Se lanza un cuerpo desde el origen con velocidad horizontal de 40 m/s, y con un ángulo de 60º.

) Find the coordinates of the point A' which is the symmetric point

soldier1979 [14.2K]

Let, coordinate of point A' is (x,y).

Since, A' is the symmetric point A(3, 2) with respect to the line 2x + y - 12 = 0.

So, slope of line containing A and A' will be perpendicular to the line 2x + y - 12 = 0 and also their center lies in the line too.

Now, their center is given by :

$C( \dfrac{x+3}{2}, \dfrac{y+2}{2})$

Also, product of slope will be -1 .( Since, they are parallel )

$\dfrac{y-2}{x-3} \times -2 = -1\\\\2y - 4 = x - 3\\\\2y - x = 1$

x = 2y - 1

So, $C( \dfrac{2y -1 +3}{2}, \dfrac{y+2}{2})\\\\C( \dfrac{2y + 2}{2}, \dfrac{y+2}{2})$

Also, C satisfy given line :

$2\times ( \dfrac{2y + 2}{2}) + \dfrac{y+2}{2} = 12\\\\4y + 4 + y + 2 = 24\\\\5y = 18\\\\y = \dfrac{18}{5}$

Also,

$x = 2\times \dfrac{18}{5 } - 1\\\\x = \dfrac{31}{5}$

Therefore, the symmetric points is $A'(\dfrac{31}{5}, \dfrac{18}{5})$ .

7 0

2 years ago

Allen's daily regular wage is P1,420. This coming Christmas, his boss is asking him to work for 14 hours. How much is his DAILY

VMariaS [17]

Using proportions, it is found that his DAILY RATE on that particular day is of Php 2,420 .

This question is solved by proportions, using a rule of three.

In a regular 8-hour day, Allen earns Php 1420. How much will he earn on a 14 hour day?

The rule of three is:

8 hours - Php 1420

14 hours - x

Applying cross multiplication:

$8x = 1420 \times 14$

$x = \frac{1420 \times 14}{8}$

$x = 2485$

The closest option is Php 2,420.

A similar problem is given at brainly.com/question/24372153

7 0

2 years ago

Letang Industrial Systems Company (LISC) is trying to decide between two different conveyor belt systems. System A costs $280,00

Vanyuwa [196]

We have to calculate the EAC for both the conveyor belt system.

Solutions :

Equivalent Annual Cost or (EAC) for the SYSTEM-A

$\text{Operating cash flow } = \text{Pre-tax annual operating cost (1-tax rate )} + (\text{depreciation expense} \times$ tax rate )

$=-855,000(1-0.23)+\left[\left(\frac{280,000}{4}\right)\times 0.23\right]$

$= (-85,000 \times 0.77 ) + (70,000 \times 0.23)$

= $ 49,350

Year Annual Cost flow Present value factor Present Value of Annual

at 10% cash flow

1 -49,350 0.909091 -44,863.64

2 -49,350 0.826446 -40,785.12

3 -49,350 0.751315 -37,077.39

4 -49,350 0.683013 -33,706.71

Total $ 3.169865 $ -156,432.86

Therefore, Net Present value = present value of the annual cash flow - initial investment.

= 156,432.86 - 280,000

= $ 436,432.86 (negative)

Now the EAC or the Equivalent Annual Cost for System A :

$\text{EAC}= \text{Net present value / (PVIFA 10 percent, 4 years)}$

$=\frac{436,432.86}{3.169865}$

$= 137,681.86$ dollar (negative)

$\text{Operating cash flow } = \text{Pre-tax annual operating cost (1-tax rate )} + (\text{depreciation expense} \times$ tax rate)

$=79,000(1-0.23)+\left[\left(\frac{360,000}{6}\right) \times 0.23\right]$

$=(-78,000 \times 0.77)+(60,000 \times 0.23)$

= -$ 47,030

Year Annual Cost flow Present value factor Present Value of Annual

at 10% cash flow

1 -47,030 0.909091 -42,754.55

2 -47,030 0.826446 -38,867.77

3 -47,030 0.751315 -35,334.34

4 -47,030 0.683013 -32,122.12

5 -47,030 0.620921 -29,201.93

6 -47,030 0.564474 -26,547.21

Total $ 4.355261 $ -204,827.91

Net Present Value = Present Value of annual cash inflows – Initial Investment

$= 204,827.91 - 360,000$

= -$ 564,827.91 (negative)

EAC for system B:

Equivalent Annual Cost for system B $=\frac{\text{net present value}}{\text{PVIFA 10 \text percent, 6 years}}$

$=\frac{-564,827.91}{4.355261}$

= -$129,688.66 (negative)

7 0

2 years ago

What is the number .2121 as a fraction

brilliants [131]

Are you being funny sayings 21

4 0

3 years ago

PLEASE HELP FOR BRAINLIEST!!!

SOVA2 [1]

The measure of an angle formed by a secant and a tangent drawn from a point outside the circle is half the the difference of the intercepted arcs.

 $21^o= \frac{arc \ RU-arc \ SU}{2}= \frac{119^o-arc \ SU}{2} \ \to \ \\ \\ 42= 119-arcSU \\ arc \ SU=119-42=77^o$

5 0

3 years ago