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Georgia [21]
3 years ago
12

[tex]23 1/5 in to percentAGE

Mathematics
1 answer:
MariettaO [177]3 years ago
5 0
20% just divide 1/5 bro I hate the 20 character rule
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Find the missing number to make these fractions equal.
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Is there any picture or something ?
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SIMPLY FIND THE DERIVATIVE. I'M LAZY.
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3, 2, 8, 3, -12

Step-by-step explanation:

d(2x⁴ - 6x²)³/dx

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Explain how you know that 0.38 is a rational number?
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Answer:

0.38 is a rational number because it can be expressed as fraction, and it isn't a repeating decimal.

Step-by-step explanation:

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Answer:

  see attached

Step-by-step explanation:

Most of this exercise is looking at different ways to identify the slope of the line. The first attachment shows the corresponding "run" (horizontal change)  and "rise" (vertical change) between the marked points.

In your diagram, these values (run=1, rise=-3) are filled in 3 places. At the top, the changes are described in words. On the left, they are described as "rise" and "run" with numbers. At the bottom left, these same numbers are described by ∆y and ∆x.

The calculation at the right shows the differences between y (numerator) and x (denominator) coordinates. This is how you compute the slope from the coordinates of two points.

If you draw a line through the two points, you find it intersects the y-axis at y=4. This is the y-intercept that gets filled in at the bottom. (The y-intercept here is 1 left and 3 up from the point (1, 1).)

4 0
4 years ago
write a polynomial function of least degree with integral coefficients that has the given zeros. -(1/3), -i
inessss [21]

Answer:

f(x)=3x^3+x^2+3x+1

Step-by-step explanation:

If a real number -\frac{1}{3} is a zero of polynomial function, then

x-\left(-\dfrac{1}{3}\right)=x+\dfrac{1}{3}

is the factor of this function.

If a complex number -i is a xero of the polynomial function, then the complex number i is also a zero of this function and

x-(-i)=x+i\ \text{ and }\ x-i

are two factors of this function.

So, the function of least degree is

f(x)=\left(x+\dfrac{1}{3}\right)(x+i)(x-i)=\left(x+\dfrac{1}{3}\right)(x^2-i^2)=\\ \\ =\left(x+\dfrac{1}{3}\right)(x^2+1)=\dfrac{1}{3}(3x+1)(x^2+1)=\dfrac{1}{3}(3x^3+x^2+3x+1)

If the polynomial function must be with integer coefficients, then it has a form

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