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Harrizon [31]
3 years ago
14

3(x - 4) + 8(x + 2).

Mathematics
1 answer:
FromTheMoon [43]3 years ago
3 0
3(x - 4) + 8(x + 2) = 3x - 12 + 8x + 16 = 11x + 4
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How many cups of sauce are needed for 8 servings of spaghetti?
ioda

Answer:

14

Step-by-step explanation:

8 0
2 years ago
What is the value of the expression below <br> (-64)^2/3
damaskus [11]

Answer:

16

Step-by-step explanation:

(-64)^2/3

64 = 2^{6}

2^{6} ^{\frac{1}{3} } ^{2}

2^{ 2^{2} }

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7 0
3 years ago
Which ordered pairs lie on the graph of the exponential function f(x)=−3^2x+5 ? Select each correct answer. ​ (−2,  76) ​ ​ ​​ (
In-s [12.5K]

Plugging in values, we get that (1,-4) works.

f(1)=-3^{2*1}+5=-3^2+5=-9+5=-4


3 0
3 years ago
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Which of the following expressions are equivalent 6^-3
Lilit [14]

Answer:

6^{-3} = \frac{1}{6^3} =\frac{1}{6*6*6}=\frac{1}{216}

Step-by-step explanation:

The expression 6^{-3} can be simplified using rules for negative exponents - move the base to the denominator. After following this rule, simplify.

6^{-3} = \frac{1}{6^3} =\frac{1}{6*6*6}=\frac{1}{216}

5 0
3 years ago
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Derive the first three (non-zero) terms of Taylor's series expansion for the function
mestny [16]

Answer:

We want to find the first 3 terms of the Taylor's series expansion for f(x) = sin(x) around x = 0.

Remember that a Taylor's series expansion of a function f(x) around the point x₀ is given by:

f(x) = f(x_0) + \frac{1}{2!}f'(x_0)*(x - x_0) + \frac{1}{3!}*f''(x_0)*(x - x0)^2 + ...

Where in the formula we have the first 3 terms of the expansion (but there are a lot more).

So, if:

f(x) = sin(x)

x₀ = 0

The terms are:

f(x_0) = sin(0) = 0

\frac{1}{2!}*f(x_0)'*(x - x_0) = \frac{1}{2} cos(0)*(x - 0) = x/2

\frac{1}{3!}*f(x_0)''*(x - x_0)^2 = \frac{1}{6}*-sin(0)*(x - 0)^2 = 0

\frac{1}{4!}*f(x_0)'''*(x - x_0)^3 = \frac{1}{24}*-cos(0)*(x- 0)^3 = -\frac{x^3}{24}

We already can see that the next term is zero (because when we derive the cos part, we will get a sin() that is zero when evaluated in x = 0), then the next non zero term is:

\frac{1}{6!}*f(x_0)''''*(x - x_0)^5 = \frac{1}{2*3*4*5*6} *(x - 0)^5 = \frac{x^5}{720}

Then we can write:

sin(x) = \frac{x}{2} - \frac{x^3}{24} + \frac{x^5}{720}

Evaluating this in x = 0.2, we get:

sin(0.2) = \frac{0.2}{2} - \frac{0.2^3}{24} + \frac{0.2^5}{720} = 0.099667

7 0
3 years ago
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