]
Given: 1 - sin C[cos (A - B) - sin (90° - (A + B))]
Cofunction: 1 - sin C[cos (A - B) - cos (A + B)]
Sum to Product: 1 - sin C [2 sin A · sin B]
= 1 - 2 sin A · sin B · sin C
LHS = RHS: 1 - 2 sin A · sin B · sin C = 1 - 2 sin A · sin B · sin C 
Answer:
the first one you have is a negative and 5-5-5-5-5
then same for other but oppoiste
Step-by-step explanation: