When roots of polynomials occur in radical form, they occur as two conjugates.
That is,
The conjugate of (a + √b) is (a - √b) and vice versa.
To show that the given conjugates come from a polynomial, we should create the polynomial from the given factors.
The first factor is x - (a + √b).
The second factor is x - (a - √b).
The polynomial is
f(x) = [x - (a + √b)]*[x - (a - √b)]
= x² - x(a - √b) - x(a + √b) + (a + √b)(a - √b)
= x² - 2ax + x√b - x√b + a² - b
= x² - 2ax + a² - b
This is a quadratic polynomial, as expected.
If you solve the quadratic equation x² - 2ax + a² - b = 0 with the quadratic formula, it should yield the pair of conjugate radical roots.
x = (1/2) [ 2a +/- √(4a² - 4(a² - b)]
= a +/- (1/2)*√(4b)
= a +/- √b
x = a + √b, or x = a - √b, as expected.
Answer:
yes, triangle DEF is similar to triangle DBC, BC corresponds to EF, and angle DCB corresponds to angle F.
Step-by-step explanation:
Part A: Angle D is congruent to angle D by the reflexive property. Since line BC is parallel to line EF then angle DCB = angle DFE by corresponding angles. Hence triangle DEF is similar to triangle DBS by the AA Similarity Postulate.
Part B: BC corresponds to EF because they are in the same order and the triangles listed DEF and DBC
Part C: Angle DCB corresponds to angle F since they are corresponding angles with the two given parallel lines BC and EF.
Answer: <em>Multiply the original price by the decimal form of the percent and subtract that from the original price.
</em>
<em>37*0.85=31.45
</em>
<em>37-31.45=5.55
</em>
<u><em>Answer: New price is $5.55</em></u>
Answer:
80 feet
Step-by-step explanation:
Given:
Initial speed of the car (
) = 40 ft/sec
Deceleration of the car (
) = -10 ft/sec²
Final speed of the car (
) = 0 ft/sec
Let the distance traveled by the car be 'x' at any time 't'. Let 'v' be the velocity at any time 't'.
Now, deceleration means rate of decrease of velocity.
So, 
Negative sign means the velocity is decreasing with time.
Now, 
using chain rule of differentiation. Therefore,
Integrating both sides under the limit 40 to 0 for 'v' and 0 to 'x' for 'x'. This gives,
![\int\limits^0_{40} {v} \, dv=\int\limits^x_0 {-10} \, dx\\\\\left [ \frac{v^2}{2} \right ]_{40}^{0}=-10x\\\\-10x=\frac{0}{2}-\frac{1600}{2}\\\\10x=800\\\\x=\frac{800}{10}=80\ ft](https://tex.z-dn.net/?f=%5Cint%5Climits%5E0_%7B40%7D%20%7Bv%7D%20%5C%2C%20dv%3D%5Cint%5Climits%5Ex_0%20%7B-10%7D%20%5C%2C%20dx%5C%5C%5C%5C%5Cleft%20%5B%20%5Cfrac%7Bv%5E2%7D%7B2%7D%20%5Cright%20%5D_%7B40%7D%5E%7B0%7D%3D-10x%5C%5C%5C%5C-10x%3D%5Cfrac%7B0%7D%7B2%7D-%5Cfrac%7B1600%7D%7B2%7D%5C%5C%5C%5C10x%3D800%5C%5C%5C%5Cx%3D%5Cfrac%7B800%7D%7B10%7D%3D80%5C%20ft)
Therefore, the car travels a distance of 80 feet before stopping.
