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dimulka [17.4K]
2 years ago
11

Determine which binomial is a factor of 4x^3+17x^2+7x+12 A. X+12 B. x +4 C. x +7 D. X-4

Mathematics
1 answer:
slega [8]2 years ago
7 0

Answer:

D is correct Answer. x-4. THANK YOU.

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Let f(x)= x^2-4x-c. Find a nonzero value of c such that f(c)=c.​
Dimas [21]

Answer:

The nonzero value of c will be:

  • c = 6

Step-by-step explanation:

Given the function

f\left(x\right)=\:x^2-4x-c

f\left(c\right)=\:c^2-4c-c

as

f(c) = c

so

c=\:c^2-4c-c

switching the sides

c^2-4c-c=c

subtract c from both sides

c^2-4c-c-c=c-c

c^2-6c=0

c\left(c-6\right)=0

Using the zero factor principle

\:If}\:ab=0\:\mathrm{then}\:a=0\:\mathrm{or}\:b=0\:\left(\mathrm{or\:both}\:a=0\:\mathrm{and}\:b=0\right)

c=0\quad \mathrm{or}\quad \:c-6=0

so, the solutions to the quadratic equations are:

c=0,\:c=6

Therefore, a nonzero value of c will be:

  • c = 6
5 0
3 years ago
PLZ HELP!!!! WILL GIVE BRAINLIEST ANSWER!!!
Shalnov [3]

the correct answer is - 1/2

7 0
3 years ago
Read 2 more answers
Which statements are true about the place values in the number 22.222? Choose all answers that are correct. A. The 2 in the tent
Vinvika [58]
B. THE 2 IN THE TENTHS PLACE IS 10 TIMES THE VALUE OF THE 2 IN THE HUNDREDTHS PLACE.

     2       2             .                          2          2                    2
tens ones      decimal point      tenths  hundredths  thousandths

2 in the tenths place = 0.20
2 in the hundredths place = 0.02

0.02 x 10 = 0.20

6 0
3 years ago
What is the us as a decimal number
PolarNik [594]

Answer:

-86.07

Step-by-step explanation:

Theres a negative 86 and a 7 over 100.

3 0
2 years ago
Read 2 more answers
Evaluate the given integral by changing to polar coordinates. 8xy dA D , where D is the disk with center the origin and radius 9
BabaBlast [244]

Answer:

0

Step-by-step explanation:

∫∫8xydA

converting to polar coordinates, x = rcosθ and y = rsinθ and dA = rdrdθ.

So,

∫∫8xydA = ∫∫8(rcosθ)(rsinθ)rdrdθ = ∫∫8r²(cosθsinθ)rdrdθ = ∫∫8r³(cosθsinθ)drdθ

So we integrate r from 0 to 9 and θ from 0 to 2π.

∫∫8r³(cosθsinθ)drdθ = 8∫[∫r³dr](cosθsinθ)dθ

= 8∫[r⁴/4]₀⁹(cosθsinθ)dθ

= 8∫[9⁴/4 - 0⁴/4](cosθsinθ)dθ

= 8[6561/4]∫(cosθsinθ)dθ

= 13122∫(cosθsinθ)dθ

Since sin2θ = 2sinθcosθ, sinθcosθ = (sin2θ)/2

Substituting this we have

13122∫(cosθsinθ)dθ = 13122∫(1/2)(sin2θ)dθ

= 13122/2[-cos2θ]/2 from 0 to 2π

13122/2[-cos2θ]/2 = 13122/4[-cos2(2π) - cos2(0)]

= -13122/4[cos4π - cos(0)]

= -13122/4[1 - 1]

= -13122/4 × 0

= 0

5 0
3 years ago
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