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AleksandrR [38]
3 years ago
13

Complete the following statements.

Mathematics
1 answer:
masha68 [24]3 years ago
8 0
  • A conjunction is true only if both statements are true.

The conjunction is the AND logical operator, and the result of A AND B is true only when both A and B are true. For example, "We are on planet earth, and elephants fly" is false, even if the first part is actually true.

  • A disjunction is true if at least one statement is true

In fact, the disjunction is the OR operator, and this time one true statement is sufficient to make the whole expression true: "You are reading this answer on brainly or I am 4 meters tall" is true, although I'm not 4 meters tall.

  • For a conditional statement, a true hypothesis cannot imply a false conclusion

In fact, the latin expression "ex falso sequitur quodlibet" translates as "you can deduce whatever you want from a false premise". Using formulas, this means that the implication A \implies B is always true, no matter the truth value of B. So, both these implications

1+2 = 4 \implies \pi = 8

1+2 = 4 \implies 10-3=7

are true.

Conversely, if the hypothesis is true, then the implication is true only if the conclusion is true as well: if we change the previous example to

1+2 = 3 \implies \pi = 8

1+2 = 3 \implies 10-3=7

Only the second implication will hold.

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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.

8 0
3 years ago
Hannah does push-ups in sets of three she did 10 sets of push-ups today then her coach asked her to do 15 more push-ups how many
scoray [572]

Answer:

30+ 15 =45

Step-by-step explanation:

(3×10)+15 =45

4 0
3 years ago
5x - 2 = 8<br> What the answer
Montano1993 [528]

Answer:

x = 2

Step-by-step explanation:

5x - 2 = 8

5x = 8 + 2

5x = 10

x = 10/5

x = 2

7 0
3 years ago
Read 2 more answers
Tickets to a hockey game cost $45. You and 3 of your friends decide to go together. how much will your tickets cost all together
MAXImum [283]

Answer:

135

Step-by-step explanation:

Well 3 friends and 45 dollars each

45*3=135

3 0
3 years ago
Read 2 more answers
The length of the observable universe is.
alekssr [168]

Answer:

Kindly check explanation

Step-by-step explanation:

Organizations like NASA who explore space and planets often have to deal with measurement of very long distances such as the one stated above. The Major challenge with the distance written in the format expressed above is the difficulty to read, state or use in mathematical calculations as it is too explicit. A more effective method is to express this distance in standard format and more suitable long distant units such as miles.

For instance;

880,000,000,000,000,000,000,000 kilometers could be expressed as in standard form as ;

8.8 * 10^23 km

5 0
3 years ago
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