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3 years ago

9

Persons who work as sales managers, therapists, and teachers may be high in ____ intelligence.

1 answer:

Basile [38]3 years ago

7 0

These are all service professions and ones in which the individual does well if they are in tune with how their clients, patients and students are feeling/understanding. Therefore, it is likely that people who excel in these fields are high in "emotional intelligence."

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Someone help me with this. I forgot my notes in class

Nonamiya [84]

Answer:

see explanation

Step-by-step explanation:

Since AB = BC the the triangle is isosceles and thus

∠ A = ∠ C = 8x

The sum of the 3 angles in a triangle = 180°

Sum the 3 angles and equate to 180

44x + 8x + 8x = 180, that is

60x = 180 ( divide both sides by 60 )

x = 3

Thus

∠ A = ∠ C = 8x = 8(3) = 24°

∠ B = 44x = 44(3) = 132°

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3 years ago

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Use the data thing to answer all the questions plz

KonstantinChe [14]

Where is the data chart to answer the questions?

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4 years ago

The volume of this cylinder is 1,909.92384 cubic millimeters. What is the radius? Use n=3.14 and round your answer to the neares

klemol [59]

Remember that the volume of a cylinder is

$\begin{gathered} V=\pi\cdot r^2\cdot h \\ \end{gathered}$

so we have that

$r=\sqrt[]{\frac{V}{\pi\cdot h}}=\sqrt[]{\frac{1909.92384}{3.14\cdot6.6}}=\sqrt[]{92.12}=9.6$

so the radius is 9.6 milimeters

6 0

1 year ago

Let a1, a2, a3, ... be a sequence of positive integers in arithmetic progression with common difference

Bezzdna [24]

Since $a_1,a_2,a_3,\cdots$ are in arithmetic progression,

$a_2 = a_1 + 2$

$a_3 = a_2 + 2 = a_1 + 2\cdot2$

$a_4 = a_3+2 = a_1+3\cdot2$

$\cdots \implies a_n = a_1 + 2(n-1)$

and since $b_1,b_2,b_3,\cdots$ are in geometric progression,

$b_2 = 2b_1$

$b_3=2b_2 = 2^2 b_1$

$b_4=2b_3=2^3b_1$

$\cdots\implies b_n=2^{n-1}b_1$

Recall that

$\displaystyle \sum_{k=1}^n 1 = \underbrace{1+1+1+\cdots+1}_{n\,\rm times} = n$

$\displaystyle \sum_{k=1}^n k = 1 + 2 + 3 + \cdots + n = \frac{n(n+1)}2$

It follows that

$a_1 + a_2 + \cdots + a_n = \displaystyle \sum_{k=1}^n (a_1 + 2(k-1)) \\\\ ~~~~~~~~ = a_1 \sum_{k=1}^n 1 + 2 \sum_{k=1}^n (k-1) \\\\ ~~~~~~~~ = a_1 n + n(n-1)$

so the left side is

$2(a_1+a_2+\cdots+a_n) = 2c n + 2n(n-1) = 2n^2 + 2(c-1)n$

Also recall that

$\displaystyle \sum_{k=1}^n ar^{k-1} = \frac{a(1-r^n)}{1-r}$

so that the right side is

$b_1 + b_2 + \cdots + b_n = \displaystyle \sum_{k=1}^n 2^{k-1}b_1 = c(2^n-1)$

Solve for $c$ .

$2n^2 + 2(c-1)n = c(2^n-1) \implies c = \dfrac{2n^2 - 2n}{2^n - 2n - 1} = \dfrac{2n(n-1)}{2^n - 2n - 1}$

Now, the numerator increases more slowly than the denominator, since

$\dfrac{d}{dn}(2n(n-1)) = 4n - 2$

$\dfrac{d}{dn} (2^n-2n-1) = \ln(2)\cdot2^n - 2$

and for $n\ge5$ ,

$2^n > \dfrac4{\ln(2)} n \implies \ln(2)\cdot2^n - 2 > 4n - 2$

This means we only need to check if the claim is true for any $n\in\{1,2,3,4\}$ .

$n=1$ doesn't work, since that makes $c=0$ .

If $n=2$ , then

$c = \dfrac{4}{2^2 - 4 - 1} = \dfrac4{-1} = -4 < 0$

If $n=3$ , then

$c = \dfrac{12}{2^3 - 6 - 1} = 12$

If $n=4$ , then

$c = \dfrac{24}{2^4 - 8 - 1} = \dfrac{24}7 \not\in\Bbb N$

There is only one value for which the claim is true, $c=12$ .

3 0

2 years ago

Brainiest if correct

Mademuasel [1]

Answer:one solution

Step-by-step explanation:

6 0

3 years ago