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Roman55 [17]
3 years ago
12

Use the Factor Theorem to determine if the given binomial is a factor of the given polynomial function:f(x)=4x^4-20x^3+33x^2-22x

+5;x-5/2
A:f(-5/2)≠0; the binomial is not a factor of the polynomial

B: f(5/2)≠0; the binomial is not a factor of the polynomial

C:f(-5/2)=0; yes, the binomial is a factor of the polynomial

D: f(5/2)=0; yes, the binomial is a factor of the polynomial
Mathematics
1 answer:
Xelga [282]3 years ago
6 0

Answer:

D: f(5/2)=0; yes, the binomial is a factor of the polynomial

Step-by-step explanation:

The easy way to determine is to plug in the potential zero into the polynomial given the binomial divisor and see if it results in 0. If so, then the binomial is a factor of the polynomial:

Because f(5/2)=0, then the binomial is indeed a factor of the polynomial!

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A stadium has 39 sections. there are 18 sections that each have 478 seats. The remaining sections each have 292 seats. If 13,528
vesna_86 [32]

Answer: 1,208 seats

Step-by-step explanation:

39-18= 21

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Step-by-step explanation:

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3 years ago
Find the components of the vertical force Bold Upper FFequals=left angle 0 comma negative 10 right angle0,−10 in the directions
quester [9]

Solution :

Let $v_0$ be the unit vector in the direction parallel to the plane and let $F_1$ be the component of F in the direction of v_0 and F_2 be the component normal to v_0.

Since, |v_0| = 1,

$(v_0)_x=\cos 60^\circ= \frac{1}{2}$

$(v_0)_y=\sin 60^\circ= \frac{\sqrt 3}{2}$

Therefore, v_0 = \left

From figure,

|F_1|= |F| \cos 30^\circ = 10 \times \frac{\sqrt 3}{2} = 5 \sqrt3

We know that the direction of F_1 is opposite of the direction of v_0, so we have

$F_1 = -5\sqrt3 v_0$

    $=-5\sqrt3 \left$

    $= \left$

The unit vector in the direction normal to the plane, v_1 has components :

$(v_1)_x= \cos 30^\circ = \frac{\sqrt3}{2}$

$(v_1)_y= -\sin 30^\circ =- \frac{1}{2}$

Therefore, $v_1=\left< \frac{\sqrt3}{2}, -\frac{1}{2} \right>$

From figure,

|F_2 | = |F| \sin 30^\circ = 10 \times \frac{1}{2} = 5

∴  F_2 = 5v_1 = 5 \left< \frac{\sqrt3}{2}, - \frac{1}{2} \right>

                   $=\left$

Therefore,

$F_1+F_2 = \left< -\frac{5\sqrt3}{2}, -\frac{15}{2} \right> + \left< \frac{5 \sqrt3}{2}, -\frac{5}{2} \right>$

           $= = F$

3 0
3 years ago
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