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mariarad [96]
3 years ago
9

Solve pls for the function -2x²+x+1 find f(3)

Mathematics
1 answer:
castortr0y [4]3 years ago
3 0

Answer:

f(3) = -14

General Formulas and Concepts:

<u>Pre-Algebra</u>

Order of Operations: BPEMDAS

  1. Brackets
  2. Parenthesis
  3. Exponents
  4. Multiplication
  5. Division
  6. Addition
  7. Subtraction
  • Left to Right

<u>Algebra I</u>

  • Function Notation

Step-by-step explanation:

<u>Step 1: Define</u>

f(x) = -2x² + x + 1

f(3) is x = 3

<u>Step 2: Evaluate</u>

  1. Substitute:                    f(3) = -2(3)² + 3 + 1
  2. Exponents:                   f(3) = -2(9) + 3 + 1
  3. Multiply:                        f(3) = -18 + 3 + 1
  4. Add:                              f(3) = -15 + 1
  5. Add:                              f(3) = -14
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To find 8x397, I used the stack method.

a.)   397
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      3176

b.)   This method works best because it's easy to do, and the factors aren't too big to multiply by hand, unless it was like a billion times a million. But in this case, it's not, so it's perfectly fine to use the stack method.

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4 0
4 years ago
A brand of cereal had 1.2 milligrams of iron per serving. Then they changed their recipe so they had 1.8 mg of iron per serving.
Anvisha [2.4K]

Using it's concept, it is found that the percent increase in the amount of iron per serving was of 50%.

<h3>What is the percentage increase of a value?</h3>

It is given by the increase divided by the initial value, and subtracted by 100%.

In this problem:

  • The initial value is of 1.2 mg.
  • The increase was of 1.8 mg - 1.2 mg = 0.6 mg.

Hence the percent increase is given by:

P = 0.6/1.2 x 100% = 50%.

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8 0
2 years ago
Which statement describes the inverse of m(x) = x2 – 17x?
stealth61 [152]

Answer:

The correct option is;

The \ domain \ restriction \ x \geq \dfrac{17}{2} \ results \ in \ m^{-1}(x) = \dfrac{17}{2} \pm \sqrt{x + \dfrac{289}{4} }}

Step-by-step explanation:

The given information is that m(x) = x² - 17·x

The above equation can be written in the form;

y = x² - 17·x

Therefore;

0 = x² - 17·x - y

From the general solution of a quadratic equation, 0 = a·x² + b·x + c we have;

x = \dfrac{-b\pm \sqrt{b^{2}-4\cdot a\cdot c}}{2\cdot a}

By comparison to the equation,0 = x² - 17·x - y, we have;

a = 1, b = -17, and c = -y

Substituting the values of a, b and c into the formula for the general solution of a quadratic equation, we have;

x = \dfrac{-(-17)\pm \sqrt{(-17)^{2}-4\times (1) \times (-y)}}{2\times (1)} = \dfrac{17\pm \sqrt{289+4\cdot y}}{2}

Which can be simplified as follows;

x =  \dfrac{17\pm \sqrt{289+4\cdot y}}{2}= \dfrac{17}{2} \pm \dfrac{1}{2}  \times \sqrt{289+4\cdot y}} = \dfrac{17}{2} \pm \sqrt{\dfrac{289}{4} +\dfrac{4\cdot y}{4} }}

And further simplified as follows;

x = \dfrac{17}{2} \pm \sqrt{\dfrac{289}{4} +y }} = \dfrac{17}{2} \pm \sqrt{y + \dfrac{289}{4} }}

Interchanging x and y in the function of the inverse, m⁻¹(x), we have;

m^{-1}(x) = \dfrac{17}{2} \pm \sqrt{x + \dfrac{289}{4} }}

We note that the maximum or minimum point of the function, m(x) = x² - 17·x found by differentiating the function and equating the result to zero, gives;

m'(x) = 2·x - 17 = 0

x = 17/2

Similarly, the second derivative is taken to determine if the given point is a maximum or minimum point as follows;

m''(x) = 2 > 0, therefore, the point is a minimum point on the graph

Therefore, as x increases past the minimum point of 17/2, m⁻¹(x) increases to give;

The \ domain \ restriction \ x \geq \dfrac{17}{2} \ results \ in \ m^{-1}(x) = \dfrac{17}{2} \pm \sqrt{x + \dfrac{289}{4} }} to increase m⁻¹(x) above the minimum.

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How can 26n−7m+4(10n−6m) be rewritten?
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Step-by-step explanation:

..................

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