Please solve and show step by step

Solve for a side in right triangles help

Pani-rosa [81]

Answer:

The length of AB is 7.45 units.

Step-by-step explanation:

You can find the length of AB using Cosine Rule, cosθ = adj./hypo. :

θ = 20°

adj. = 7 units

hypo. = AB

cos 20° = 7/AB

AB = 7/cos 20°

= 7.45 units (3s.f)

7 0

3 years ago

Read 2 more answers

The equation of line GH is Y=4x-3. What's the equation of a line GH in slope-intercept form contains point (2-3)

Nonamiya [84]

This question is a little unclear. y= 4x - 3 is already in slope-intercept form. If you use (2,-3), then it would look like -3 = 4(2) - 3...

8 0

3 years ago

Yall whats better anime or real life

irga5000 [103]

Answer:

anime you can do or be what you want to be so ya hope you have a nice day

Step-by-step explanation:

3 0

3 years ago

As a recruiter, you are asked to recruit 65 new sales associates for an organization. Traditionally, for the organization, 25% o

Lorico [155]

Answer: 1733.33 ≈ 1734 applicants

Step-by-step explanation:

Let x be the number of job applicants being looked at.

If 25% of applicants become job candidates, then number of job candidates = 0.25x

If 20% of job candidates receive job offers, then the number of job offers = 0.20 × 0.25x = 0.05x

If 75% of job offers are accepted, then number of accepted recruits = 0.75 × 0.05x = 0.0375x

This 0.0375x shows the number of new recruits which is equivalent to the 65 that is needed by the client if the recruiter.

By equating this 0.0375x to 65,we have:

0.0375x = 65

x = 65/0.0375

x= 1733.33 applicants, for whole number sake because we're dealing with humans, we approximate to 1734

5 0

3 years ago

Prove that if x is an positive real number such that x + x^-1 is an integer, then x^3 + x^-3 is an integer as well.

Shkiper50 [21]

Answer:

By closure property of multiplication and addition of integers,

If $x + \dfrac{1}{x}$ is an integer

∴ $\left ( x + \dfrac{1}{x} \right) ^3 = x^3 + \dfrac{1}{x^3} +3\cdot \left (x + \dfrac{1}{x} \right )$ is an integer

From which we have;

$x^3 + \dfrac{1}{x^3}$ is an integer

Step-by-step explanation:

The given expression for the positive integer is x + x⁻¹

The given expression can be written as follows;

$x + \dfrac{1}{x}$

By finding the given expression raised to the power 3, sing Wolfram Alpha online, we we have;

$\left ( x + \dfrac{1}{x} \right) ^3 = x^3 + \dfrac{1}{x^3} +3\cdot x + \dfrac{3}{x}$

By simplification of the cube of the given integer expressions, we have;

$\left ( x + \dfrac{1}{x} \right) ^3 = x^3 + \dfrac{1}{x^3} +3\cdot \left (x + \dfrac{1}{x} \right )$

Therefore, we have;

$\left ( x + \dfrac{1}{x} \right) ^3 - 3\cdot \left (x + \dfrac{1}{x} \right )= x^3 + \dfrac{1}{x^3}$

By rearranging, we get;

$x^3 + \dfrac{1}{x^3} = \left ( x + \dfrac{1}{x} \right) ^3 - 3\cdot \left (x + \dfrac{1}{x} \right )$

Given that $x + \dfrac{1}{x}$ is an integer, from the closure property, the product of two integers is always an integer, we have;

$\left ( x + \dfrac{1}{x} \right) ^3$ is an integer and $3\cdot \left (x + \dfrac{1}{x} \right )$ is also an integer

Similarly the sum of two integers is always an integer, we have;

$\left ( x + \dfrac{1}{x} \right) ^3 + \left(- 3\cdot \left (x + \dfrac{1}{x} \right ) \right )$ is an integer

$\therefore x^3 + \dfrac{1}{x^3} = \left ( x + \dfrac{1}{x} \right) ^3 - 3\cdot \left (x + \dfrac{1}{x} \right )= \left ( x + \dfrac{1}{x} \right) ^3 + \left(- 3\cdot \left (x + \dfrac{1}{x} \right ) \right )$ is an integer

From which we have;

$x^3 + \dfrac{1}{x^3}$ is an integer.

4 0

3 years ago