In geometry, the length of JK in terms of a, b, and x will be D. abx
<h3>How to illustrate the information?</h3>
It should be noted that from the information given, it was stated that in circle M, EMF measures the radians allocated, the length of radius FM is and the length of EF is x.
In the other circle, JNK also measures the radians allocated and the length of radius KN is b.
Therefore, the length of JK in terms of a, b, and x will be:
= a × b × x
= abx
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Function is p(x)=(x-4)^5(x^2-16)(x^2-5x+4)(x^3-64)
first factor into (x-r1)(x-r2)... form
p(x)=(x-4)^5(x-4)(x+4)(x-4)(x-1)(x-4)(x^2+4x+16)
group the like ones
p(x)=(x-4)^8(x+4)^1(x-1)^1(x^2+4x+16)
multiplicity is how many times the root repeats in the function
for a root r₁, the root r₁ multiplicity 1 would be (x-r₁)^1, multility 2 would be (x-r₁)^2
so
p(x)=(x-4)^8(x+4)^1(x-1)^1(x^2+4x+16)
(x-4)^8 is the root 4, it has multiplicity 8
(x-(-4))^1 is the root -4 and has multiplicity 1
(x-1)^1 is the root 1 and has multiplity 1
(x^2+4x+16) is not on the real plane, but the roots are -2+2i√3 and -2-2i√3, each multiplicity 1 (but don't count them because they aren't real
baseically
(x-4)^8 is the root 4, it has multiplicity 8
(x-(-4))^1 is the root -4 and has multiplicity 1
(x-1)^1 is the root 1 and has multiplity 1
Good luck on your test love ya also I have bad period cramps send chocolate please
Answer:
1
Step-by-step explanation:
3^3/5-3-(4-2)^2/2
3^3/3-2^2/2
3-2
=1
Answer:
7 1/2+ 7 1/2=146 of 146 Kerry and Eric ran the same amount each
