What is the product of (7x+2) and 20-3x/4
1 answer:
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This can be solved by factoring.
First, set the expression equal to zero.
Then, find two the factors of
whose sum is 
.
Split into these two factors.
Next, factor by grouping.
By the Zero Product Property, set each factor equal to zero.
These are the solutions. The Complex Conjugate Root Theorem and the Fundamental Theorem of Algebra both state that, in essence, real and imaginary solutions come in pairs of two and every polynomial of degree
has exactly complex roots, but real roots are also complex roots. That sounds confusing, but this just means that you're done. Your answers are -2 and 1/3. There are two real roots.
First, set the expression equal to zero.
Then, find two the factors of
Split
Next, factor by grouping.
By the Zero Product Property, set each factor equal to zero.
These are the solutions. The Complex Conjugate Root Theorem and the Fundamental Theorem of Algebra both state that, in essence, real and imaginary solutions come in pairs of two and every polynomial of degree
Answer:
The angular velocity is 6.72 π radians per second
Step-by-step explanation:
The formula of the angular velocity is ω = , where v is the linear velocity and r is the radius of the circle
The unit of the angular velocity is radians per second
∵ The diameter of the tire is 25 inches
∵ The linear velocity is 15 miles per hour
- We must change the mile to inch and the hour to seconds
∵ 1 mile = 63360 inches
∵ 1 hour = 3600 second
∴ 15 miles/hour = 15 ×
∴ 15 miles/hour = 264 inches per second
Now let us find the angular velocity
∵ ω =
∵ v = 264 in./sec.
∵ d = 25 in.
- The radius is one-half the diameter
∴ r = × 25 = 12.5 in.
- Substitute the values of v and r in the formula above to find ω
∴ ω =
∴ ω = 21.12 rad./sec.
- Divide it by π to give the answer in terms of π
∴ ω = 6.72 π radians per second
The angular velocity is 6.72 π radians per second
We know that
(ad)/(bd)=d/d time a/b=a/b since d's cancel
also
if a/b=c/d in simplest form, then a=c and b=d
we have
p/(x^2-5x+6)=(x+4)/(x-2)
therefor
p/(x^2-5x+6)=d/d times (x+4)/(x-2)
p/(x^2-5x+6)=d(x+4)/d(x-2)
therefor
p=d(x+4) and
x^2-5x+6=d(x-2)
we can solve last one
factor
(x-6)(x+1)=d(x-2)
divide both sides by (x-2)
[(x-6)(x+1)]/(x-2)=d
sub
p=d(x+4)
p=([(x-6)(x+1)]/(x-2))(x+4)
(ad)/(bd)=d/d time a/b=a/b since d's cancel
also
if a/b=c/d in simplest form, then a=c and b=d
we have
p/(x^2-5x+6)=(x+4)/(x-2)
therefor
p/(x^2-5x+6)=d/d times (x+4)/(x-2)
p/(x^2-5x+6)=d(x+4)/d(x-2)
therefor
p=d(x+4) and
x^2-5x+6=d(x-2)
we can solve last one
factor
(x-6)(x+1)=d(x-2)
divide both sides by (x-2)
[(x-6)(x+1)]/(x-2)=d
sub
p=d(x+4)
p=([(x-6)(x+1)]/(x-2))(x+4)