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avanturin [10]
3 years ago
12

Nd the value of the expression. Type a numerical answer in the space provided. 4 ∙ 1 - 3^2 ÷ 3

Mathematics
1 answer:
Gemiola [76]3 years ago
4 0
This is a problem of Arithmetic that is a branch of mathematics that consists of the study of numbers. Its main goal is the study of the traditional operations on them just as: <em>addition, subtraction, multiplication and</em><span><em> division</em>. </span>There is an order of operation to solve problems in arithmetic that is the way you choose to simplify expressions. This order is like this:

1. First. Solve Parentheses.
2. Second. Solve Exponents (and root).
3. Third. Solve Multiplication and Division
3. Fourth. Solve addition and subtraction.  

So let's apply these concepts to our problem:

4 \times 1-3^2 \div 3

We don't have parentheses, so let's apply second step:

4 \times 1-9 \div 3

Applying third step:

4-3

Finally, applying fourth step, the result is:

1

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7 0
3 years ago
Write a polynomial function of least degree with the given zero. -1+ √ 2, √ 3
kirill115 [55]

Answer:

f(x) = x^{4} + 2x³ - 4x² - 6x + 3

Step-by-step explanation:

Note that radical zeros occur in conjugate pairs, thus

- 1 + \sqrt{2} is a zero then - 1 - \sqrt{2} is also a zero

\sqrt{3} is a zero then - \sqrt{3} is also a zero

Thus the corresponding factors are

(x - (- 1 + \sqrt{2}) ), (x - (- 1 - \sqrt{2}) ), (x - \sqrt{3}), (x - (- \sqrt{3})), that is

(x + 1 - \sqrt{2}), (x + 1 + \sqrt{2}), (x - \sqrt{3}), (x + \sqrt{3})

The polynomial is then the product of the roots

f(x) = (x + 1 - \sqrt{2})(x + 1 + \sqrt{2})(x - \sqrt{3})(x + \sqrt{3})

     = ((x + 1)² - (\sqrt{2})²)((x² - (\sqrt{3})²)

     = (x² + 2x + 1 - 2)(x² - 3)

     = (x² + 2x - 1)(x² - 3) ← distribute

     = x^{4} - 3x² + 2x³ - 6x - x² + 3

     = x^{4} + 2x³ - 4x² - 6x + 3

6 0
4 years ago
Evaluate (picture shows equation)
PilotLPTM [1.2K]
If we try to evaluate this expression, we arrive at an indeterminate limit of the form ∞ · 0. So we rewrite this limit as
                      lim of [ e^(-x/5) / (1 / x^3) ] as x→∞
This results in an determinate form of 0/0, so we apply L'Hospital's rule by differentiating the numerator and denominator.
                      lim of [ -1/5e^(-x/5) / (-3/x^4) ] as x→∞
Multiply the numerator and denominator by x^4
                      lim of [ -1/5 * x^4 * e^(-x/5) / (-3) ] as x→∞
Since e^(-x/5) approaches 0 as x→∞, the limit evaluates to 0.
The answer to this question is 0.
7 0
3 years ago
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