Answer:
I would rather use their multiples.
Step-by-step explanation:
With 13 and 14, neither number has many factors (and few prime factors, for that matter) so factoring would be pretty much useless. If you wanted to use the prime factors of 14 (7 and 2) multiplying either of them by 13 would not give you the LCM. Actually the LCM is just 13*14 which is 182.
Good luck!
The roots of the polynomial <span><span>x^3 </span>− 2<span>x^2 </span>− 4x + 2</span> are:
<span><span>x1 </span>= 0.42801</span>
<span><span>x2 </span>= −1.51414</span>
<span><span>x3 </span>= 3.08613</span>
x1 and x2 are in the desired interval [-2, 2]
f'(x) = 3x^2 - 4x - 4
so we have:
3x^2 - 4x - 4 = 0
<span>x = ( 4 +- </span><span>√(16 + 48) </span>)/6
x_1 = -4/6 = -0.66
x_ 2 = 2
According to Rolle's theorem, we have one point in between:
x1 = 0.42801 and x2 = −1.51414
where f'(x) = 0, and that is <span>x_1 = -0.66</span>
so we see that Rolle's theorem holds in our function.
Answer:
22.5 ft
Step-by-step explanation:
1/1.5=15/x
cross multiply and solve
22.5
Answer: D
Step-by-step explanation:
Consider the first equation. Subtract 3x from both sides.
y−3x=−2
Consider the second equation. Subtract x from both sides.
y−2−x=0
Add 2 to both sides. Anything plus zero gives itself.
y−x=2
To solve a pair of equations using substitution, first solve one of the equations for one of the variables. Then substitute the result for that variable in the other equation.
y−3x=−2,y−x=2
Choose one of the equations and solve it for y by isolating y on the left hand side of the equal sign.
y−3x=−2
Add 3x to both sides of the equation.
y=3x−2
Substitute 3x−2 for y in the other equation, y−x=2.
3x−2−x=2
Add 3x to −x.
2x−2=2
Add 2 to both sides of the equation.
2x=4
Divide both sides by 2.
x=2
Substitute 2 for x in y=3x−2. Because the resulting equation contains only one variable, you can solve for y directly.
y=3×2−2
Multiply 3 times 2.
y=6−2
Add −2 to 6.
y=4
The system is now solved.
y=4,x=2
