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givi [52]
3 years ago
14

What term describes 3y-5

Mathematics
1 answer:
Arisa [49]3 years ago
4 0
Please provide more information.
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What would be the expression after distributing, (2m+6)2?
MissTica

Answer:

Step-by-step explanation:

(2m + 6)2 = 2m*2 + 6*2

               = 4m + 12

5 0
4 years ago
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14, 23, 31, 29, 33 what is the standard deviation for this set of population data?
Readme [11.4K]
31 would be the standard 
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3 years ago
State the number of complex zeros, the possible number of real and imaginary zeros, and the possible rational zeros for each fun
ArbitrLikvidat [17]
Since the degree of this polynomial is 5, there will be 5 possible zeros. To find the possible rational 0s, use the rational root theorem (p/q). P is the last, non x value, which here it is the four on the end. The q is the leading coefficient, which is also q. Next, find all of the factors of q and p, which since they are both 4, are ±1, ±2, and ±4. Next do all possible values of p/q, which are ±1, ±2, ±4, ±1/2, and ±1/4. These are all your possible rational zeros. complex 0s only come in pairs, so the maximum there can be is 4 complex zeros, meaning there is at least one rational, real 0. (i graphed it it is -1/2, so all others must be rational or imaginary)
4 0
4 years ago
Find the roots of the polynomial equation.<br><br> 2x3 + 2x2 – 19x + 20 = 0
d1i1m1o1n [39]
First, find any zero of the polynomial. Since you didn't ask for work, I'll assume it's okay if I use my calculator. Your given polynomial has only one real root which is x=-4.

Now we use the rule that x-a divides the polynomial where a is a zero of said polynomial.

So x+4 divides 2x^3+2x^2-19x+20. 
<span>(2x^3+2x^2-19x+20) / (x+4 equals 2x^2-6x+5).
 
If we factor out a two, we can use the quadratic formula.

2(x^2-3x+2.5) so we have x = (-(-3)+/-(9-4*1*2.5)^(1/2))/2*1)=(3+i)... or (3-i)/2 Where i is the square root of negative one. final answer:
2x^3+2x^2-19x+20=0
 
then x=-4, (3+i)/2, or (3-i)/2
</span>we have two imaginary number.
I hope it helped you
5 0
3 years ago
Read 2 more answers
Use (a) the midpoint rule and (b) simpson's rule to approximate the below integral. ∫ x^2sin(x) dx with n = 8.
MaRussiya [10]

Answer:

midpoint rule =  5.93295663

simpson's rule = 5.869246855

Step-by-step explanation:

a) midpoint rule

\int\limits^b_a {(x)} \, dx≈ Δ x (f(x₀+x₁)/2 + f(x₁+x₂)/2 + f(x₂+x₃)/2 +...+ f(x_{n}_₂+x_{n}_₁)/2 +f(x_{n}_₁+x_{n})/2)

Δx = (b − a) / n

We have that a = 0, b = π, n = 8

Therefore

Δx = (π − 0) / 8 = π/8

Divide the interval [0,π] into n=8 sub-intervals of length Δx = π/8 with the following endpoints:

a=0, π/8, π/4, 3π/8, π/2, 5π/8, 3π/4, 7π/8, π = b

Now, we just evaluate the function at these endpoints:

f(\frac{x_{0}+x_{1}  }{2} ) = f(\frac{0+\frac{\pi}{8}   }{2} ) = f(\frac{\pi }{16})=\frac{\pi^{2}sin(\frac{\pi }{16})  }{256} = 0.00752134

f(\frac{x_{1}+x_{2}  }{2} ) = f(\frac{\frac{\pi }{8} +\frac{\pi}{4}   }{2} ) = f(\frac{3\pi }{16})=\frac{9\pi ^{2} sin(\frac{3\pi }{16}) }{256} = 0.19277080

f(\frac{x_{2}+x_{3}  }{2} ) = f(\frac{\frac{\pi }{4} +\frac{3\pi}{8}   }{2} ) = f(\frac{5\pi }{16})=\frac{25\pi ^{2} sin(\frac{5\pi }{16}) }{256} = 0.80139415

f(\frac{x_{3}+x_{4}  }{2} ) = f(\frac{\frac{3\pi }{8} +\frac{\pi}{2}   }{2} ) = f(\frac{7\pi }{16})=\frac{49\pi ^{2} sin(\frac{7\pi }{16}) }{256} = 1.85280536

f(\frac{x_{4}+x_{5}  }{2} ) = f(\frac{\frac{\pi }{2} +\frac{5\pi}{8}   }{2} ) = f(\frac{9\pi }{16})=\frac{81\pi ^{2} sin(\frac{7\pi }{16}) }{256} = 3.062800704

f(\frac{x_{5}+x_{6}  }{2} ) = f(\frac{\frac{5\pi }{8} +\frac{3\pi}{4}   }{2} ) = f(\frac{11\pi }{16})=\frac{121\pi ^{2} sin(\frac{5\pi }{16}) }{256} = 3.878747709

f(\frac{x_{6}+x_{7}  }{2} ) = f(\frac{\frac{3\pi }{4} +\frac{7\pi}{8}   }{2} ) = f(\frac{13\pi }{16})=\frac{169\pi ^{2} sin(\frac{3\pi }{16}) }{256} = 3.61980731

f(\frac{x_{7}+x_{8}  }{2} ) = f(\frac{\frac{7\pi }{8} +\pi    }{2} ) = f(\frac{15\pi }{16})=\frac{225\pi ^{2} sin(\frac{\pi }{16}) }{256} = 1.69230261

Finally, just sum up the above values and multiply by Δx = π/8:

π/8 (0.00752134 +0.19277080+ 0.80139415 + 1.85280536 + 3.062800704 + 3.878747709 + 3.61980731 + 1.69230261) = 5.93295663

b) simpson's rule

\int\limits^b_a {(x)} \, dx  ≈ (Δx)/3 (f(x₀) + 4f(x₁) + 2f(x₂) + 4f(x₃) + 2f(x₄) + ... + 2f(x_{n-2}) + 4f(x_{n-1}) + f(x_{n}))

where Δx = (b−a) / n

We have that a = 0, b = π, n = 8

Therefore

Δx = (π−0) / 8 = π/8

Divide the interval [0,π] into n = 8 sub-intervals of length Δx = π/8, with the following endpoints:

a = 0, π/8, π/4, 3π/8, π/2, 5π/8, 3π/4, 7π/8 ,π = b

Now, we just evaluate the function at these endpoints:  

f(x₀) = f(a) = f(0) = 0 = 0

4f(x_{1} ) = 4f(\frac{\pi }{8} )=\frac{\pi^{2}\sqrt{\frac{1}{2}-\frac{\sqrt{2} }{4}   }  }{16} = 0.23605838

2f(x_{2} ) = 2f(\frac{\pi }{4} )=\frac{\sqrt{2\pi^{2}  } }{16} = 0.87235802

4f(x_{3} ) = 4f(\frac{3\pi }{8} )=\frac{9\pi^{2}\sqrt{\frac{\sqrt{2} }{4}-\frac{{1} }{2}   }  }{16} = 5.12905809

2f(x_{4} ) = 2f(\frac{\pi }{2} )=\frac{\pi ^{2} }{2} = 4.93480220

4f(x_{5} ) = 4f(\frac{5\pi }{8} )=\frac{25\pi^{2}\sqrt{\frac{\sqrt{2} }{4}-\frac{{1} }{2}   }  }{16} = 14.24738359

2f(x_{6} ) = 2f(\frac{3\pi }{4} )=\frac{9\sqrt{2\pi^{2}  } }{16} = 7.85122222

4f(x_{7} ) = 4f(\frac{7\pi }{8} )=\frac{49\pi^{2}\sqrt{\frac{1}{2}-\frac{\sqrt{2} }{4}   }  }{16} = 11.56686065

f(x₈) = f(b) = f(π) = 0 = 0

Finally, just sum up the above values and multiply by Δx/3 = π/24:

π/24 (0 + 0.23605838 + 0.87235802 + 5.12905809 + 4.93480220 + 14.24738359 + 7.85122222 + 11.56686065 = 5.869246855

7 0
3 years ago
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