Authority is a mixture of conviction and _____. intelligence hardheadedness acuteness business skills

Answers

Answer 1
Answer: Intelligence is the answer

Answer 2
Answer:

Answer:

intelligence

Step-by-step explanation:


Related Questions

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Which second degree polynomial function has a leading coefficient of –1 and root 4 with multiplicity 2f(x) = x3 – x2 – 4x + 4 f(x) = x4 – 3x2 – 4 f(x) = x4 + 3x2 – 4 f(x) = x3 + x2 – 4x – 4
The lengths of the sides of a rectangular window have the ratio 1.6 to 1. The area of the window is 2,560 square inches. What are the dimensions of the window? A. 50.6 inches by 50.6 inches B. 40 inches by 64 inches C. 40 inches by 80 inches D. 40 inches by 102.4 inches
Solve 4^(2x) = 7^(x−1).
What is 9x > -18 simplified

A store makes a $0.45 profit on every toy they sell. If they made $237.15 in profit last month, how many toys did they sell?

Answers

Answer:

527

Step-by-step explanation:

237.15/0.45 = 527

Answer:

They sold 527 toys.

Step-by-step explanation:

If they make 0.45$ profit on every toy.

Total profit of a month is $237.15.

then no of toys sold = total profit divided by individual profit

Number of toys= 237.15/0.45 =527

Pls help I need help why are

Answers

I believe they are about the similar in shape unless you have answers to them?

Answer:

Both of their bases are parallel

Step-by-step explanation:

Find three consecutive ODD integers such that the sum of 7 times the smallest and twice the largest is -91. Name the smallest integer..

Answers

Answer:

              -11, -9, -7

the smallest:   2x+1 = -11

Step-by-step explanation:

{z - some integer}

2z+1  - the smallest number

7(2z+1)  - seven times the smallest number

2z+1+2=2z+3  - the middle number

2z+3+2 = 2z+5 - the largest number

2(2z+5) - twice the largest number

7(2z+1) + 2(2z+5)  - the sum of 7 times the smallest and twice the largest

7(2z+1) + 2(2z+5)   = -91

14z + 7 + 4z + 10 = -91

              -17              -17

      18z   = -108

       ÷18        ÷18

        z = -6

2z+1 = 2(-6)+1 = -12 + 1 = -11  

2z+3 = 2(-6)+3 = -12 + 3 = -9

2z+5 = 2(-6)+5 = -12 + 5 = -7        

Final answer:

To solve this problem, we denote the smallest odd integer as 'x' and set up the equation 7x + 2(x+4) = -91. By solving this equation, we find that the smallest integer is -15.

Explanation:

To find the three consecutive odd integers, let us denote the smallest odd integer as x; therefore, the next two consecutive odd integers would be x + 2 and x + 4, respectively. The problem states that the sum of seven times the smallest integer and twice the largest integer equals to -91. So, we can translate this into the equation, 7x + 2(x+4) = -91. Solving this gives us x = -15. Hence the smallest integer is -15. The full solution is as follows:

  1. Let the smallest integer be x.
  2. Then the other two consecutive odd numbers are x + 2 and x + 4.
  3. According to the question, 7x + 2(x + 4) = -91.
  4. Solving for x, we get x = -15.
  5. Hence, the smallest odd integer sought is -15.

Learn more about Consecutive Odd Integers here:

brainly.com/question/32448319

#SPJ3

Helppp plzz!! A.) Side - Side - Side
B.) Side - Angle - Side
C.) Angle - Side - Angle
D.) Angle - Angle - Side
E.) Hypotenuse - Leg
F.) Not enough information.

Answers

Answer:

F. not enough information

Step-by-step explanation:

there are no markings as to the lengths and there is no way of knowing

A Pythagorean triple is a triple of natural numbers satisfying the equation a^2+b^2+c^2.One way to produce a Pythagorean triple is to add the reciprocals of any two consecutive even or odd numbers. For example, 1/5+1/7=12/35. Now 12^2+35^2=1369. This is a Pythagorean triple if 1369 is a perfect square, which it is since 1369=37^2. So 12, 35, 37 is a Pythagorean triple. Prove that this method always works.

Answers

x, x+2 - two consecutive odd or even numbers
Add the reciprocals of these numbers.
(1)/(x)+(1)/(x+2)=(x+2)/(x(x+2))+(x)/(x(x+2))=(x+2+x)/(x^2+2x)=(2x+2)/(x^2+2x)

Now add the squares of the numerator and denominator, as in the example.
(2x+2)^2+(x^2+2x)^2= \n 4x^2+8x+4+x^4+4x^3+4x^2= \n x^4+4x^3+8x^2+8x+4

So this number has to be a perfect square.
x^4+4x^3+8x^2+8x+4= \nx^4+2x^3+2x^2+2x^3+4x^2+4x+2x^2+4x+4= \nx^2(x^2+2x+2)+2x(x^2+2x+2)+2(x^2+2x+2)= \n(x^2+2x+2)(x^2+2x+2)= \n(x^2+2x+2)^2
It is a perfect square, so this method always works.

The numbers 2x+2, \ x^2+2x, \ (x^2+2x+2)^2 are a Pythagorean triple for any x \in \mathbb{N^+}.

Answer:

even tho this has nothing to do with the answer ;-;

Step-by-step explanation:First a definition: A Pythagorean Triple are three natural numbers 1 <= a <= b <= c, such that a2 + b2 = c2 holds. For example 3, 4, 5 is such a triple, since 32 + 42 = 9 + 16 = 25 = 52. While 2, 3, 4 is not such a triple, since 22 + 32 = 4 + 9 = 13 and 42 = 16. We note here that only natural numbers are considered, and thus 2, 3 can not be extended to Pythagorean triple (since 13 is not the square of some integer).

Now the question: Can we colour the natural numbers 1, 2, 3, ... with two colours, say blue and red, such that there is no monochromatic Pythagorean triple? In other words, is it possible to give every natural number one of the colours blue or red, such that for every Pythagorean triple a, b, c at least one of a, b, c is blue, and at least one of a, b, c is red ? We prove: The answer is No. That is easier to express positively: Whenever we colour the natural numbers blue or red, there must exist a monochromatic triple (one blue triple or one red triple).

More precisely we prove, using "bi-colouring" for colouring blue or red: 1) However we bi-colour the numbers 1, ..., 7825, there must exist a monochromatic Pythagorean triple. 2) While there exists a bi-colouring of 1, ..., 7824, such that no Pythagorean triple is monochromatic. Part 2) is relatively easy. Part 1) is the real hard thing -- every number from 1, ..., 7825 gets one of two possible colours, so altogether there are 27825 possible colourings, which all in a sense need to be considered, and need to be excluded. What is 27825? It is approximately 3.63 * 102355, that is, a number with 2356 decimal places. The number of particles in the universe is at most 10100, a tiny number with just 100 decimal places (in comparison).

Now let's perform real brute-force, running through all the possibilities, one after another: Even if we could place on every particle in the universe a super-computer, and they all would work perfectly together for the whole lifetime of the universe -- by far not enough. Even not if inside every particle we could place a whole universe. Even if each particle in the inner universe becomes again itself a universe, with every particle carrying a super-computer, still

by far not enough. Hope you get the idea -- the $100 we got wouldn't pay that energy bill.

Fortunately there comes SAT solving to the rescue, which actually is really good with such tasks -- it can solve some such task and even more monstrous tasks. Our ``brute-reasoning'' approach solved the problem and resulted into a 200 terabytes proof -- the largest math proof ever. Though we must emphasise that this is in no way guaranteed, and possibly it will take aeons! SAT solving uses propositional logic, in the special form of CNF (conjunctive normal form). Fortunately, in this case it is easy to represent our problem in this form.

Two positive integers are 3 units apart on a number line. Their product is 108.Which equation can be used to solve for m, the greater integer?

m(m – 3) = 108

m(m + 3) = 108
(m + 3)(m – 3) = 108
(m – 12)(m – 9) = 108

Answers

Equation that can be used to solve m, if m is the greater integer is:
m(m+3)=108
Solution is: m=12 ( 12*9=108)
Answer: A) m(m-3)=108

Answer:

A

Step-by-step explanation: