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Anton [14]
3 years ago
12

Which expression can you simplify by combining like terms?

Mathematics
2 answers:
almond37 [142]3 years ago
8 0
12b^2+4ab-3ba-6
=12b^2+ab-6
pashok25 [27]3 years ago
5 0
You can reduce the first one by combining the ab and the ba term
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Use the mid-point rule with n = 4 to approximate the area of the region bounded by y = x3 and y = x. (10 points)
USPshnik [31]
See the graph attached.

The midpoint rule states that you can calculate the area under a curve by using the formula:
M_{n} = \frac{b - a}{2} [ f(\frac{x_{0} + x_{1} }{2}) +  f(\frac{x_{1} + x_{2} }{2}) + ... +  f(\frac{x_{n-1} + x_{n} }{2})]

In your case:
a = 0
b = 1
n = 4
x₀ = 0
x₁ = 1/4
x₂ = 1/2
x₃ = 3/4
x₄ = 1

Therefore, you'll have:
M_{4} = \frac{1 - 0}{4} [ f(\frac{0 +  \frac{1}{4} }{2}) +  f(\frac{ \frac{1}{4} + \frac{1}{2} }{2}) +  f(\frac{\frac{1}{2} + \frac{3}{4} }{2}) + f(\frac{\frac{3}{4} + 1} {2})]
M_{4} = \frac{1}{4} [ f(\frac{1}{8}) +  f(\frac{3}{8}) +  f(\frac{5}{8}) + f(\frac{7}{8})]

Now, to evaluate your f(x), you need to look at the graph and notice that:
f(x) = x - x³

Therefore:
M_{4} = \frac{1}{4} [(\frac{1}{8} - (\frac{1}{8})^{3}) + (\frac{3}{8} - (\frac{3}{8})^{3}) + (\frac{5}{8} - (\frac{5}{8})^{3}) + (\frac{7}{8} - (\frac{7}{8})^{3})]

M_{4} = \frac{1}{4} [(\frac{1}{8} - \frac{1}{512}) + (\frac{3}{8} - \frac{27}{512}) + (\frac{5}{8} - \frac{125}{512}) + (\frac{7}{8} - \frac{343}{512})]

M₄ = 1/4 · (2 - 478/512)
     = 0.2666

Hence, the <span>area of the region bounded by y = x³ and y = x</span> is approximately 0.267 square units.

6 0
3 years ago
Simplify: 12xy - 6x + 3y -10xy - 5y
Darina [25.2K]

Answer:

2xy - 6x +2y

Step-by-step explanation:

12xy - 10xy = 2xy

5y - 3y = 2y

2xy - 6x + 2y

5 0
3 years ago
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X = 69o
2x + 42 = 180
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Kobotan [32]
Both angles OML and OPL are right angles (90 degrees) because they are located at the point of intersection of a tangent line to a circle.
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A sequence is defined recursively using the formula f(n + 1) = f(n) – 5 . Which sequence could be generated using this formula?
kramer

In mathematics, the logarithm is the inverse function to exponentiation. That means the logarithm of a given number x is the exponent to which another fixed number, the base b, must be raised, to produce that number x.

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