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4 years ago

Arrange the given steps in the career-planning process in the correct order.

SAT

1 answer:

Ksju [112]4 years ago

4 0

Answer:

Step 1: Self-Assessment

Step 2: Research

Step 3: Experimentation

Step 4: Decision-Making

Step 5: Job Search

Step 6: Acceptance

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X³ - 19x + 30 = (x - a)(x - b) (x - c)

oksano4ka [1.4K]

Answer:

and u will find it

Explanation:

all u have to do is sumtaxk all of it

3 0

1 year ago

A: Explain right-of-way rules for navigating this traffic circle.

sesenic [268]

As a driver that is approaching an about, it is very important to respect the drivers that are already in the about.

<h3>What is an About?</h3>

An about is basically a junction that is circular in shape, in an about the vehicle at the left is given very high priority.

The driver of the SUV should wait for the oncoming vehicle to clear before he/she enter

Learn more about Round About here:

brainly.com/question/12811263

#SPJ1

5 0

2 years ago

Evaluate the line integral, where c is the given curve. C (x/y) ds, c: x = t3, y = t4, 1 ≤ t ≤ 4.

skelet666 [1.2K]

We have

x = t³ ===> dx/dt = 3t²

y = t⁴ ===> dy/dt = 4t³

Then with the given parameteriztion, the line integral along C of x/y is

$\displaystyle \int_C \frac xy \, ds = \int_1^4 \frac{t^3}{t^4} \sqrt{(3t^2)^2 + (4t^3)^2} \, dt$

$\displaystyle \int_C \frac xy \, ds = \int_1^4 \frac1t \sqrt{9t^4 + 16t^6} \, dt$

$\displaystyle \int_C \frac xy \, ds = \int_1^4 \frac{\sqrt{t^4}}t \sqrt{9 + 16t^2} \, dt$

$\displaystyle \int_C \frac xy \, ds = \int_1^4 \frac{t^2}t \sqrt{9 + 16t^2} \, dt$

$\displaystyle \int_C \frac xy \, ds = \int_1^4 t \sqrt{9 + 16t^2} \, dt$

$\displaystyle \int_C \frac xy \, ds = \frac1{32} \int_1^4 32t \sqrt{9 + 16t^2} \, dt$

$\displaystyle \int_C \frac xy \, ds = \frac1{32} \int_1^4 \sqrt{9 + 16t^2} \, d\left(9+16t^2\right)$

$\displaystyle \int_C \frac xy \, ds = \frac1{32} \cdot \frac23 \left(9+16t^2\right)^{\frac32}\bigg|_1^4$

$\displaystyle \int_C \frac xy \, ds = \frac1{48} \left(265^{\frac32} - 25^{\frac32}\right) = \boxed{\frac{265\sqrt{265}-125}{48}}$

6 0

3 years ago

Two blocks connected by a cord passing over a small, frictionless pulley rest on frictionless planes (the figure (figure 1)).

nalin [4]

We have that the acceleration when going left is mathematically given as

a = 0.671 m/s^2

<h3> Acceleration of Pulley</h3>

Let the system move leftward in direction

Hence we have tension to be

T - m1gsin53 = m1a

m2gsin30 - T = m2a

Therefore

m2gsin30 - m1gsin53.1 = a(m1+m2)

a=g(m2sin30-m1sin53.1) / (m1+m2)

a = 0.671 m/s^2

For more information on acceleration visit

brainly.com/question/605631

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Complete Question Attached below

8 0

3 years ago

How did Dr.Tierno find the answer to his question ?

nydimaria [60]

What is this referring to?

4 0

3 years ago

Read 2 more answers