Answer: M= 3/5
Step-by-step explanation: 3/5x=3= M=3/5
Only two integers can have the same distance from 0, so 2
Answer:
Step-by-step explanation:
Answer: x = -3/4 can not be a rational zero of the polynomial.
Step-by-step explanation:
We have the polynomial:
6x^5 + ax^3 -bx -12 = 0.
The theorem says that:
If P(x) is a polynomial with integer coefficients, and p/q is a zero of P(x) then p is a factor of the constant term (in this case the constant term is -12) and q is a factor of the leading coefficient (in this case the leading coefficient is 6.).
The factors of -12 (different than itself) are (independent of the sign).
1, 2, 3, 4 and 6.
So p can be: 1, -1, 2, -2, 3, -3, 4, -4, 6, -6.
The factors of 6 are:
1, 2 and 3, so q can be 1, -1, 2, -2, 3, -3.
Then the option that can not be a zero of the polynomial is
x = -3/4
because the number in the denominator must be a factor of the leading coefficient, and 4 is not a factor of six.
Answer/Step-by-step explanation:
1. Side CD and side DG meet at endpoint D to form <4. Therefore, the sides of <4 are:
Side CD and side DG.
2. Vertex of <2 is the endpoint at which two sides meet to form <2.
Vertex of <2 is D.
3. Another name for <3 is <EDG
4. <5 is less than 90°. Therefore, <5 can be classified as an acute angle.
5. <CDE is less than 180° but greater than 90°. Therefore, <CDE is classified as an obtuse angle.
6. m<5 = 42°
m<1 = 117°
m<CDF = ?
m<5 + m<1 = m<CDF (angle addition postulate)
42° + 117° = m<CDF (Substitution)
159° = m<CDF
m<CDF = 159°
7. m<3 = 73°
m<FDE = ?
m<FDG = right angle = 90°
m<3 + m<FDE = m<FDG (Angle addition postulate)
73° + m<FDE = 90° (Substitution)
73° + m<FDE - 73° = 90° - 73°
m<FDE = 17°
