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2 years ago

7

Consider the roots of a cubic equation with integral coefficients: −1, −4, and 3. Which choice is a factor of the cubic equation

? A) x + 3 B) x − 1 C) x − 4 D) x − 3

1 answer:

mash [69]2 years ago

5 0

Answer:

D

Step-by-step explanation:

Lets go case by case.

Given the roots, a factor will be part of the equation if for some of the roots the factor becomes null, i.e., equal to 0.

Is there any root that makes (x+3)=0? No, as it only becomes 0 for x = - 3 and -3 is not a root. So A NO!

Is there a root that makes (x-1)=0? No, as it only becomes 0 for x=1 and 1 is not a root. So B NO!

(x-4)=0 only for x=4, and as 4 is not a root, C NO!

The last, (x-3)=0 if x=3. As 3 is one of the roots, (x-3) is a factor of our equation!

D is the only correct option!

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5 2/3 divided by 3/2

Zinaida [17]

Answer: 34/9

Step-by-step explanation:

5 2/3 / 3/2

= 17/3 / 3/2

=17/3 x 2/3 [reciprocal]

=34/9

Hope it helped u,

pls mark as the brainliest

^_^

7 0

2 years ago

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In a shipment of 56 vials, only 13 do not have hairline cracks. If you randomly select 3 vials from the shipment, in how many wa

Elden [556K]

Answer: 27434

Step-by-step explanation:

Given : Total number of vials = 56

Number of vials that do not have hairline cracks = 13

Then, Number of vials that have hairline cracks =56-13=43

Since , order of selection is not mattering here , so we combinations to find the number of ways.

The number of combinations of m thing r things at a time is given by :-

$^nC_r=\dfrac{n!}{r!(n-r)!}$

Now, the number of ways to select at least one out of 3 vials have a hairline crack will be :-

$^{13}C_2\cdot ^{43}C_{1}+^{13}C_{1}\cdot ^{43}C_{2}+^{13}C_0\cdot ^{43}C_{3}\\\\=\dfrac{13!}{2!(13-2)!}\cdot\dfrac{43!}{1!(42)!}+\dfrac{13!}{1!(12)!}\cdot\dfrac{43!}{2!(41)!}+\dfrac{13!}{0!(13)!}\cdot\dfrac{43!}{3!(40)!}\\\\=\dfrac{13\times12\times11!}{2\times11!}\cdot (43)+(13)\cdot\dfrac{43\times42\times41!}{2\times41!}+(1)\dfrac{43\times42\times41\times40!}{6\times40!}\\\\=3354+11739+12341=27434$

Hence, the required number of ways =27434

5 0

2 years ago

6-5 solving rational equations 8/x+3=1/x+1

AfilCa [17]

Answer:

x = -5/7

Step-by-step explanation:

8/(x+3)=1/(x+1)

Using cross products

8*(x+1) = 1*(x+3)

Distribute

8x+8 = x+3

Subtract x from each side

8x-x +8 = x+3-x

7x +8 =3

Subtract 8 from each side

7x+8-8 = 3-8

7x = -5

Divide by 7

7x/7 = -5/7

x = -5/7

7 0

2 years ago

Simplify the following

den301095 [7]

Answer:

1) 11 $\sqrt{3}$

2) 2 $\sqrt{2}$

3) $20\sqrt{3} + 15\sqrt{2}$

4) $53 + 12\sqrt{10}$

5) -2

6) $7\sqrt{2} - 5\sqrt{3}$

Step-by-step explanation:

1) 2 $\sqrt{12}$ + 3 $\sqrt{48}$ - $\sqrt{75}$

=(2 × 2 $\sqrt{3}$ )+ (3 × 4 $\sqrt{3}$ ) - 5 $\sqrt{3}$

= 4 $\sqrt{3}$ + 12 $\sqrt{3}$ - 5 $\sqrt{3}$

= 11 $\sqrt{3}$

2) 4 $\sqrt{8}$ -2 $\sqrt{98}$ + $\sqrt{128}$

= (4 × 2 $\sqrt{2}$ ) - (2 × 7 $\sqrt{2}$ ) + 8 $\sqrt{2}$

= 8 $\sqrt{2}$ - 14 $\sqrt{2}$ +8 $\sqrt{2}$

= 2 $\sqrt{2}$

3) 5 $\sqrt{12\\}$ - 3 $\sqrt{18} + 4 \sqrt{72} +2\sqrt{75}$

= 5× $2\sqrt{3}$ - 3× $3\sqrt{2}$ + 4× $6\sqrt{2}$ + 2× $5\sqrt{3}$

= $10\sqrt{3} - 9\sqrt{2} +24\sqrt{2} +10\sqrt{3}$

= $20\sqrt{3} + 15\sqrt{2}$

4) $(2\sqrt{2} + 3\sqrt{5} )^{2}$

= $8 + 12\sqrt{10} + 45$

= $53 + 12\sqrt{10}$

5) $(1+\sqrt{3} ) (1-\sqrt{3} )$

= $1 - 3$

= -2

6) $(2\sqrt{6} -1) (\sqrt{3} -\sqrt{2} )$

= $2\sqrt{18}-2\sqrt{12} -\sqrt{3} +\sqrt{2}$

= 2× $3\sqrt{2}$ - 2× $2\sqrt{3}$ - $\sqrt{3} + \sqrt{2}$

= $6\sqrt{2} - 4\sqrt{3} -\sqrt{3} +\sqrt{2}$

= $7\sqrt{2} - 5\sqrt{3}$

Hope the working out is clear and will help you. :)

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2 years ago

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sammy [17]

Answer:

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Step-by-step explanation:

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2 years ago

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